{"id":825,"date":"2021-08-27T20:07:27","date_gmt":"2021-08-27T20:07:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=825"},"modified":"2022-10-29T01:05:12","modified_gmt":"2022-10-29T01:05:12","slug":"projectile-motion","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/projectile-motion\/","title":{"raw":"Projectile Motion","rendered":"Projectile Motion"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>State Kepler\u2019s laws of planetary motion.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169737393059\" class=\" \">Now let\u2019s look at an application of vector functions. In particular, let\u2019s consider the effect of gravity on the motion of an object as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignore the effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as <span id=\"term145\" data-type=\"term\">projectile motion<\/span>. It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs.<\/p>\r\n<p id=\"fs-id1169737818838\" class=\" \">First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose the positive <em data-effect=\"italics\">y<\/em>-axis to be up, the negative <em data-effect=\"italics\">y-<\/em>axis to be down, and the positive <em data-effect=\"italics\">x-<\/em>axis to be forward (i.e., away from the thrower of the object). The effect of gravity is in a downward direction, so Newton\u2019s second law tells us that the force on the object resulting from gravity is equal to the mass of the object times the acceleration resulting from to gravity, or [latex]F_{g}=mg[\/latex], where [latex]F_{g}[\/latex]\u00a0represents the force from gravity and\u00a0[latex]g[\/latex] represents the acceleration resulting from gravity at Earth\u2019s surface. The value of\u00a0[latex]g[\/latex] in the English system of measurement is approximately [latex]32\\ ft\/sec^{2}[\/latex] and it is approximately [latex]9.8\\ m\/sec^{2}[\/latex] in the metric system.\u00a0This is the only force acting on the object. Since gravity acts in a downward direction, we can write the force resulting from gravity in the form [latex]F_{g}=-mg\\,{\\bf{j}}[\/latex], as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"attachment_929\" align=\"aligncenter\" width=\"439\"]<img class=\"size-full wp-image-929\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162913\/3-4-5.jpeg\" alt=\"This figure is a ball falling in a vertical path. The ball is at the top at the initial position. From the ball a vector is drawn vertically downward labeled \u201cacceleration\u201d. The vertical line is labeled \u201cdistance\u201d. At the bottom of the line it is labeled \u201cfinal position\u201d. There is also grass at the bottom of the figure.\" width=\"439\" height=\"401\" \/> Figure 1.\u00a0An object is falling under the influence of gravity.[\/caption]\r\n\r\n<div class=\"textbox tryit\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Interactive<\/h3>\r\nVisit this <a href=\"http:\/\/www.openstax.org\/l\/20_projectile\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a> for a video showing projectile motion.\r\n\r\n<\/div>\r\nNewton\u2019s second law also tells us that [latex]F=m{\\bf{a}}[\/latex], where [latex]\\bf{a}[\/latex] represents the acceleration vector of the object. This force must be equal to the force of gravity at all times, so we therefore know that\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {F} &amp; =\\hfill &amp; {F_{g}} \\hfill \\\\ \\hfill {m{\\bf{a}}}&amp; =\\hfill &amp; {-mg\\,{\\bf{j}}} \\hfill \\\\ \\hfill {{\\bf{a}}}&amp; =\\hfill &amp; {-g\\,{\\bf{j}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nNow we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite the last equation in the form\r\n<p style=\"text-align: center;\">[latex]{\\bf{v}}'\\,(t)=-g\\,{\\bf{j}}.[\/latex]<\/p>\r\nBy taking the antiderivative of each side of this equation we obtain\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}\\,(t)} &amp; =\\hfill &amp; {\\displaystyle\\int_{}\\ -g\\,{\\bf{j}}\\,dt} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-gt\\,{\\bf{j}}+{\\bf{C}}_{1}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nfor some constant vector [latex]{\\bf{C}}_{1}[\/latex]. To determine the value of this vector, we can use the velocity of the object at a fixed time, say at time [latex]t=0[\/latex]. We call this velocity the <em>initial velocity<\/em>: [latex]{\\bf{v}}\\,(0)={\\bf{v}}_{0}[\/latex]. Therefore, [latex]{\\bf{v}}\\,(0)=-g\\,(0)\\,{\\bf{j}}+{\\bf{C}}_{1}={\\bf{v}}_{0}[\/latex] and [latex]{\\bf{C}}_{1}={\\bf{v}}_{0}[\/latex]. This gives the velocity vector as [latex]{\\bf{v}}\\,(t)=-gt\\,{\\bf{j}}+{\\bf{v}}_{0}[\/latex].\r\n\r\nNext we use the fact that velocity [latex]{\\bf{v}}\\,(t)[\/latex] is the derivative of position [latex]{\\bf{s}}\\,(t)[\/latex]. This gives the equation\r\n<p style=\"text-align: center;\">[latex]{\\bf{s}}'\\,(t)=-gt\\,{\\bf{j}}+{\\bf{v}}_{0}[\/latex].<\/p>\r\nTaking the antiderivative of both sides of this equation leads to\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(t)} &amp; =\\hfill &amp; {\\displaystyle\\int_{}\\ -gt\\,{\\bf{j}}+{\\bf{v}}_{0}\\,dt} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\frac{1}{2}\\,gt^{2}\\,{\\bf{j}}+{\\bf{v}}_{0}t+{\\bf{C}}_{2},}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nwith another unknown constant vector [latex]{\\bf{C}}_{2}[\/latex]. To determine the value of [latex]{\\bf{C}}_{2}[\/latex], we can use the position of the object at a given time, say at time [latex]t=0[\/latex]. We call this position the\u00a0<em>initial position<\/em>: [latex]{\\bf{s}}\\,(0)={\\bf{s}}_{0}[\/latex]. Therefore, [latex]{\\bf{s}}\\,(0)=-(1\/2)\\,g(0)^{2}\\,{\\bf{j}}+{\\bf{v}}_{0}\\,(0)+{\\bf{C}}_{2}={\\bf{s}}_{0}[\/latex] and [latex]{\\bf{C}}_{2}={\\bf{s}}_{0}[\/latex]. This gives the position of the object at any time as\r\n<p style=\"text-align: center;\">[latex]{\\bf{s}}\\,(t)=-\\frac{1}{2}\\,gt^{2}\\,{\\bf{j}}+{\\bf{v}}_{0}\\,t+{\\bf{s}}_{0}.[\/latex]<\/p>\r\nLet\u2019s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from the origin at an angle [latex]\\theta[\/latex] to the horizontal, with initial speed [latex]v_{0}[\/latex]. How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. Therefore, [latex]{\\bf{s}}_{0}={\\bf{0}}[\/latex], as shown in the following figure.\r\n\r\n[caption id=\"attachment_930\" align=\"aligncenter\" width=\"650\"]<img class=\"size-full wp-image-930\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163018\/3-4-6.jpeg\" alt=\"This figure has an upside down parabolic curve representing projectile motion. The figure is labeled \u201cconstant horizontal velocity; constant vertical acceleration\u201d. The curve is in the first quadrant beginning and ending on the x-axis. The height of the curve is labeled \u201cH\u201d. The distance on the x-axis is labeled \u201cR\u201d. Angle theta represents the direction of the projectile at the origin. Five points are labeled on the graph with vectors. The vectors are labeled \u201cv\u201d with subscripts representing directions.\" width=\"650\" height=\"352\" \/> Figure 2.\u00a0Projectile motion when the object is thrown upward at an angle [latex]\\theta[\/latex].\u00a0The horizontal motion is at constant velocity and the vertical motion is at constant acceleration.[\/caption]We can rewrite the initial velocity vector in the form [latex]{\\bf{v}}_{0}=v_{0}\\cos{\\theta}{\\bf{i}}+v_{0}\\sin{\\theta}{\\bf{j}}[\/latex]. Then the equation for the position functions [latex]{\\bf{s}}(t)[\/latex] becomes\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(t)} &amp; =\\hfill &amp; {-\\frac{1}{2}gt^{2}{\\bf{j}}+v_{0}t\\cos{\\theta{\\bf{i}}}+v_{0}t\\sin{\\theta{\\bf{j}}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {v_{0}t\\cos{\\theta{\\bf{i}}}+v_{0}t\\sin{\\theta{\\bf{j}}}-\\frac{1}{2}gt^{2}\\,{\\bf{j}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {v_{0}t\\cos{\\theta{\\bf{i}}}+(v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2})\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nThe coefficient of [latex]{\\bf{i}}[\/latex] represents the horizontal component of [latex]{\\bf{s}}(t)[\/latex] and is the horizontal distance of the object from the origin at time [latex]t[\/latex]. The maximum value of the horizontal distance (measured at the same initial and final attitude) is called the range [latex]R[\/latex]. The coefficient of [latex]{\\bf{j}}[\/latex] represents the vertical component of [latex]{\\bf{s}}(t)[\/latex] and is the altitude of the object at time [latex]t[\/latex]. The maximum value of the vertical distance is the height [latex]H[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Motion of a Cannonball<\/h3>\r\nDuring an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of [latex]30^\\circ[\/latex] above horizontal and the initial speed of the cannonball is [latex]600\\ ft\/sec[\/latex]. The cliff is [latex]100\\ ft[\/latex] above the water (Figure 7).\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find the maximum height of the cannonball.<\/li>\r\n \t<li>How long will it take for the cannonball to splash into the sea?<\/li>\r\n \t<li>How far out to sea will the cannonball hit the water?\r\n\r\n[caption id=\"attachment_931\" align=\"aligncenter\" width=\"692\"]<img class=\"size-full wp-image-931\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163109\/3-4-7.jpeg\" alt=\"This figure has a cannon at the edge of a cliff aimed upwards. There is a cannonball coming out of the cannon. The path of the cannonball is an upside down parabola represented with a broken line. The maximum height is labeled with \u201c?\u201d. The height of the cliff is 100 feet.\" width=\"692\" height=\"412\" \/> Figure 3.\u00a0The flight of a cannonball (ignoring air resistance) is projectile motion.[\/caption]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1267795055160\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1267795055160\"]We use the equation\r\n<div style=\"text-align: center;\">[latex]{\\bf{s}}\\,(t)=v_{0}t\\cos{\\theta{\\bf{i}}}+\\big(v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2}\\big){\\bf{j}}[\/latex]<\/div>\r\n<div><\/div>\r\nwith [latex]\\theta={30^\\circ}[\/latex],<span style=\"font-size: 1rem; text-align: initial;\">\u00a0[latex]g=32\\ ft\/sec^{2}[\/latex], and [latex]v_{0}=600\\ ft\/sec[\/latex]. Then the position equation becomes<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(t)} &amp; =\\hfill &amp; {600t(\\cos{30}){\\bf{i}}+(600t\\sin{30}-\\frac{1}{2}(32)t^{2}){\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {300t\\sqrt{3}\\,{\\bf{i}}+(300t-16t^{2})\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The cannonball reaches its maximum height when the vertical component of its velocity is zero, because the cannonball is neither rising nor falling at that point. The velocity vector is\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}\\,(t)} &amp; =\\hfill &amp; {{\\bf{s}}'\\,(t)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {300\\sqrt{3}\\,{\\bf{i}}+(300-32t)\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, the vertical component of velocity is given by the expression [latex]300-32t[\/latex]. Setting this expression equal to zero and solving [latex]t[\/latex] gives [latex]t=9.375\\ sec[\/latex]. The height of the cannonball at this time is given by the vertical component of the position vector, evaluated at [latex]t=9.375[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(9.375)} &amp; =\\hfill &amp; {300\\,(9.375)\\,\\sqrt{3}\\,{\\bf{i}}+(300\\,(9.375)-16\\,(9.375)^{2}){\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {4871.39\\,{\\bf{i}}+1406.25\\,{\\bf{j}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, the maximum height of the cannonball is [latex]1406.39\\text{ }ft[\/latex] above the cannon, or [latex]1506.39\\text{ }ft[\/latex] above sea level.<\/li>\r\n \t<li>When the cannonball lands in the water, is is [latex]100\\text{ }ft[\/latex] below the cannon. Therefore, the vertical component of the position vector is equal to [latex]-100[\/latex]. Setting the vertical component of [latex]{\\bf{s}}\\,(t)[\/latex] equal to [latex]-100[\/latex] and solving, we obtain\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {300t-16t^{2}} &amp; =\\hfill &amp; {-100} \\hfill \\\\ \\hfill {16t^{2}-300t-100}&amp; =\\hfill &amp; {0} \\hfill \\\\ \\hfill {4t^{2}-75t-25}&amp; =\\hfill &amp; {0} \\hfill \\\\ \\hfill {t}&amp; =\\hfill &amp; {\\frac{75\\pm\\sqrt{(-75)^{2}-4(4)(-25)}}{2(4)}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{75\\pm\\sqrt{6025}}{8}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{75\\pm5\\sqrt{241}}{8}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe positive value of [latex]t[\/latex] that solves this equation is approximately [latex]19.08[\/latex]. Therefore, the cannonball hits the water after approximately [latex]19.08\\text{ }sec[\/latex].<\/li>\r\n \t<li>To find the distance out to sea, we simply substitute the answer from part (b) into [latex]{\\bf{s}}\\,(t)[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(19.08)} &amp; =\\hfill &amp; {300\\,(19.08)\\,\\sqrt{3}\\,{\\bf{i}}+(300\\,(19.08)-16\\,(19.08)^{2})\\,{\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {9914.26\\,{\\bf{i}}-100.7424\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, the ball hits the water about [latex]9914.26\\text{ }ft[\/latex] away from the base of the cliff. Notice that the vertical component of the position vector is very close to [latex]-100[\/latex], which tells us that the ball just hit the water. Note that [latex]9914.26[\/latex] feet is not the true range of the cannon since the cannonball lands in the ocean at a location below the cannon. The range of the cannon would be determined by finding how far out the cannonball is when its height is [latex]100\\text{ }ft[\/latex] above the water (the same as the altitude of the cannon).<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nAn archer fires an arrow at an angle of [latex]40^\\circ[\/latex] above the horizontal with an initial speed of [latex]98\\text{ }m\/sec[\/latex]. The height of the archer is [latex]171.5\\text{ }cm[\/latex]. Find the horizontal distance the arrow travels before it hits the ground.\r\n\r\n[reveal-answer q=\"fs-id1168794933165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1168794933165\"]\r\n<div style=\"text-align: center;\">[latex]967.15\\ m[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949633&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=oFRnQ2QtMeY&amp;video_target=tpm-plugin-05gnio59-oFRnQ2QtMeY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.16_transcript.html\">transcript for \u201cCP 3.16\u201d here (opens in new window).<\/a><\/center>One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? To determine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level. In other words, we want to determine an equation for the range. In this case, the equation of projectile motion is\r\n<p style=\"text-align: center;\">[latex]{\\bf{s}}(t)=v_{0}t\\cos{\\theta{\\bf{i}}}+(v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2})\\,{\\bf{j}}.[\/latex]<\/p>\r\nSetting the second component equal to zero and solving for [latex]t[\/latex] yields\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2}} &amp; =\\hfill &amp; {0} \\hfill \\\\ \\hfill {t(v_{0}\\sin{\\theta}-\\frac{1}{2}gt)}&amp; =\\hfill &amp; {0}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nTherefore, either [latex]t=0[\/latex] or [latex]t=\\frac{2v_{0}\\sin{\\theta}}{g}[\/latex]. We are interested in the second value of [latex]t[\/latex], so we substitute this into [latex]{\\bf{s}}(t)[\/latex], which gives\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{{\\bf{s}}\\,\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)} &amp; =\\hfill &amp; {v_{0}\\,\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)\\cos{\\theta{\\bf{i}}}+\\Big(v_{0}\\,\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)\\sin{\\theta}-\\frac{1}{2}g\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)^{2}\\Big)\\,{\\bf{j}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\bigg(\\frac{2v^2_0\\sin{\\theta}\\cos{\\theta}}{g}\\bigg)\\,{\\bf{i}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{v^2_0\\sin{2\\theta}}{g}\\,{\\bf{i}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nThus, the expression for the range of a projectile fired at an angle [latex]\\theta[\/latex] is\r\n<p style=\"text-align: center;\">[latex]\\large{R=\\frac{v^2_0\\sin{2\\theta}}{g}\\,{\\bf{i}}}.[\/latex]<\/p>\r\nThe only variable in this expression is [latex]\\theta[\/latex]. To maximize the distance traveled, take the derivative of the coefficient of [latex]{\\bf{i}}[\/latex] with respect to [latex]\\theta[\/latex] and set it equal to zero:\r\n\r\n[latex] \\begin{align}\r\n\r\n\\hspace{10cm} \\frac{d}{d\\theta}\\left(\\frac{v_{0}^2t\\sin{2\\theta}}{g}\\right) &amp; =0 \\\\\r\n\r\n\\frac{2v_{0}\\cos{2\\theta}}{g} &amp; = 0 \\\\\r\n\r\n\\theta &amp; = 45^\\circ.\r\n\r\n\\end{align} [\/latex]\r\n<div>\r\n\r\nThis value of [latex]\\theta[\/latex] is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air resistance, the best angle to fire a projectile (to maximize the range) is at a [latex]45^\\circ[\/latex] angle. The distance it travels is given by\r\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{s}}\\bigg(\\frac{2v_{0}\\sin{45}}{g}\\bigg)=\\frac{v^2_0\\sin{90}}{g}\\,{\\bf{i}}=\\frac{v^2_0}{g}\\,{\\bf{j}}}.[\/latex]<\/p>\r\nTherefore, the range for an angle of [latex]45^\\circ[\/latex] is [latex]v^2_0\/g[\/latex].\r\n<h2>Kepler's Laws<\/h2>\r\nDuring the early 1600s, Johannes <span id=\"term146\" class=\"no-emphasis\" data-type=\"term\">Kepler<\/span> was able to use the amazingly accurate data from his mentor Tycho <span id=\"term147\" class=\"no-emphasis\" data-type=\"term\">Brahe<\/span>to formulate his three laws of planetary motion, now known as <strong><span id=\"term148\" data-type=\"term\">Kepler\u2019s laws of planetary motion<\/span><\/strong>. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley\u2019s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Kepler's Laws of Planetary Motion Theorem<\/h3>\r\n\r\n<hr \/>\r\n\r\n<ol>\r\n \t<li>The path of any planet about the Sun is elliptical in shape, with the center of the Sun located at one focus of the ellipse (the law of ellipses).<\/li>\r\n \t<li>A line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal time intervals (the law of equal areas) (Figure 8).<\/li>\r\n \t<li>The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of the lengths of their semi-major orbital axes (the law of harmonies).<\/li>\r\n<\/ol>\r\n<\/div>\r\n\r\n[caption id=\"attachment_932\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-932\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163209\/3-4-8.jpeg\" alt=\"This figure is an elliptical curve labeled \u201cplanets orbit\u201d. The sun is represented towards the left inside the ellipse, at a focal point. Along the ellipse there are points A,B,C,D,E,F. There are line segments from the sun to each point.\" width=\"417\" height=\"345\" \/> Figure 4. Kepler\u2019s first and second laws are pictured here. The Sun is located at a focus of the elliptical orbit of any planet. Furthermore, the shaded areas are all equal, assuming that the amount of time measured as the planet moves is the same for each region.[\/caption]\r\n\r\n&nbsp;\r\n\r\nKepler\u2019s third law is especially useful when using appropriate units. In particular, [latex]1\\text{ }\\textit{astronomical unit}[\/latex] is defined to be the average distance from Earth to the Sun, and is now recognized to be [latex]149,597,870,700\\text{ }m[\/latex] or, approximately [latex]93,000,000[\/latex]mi. We therefore write [latex]1\\text{ }A.U. = 93,000,000[\/latex]mi. Since the time it takes for Earth to orbit the Sun is [latex]1[\/latex] year, we use Earth years for units of time. Then, substituting [latex]1[\/latex] year for the period of Earth and [latex]1[\/latex]A.U. for the average distance to the Sun, Kepler\u2019s third law can be written as\r\n<p style=\"text-align: center;\">[latex]T^2_p=D^3_p[\/latex]<\/p>\r\nfor any planet in the solar system, where [latex]T_p[\/latex] is the period of that planet measured in Earth years and [latex]D_p[\/latex]\u00a0is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.\r\n<p id=\"fs-id1169738059916\" class=\" \">Kepler\u2019s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler\u2019s third law to other orbital systems, such as a moon orbiting around a planet. Kepler\u2019s original third law only applies to objects orbiting the Sun.<\/p>\r\n\r\n<h3>Proof<\/h3>\r\nLet\u2019s now prove Kepler\u2019s first law using the calculus of vector-valued functions. First we need a coordinate system. Let\u2019s place the Sun at the origin of the coordinate system and let the vector-valued function [latex]{\\bf{r}}\\,(t)[\/latex] represent the location of a planet as a function of time. Newton proved Kepler's law using his second law of\u00a0motion and his law of universal gravitation. Newton\u2019s second law of motion can be written as [latex]{\\bf{F}}=m{\\bf{a}}[\/latex], where [latex]{\\bf{F}}[\/latex] represents the net force acting on the planet. His law of universal gravitation can be written in the form [latex]{\\bf{F}}=-\\frac{GmM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\cdot\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}[\/latex], which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude [latex]\\frac{GmM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}[\/latex] (Figure 9).\r\n\r\n[caption id=\"attachment_933\" align=\"aligncenter\" width=\"378\"]<img class=\"size-full wp-image-933\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163250\/3-4-9.jpeg\" alt=\"This figure is an ellipse with a circle to the left on the inside at a focal point. The circle represents the sun. On the ellipse is a smaller circle representing Earth. The line segment drawn between the circles is labeled \u201cgravitational force\u201d.\" width=\"378\" height=\"286\" \/> Figure 5.\u00a0The gravitational force between Earth and the Sun is equal to the mass of the earth times its acceleration.[\/caption]\r\n\r\nSetting these two forces equal to each other, and using the fact that [latex]{\\bf{a}}\\,(t)={\\bf{v}}'\\,(t)[\/latex], we obtain\r\n<p style=\"text-align: center;\">[latex]\\large{m{\\bf{v}}'\\,(t)=-\\frac{GmM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\cdot\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}}[\/latex],<\/p>\r\nwhich can be rewritten as\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{d{\\bf{v}}}{dt}=-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}{\\bf{r}}}.[\/latex]<\/p>\r\nThis equation shows that the vectors [latex]d{\\bf{v}}\/dt[\/latex] and [latex]{\\bf{r}}[\/latex] are parallel to each other, so [latex]d{\\bf{v}}\/dt\\,\\times\\,{\\bf{r}}=0[\/latex]. Next, let's differentiate [latex]{\\bf{r}}\\,\\times\\,{\\bf{v}}[\/latex] with respect to time:\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{d}{dt}({\\bf{r}}\\,\\times\\,{\\bf{v}})=\\frac{d{\\bf{r}}}{dt}\\,\\times\\,{\\bf{v}}+{\\bf{r}}\\,\\times\\frac{d{\\bf{v}}}{dt}={\\bf{v}}\\,\\times\\,{\\bf{v}}+{\\bf{0}}={\\bf{0}}}.[\/latex]<\/p>\r\nThis proves that [latex]{\\bf{r}}\\,\\times\\,{\\bf{v}}[\/latex] is a constant vector, which we call [latex]{\\bf{C}}[\/latex]. Since [latex]{\\bf{r}}[\/latex] and [latex]{\\bf{v}}[\/latex] are both perpendicular to\u00a0[latex]{\\bf{C}}[\/latex] for all values of [latex]t[\/latex], they must lie in a plane perpendicular to [latex]{\\bf{C}}[\/latex]. Therefore, the motion of the planet lies in a plane.\r\n\r\nNext we calculate the expression [latex]d{\\bf{v}}\/dt\\,\\times\\,{\\bf{C}}[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}=-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}{\\bf{r}}\\,\\times\\,({\\bf{r}}\\,\\times\\,{\\bf{v}})=-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}[({\\bf{r}}\\cdot{\\bf{v}})\\,{\\bf{r}}-({\\bf{r}}\\cdot{\\bf{r}})\\,{\\bf{v}}].[\/latex]<\/p>\r\nThe last equality in the above equation is from the triple cross product formula (Introduction to Vectors in Space). We need an expression for [latex]{\\bf{r}}\\cdot{\\bf{v}}[\/latex]. To calculate this, we differentiate [latex]{\\bf{r}}\\cdot{\\bf{r}}[\/latex] with respect to time:\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}({\\bf{r}}\\,\\cdot\\,{\\bf{r}})=\\frac{d{\\bf{r}}}{dt}\\cdot{\\bf{r}}+{\\bf{r}}\\cdot\\frac{d{\\bf{r}}}{dt}=2{\\bf{r}}\\cdot\\frac{d{\\bf{r}}}{dt}=2{\\bf{r}}\\cdot{\\bf{v}}.[\/latex]<\/p>\r\nSince [latex]{\\bf{r}}\\,\\cdot\\,{\\bf{r}}=\\left\\Vert{\\bf{r}}\\right\\Vert^{2}[\/latex], we also have\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}({\\bf{r}}\\cdot{\\bf{r}})=\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert^{2}=2\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert[\/latex].<\/p>\r\nCombining the last two equations, we get\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {2{\\bf{r}}\\cdot{\\bf{v}}} &amp; =\\hfill &amp; {2\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert} \\hfill \\\\ \\hfill {{\\bf{r}}\\cdot{\\bf{v}}}&amp; =\\hfill &amp; {\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert{.}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nSubstituting into our expression for [latex]\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}[\/latex] gives us\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}} &amp; =\\hfill &amp; {-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}[({\\bf{r}}\\cdot{\\bf{r}})\\,{\\bf{r}}-({\\bf{r}}\\cdot{\\bf{r}})\\,{\\bf{v}}]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}\\big[\\left\\Vert{\\bf{r}}\\right\\Vert{\\big(}\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert\\big)\\,{\\bf{r}}-\\left\\Vert{\\bf{r}}\\right\\Vert^{2}{\\bf{v}}\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-GM\\Big[\\frac{1}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\big(\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert\\big)\\,{\\bf{r}}-\\frac{1}{\\left\\Vert{\\bf{r}}\\right\\Vert}{\\bf{v}}\\Big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {GM\\Big[\\frac{\\bf{v}}{\\left\\Vert{\\bf{r}}\\right\\Vert}-\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\big(\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert\\big)\\Big].}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nHowever,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\,\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}} &amp; =\\hfill &amp; {\\frac{\\frac{d}{dt}({\\bf{r}})\\left\\Vert{\\bf{r}}\\right\\Vert-{\\bf{r}}\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert}{\\left\\Vert{\\bf{r}}\\right\\Vert}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\frac{d{\\bf{r}}}{dt}}{\\left\\Vert{\\bf{r}}\\right\\Vert}-\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\bf{v}}{\\left\\Vert{\\bf{r}}\\right\\Vert}-\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{v}}\\right\\Vert^{2}}\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nTherefore, we get\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}=GM\\Big(\\frac{d}{dt}\\,\\frac{\\bf{r}}{\\left\\Vert{\\bf{r}}\\right\\Vert}\\Big)}.[\/latex]<\/p>\r\nSince [latex]\\bf{C}[\/latex] is a constant vector, we can integrate both sides and obtain\r\n<p style=\"text-align: center;\">[latex]{\\bf{v}}\\,\\times\\,{\\bf{C}}=GM\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}+{\\bf{D}},[\/latex]<\/p>\r\nwhere [latex]\\bf{D}[\/latex] is a constant vector. Our goal is to solve for [latex]\\left\\Vert{\\bf{r}}\\right\\Vert[\/latex]. Let's start by calculating [latex]{\\bf{r}}\\cdot({\\bf{v}}\\,\\times\\,{\\bf{C}})[\/latex]:\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\cdot({\\bf{v}}\\,\\times\\,{\\bf{C}})={\\bf{r}}\\cdot\\Big(GM\\frac{\\bf{r}}{\\left\\Vert{\\bf{r}}\\right\\Vert}+{\\bf{D}}\\Big)=GM\\frac{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}{\\left\\Vert{\\bf{r}}\\right\\Vert}+{\\bf{r}}\\cdot{\\bf{D}}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+{\\bf{r}}\\cdot{\\bf{D}}.[\/latex]<\/p>\r\nHowever, [latex]{\\bf{r}}\\cdot({\\bf{v}}\\,\\times\\,{\\bf{C}})=({\\bf{r}}\\,\\times\\,{\\bf{v}})\\,\\times\\,{\\bf{C}}[\/latex], so\r\n<p style=\"text-align: center;\">[latex]({\\bf{r}}\\,\\times\\,{\\bf{v}})\\,\\times\\,{\\bf{C}}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+{\\bf{r}}\\cdot{\\bf{D}}[\/latex].<\/p>\r\nSince [latex]{\\bf{r}}\\,\\times\\,{\\bf{v}}={\\bf{C}}[\/latex], we have\r\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{C}}\\right\\Vert^{2}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+{\\bf{r}}\\cdot{\\bf{D}}[\/latex].<\/p>\r\nNote that [latex]{\\bf{r}}\\cdot{\\bf{D}}=\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\left\\Vert{\\bf{D}}\\right\\Vert\\cos{\\theta}[\/latex], where [latex]\\theta[\/latex] is the angle between [latex]\\bf{r}[\/latex] and [latex]\\bf{D}[\/latex]. Therefore,\r\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{C}}\\right\\Vert^{2}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\left\\Vert{\\bf{D}}\\right\\Vert\\cos{\\theta}[\/latex].<\/p>\r\nSolving for [latex]\\left\\Vert{\\bf{r}}\\right\\Vert[\/latex],\r\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{r}}\\right\\Vert=\\frac{\\left\\Vert{\\bf{C}}\\right\\Vert^{2}}{GM+\\left\\Vert{\\bf{D}}\\right\\Vert\\cos{\\theta}}=\\frac{\\left\\Vert{\\bf{C}}\\right\\Vert^{2}}{GM}\\bigg(\\frac{1}{1+e\\cos{\\theta}}\\bigg)[\/latex],<\/p>\r\nwhere [latex]e=\\left\\Vert{\\bf{D}}\\right\\Vert\/GM[\/latex].\u00a0This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if [latex]e\\,&gt;\\,1[\/latex], a parabola if [latex]e=1[\/latex], or an ellipse if [latex]e\\,&lt;\\,1[\/latex]. Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.\r\n\r\n[latex]_\\blacksquare[\/latex]\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Kepler's Third Law for Nonheliocentric Orbits<\/h3>\r\nKepler\u2019s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler\u2019s third law becomes\r\n<p style=\"text-align: center;\">[latex]\\large{P^{2}=\\frac{4\\pi^{2}a^{3}}{G(m+M)}},[\/latex]<\/p>\r\nwhere [latex]m[\/latex] is the mass of the Moon and [latex]M[\/latex] is the mass of Earth,\u00a0[latex]a[\/latex] represents the length of the major axis of the elliptical orbit, and\u00a0[latex]P[\/latex] represents the period.\r\n\r\nGiven that the mass of the Moon is [latex]7.35\\ \\times\\ 10^{22}\\text{ kg}[\/latex], the mass of the Earth is [latex]5.97\\ \\times\\ 10^{24}\\text{ kg}[\/latex],\r\n\r\n[latex]G=6.67\\ \\times\\ 10^{-11}\\text{m}_{\\small{3}}\/\\text{kg}\\cdot\\text{sec}^{2}[\/latex], and the period of the moon is [latex]27.3[\/latex] days, let's find the length of the major axis of the orbit of the Moon around Earth.\r\n<div>[reveal-answer q=\"fs-id1177795055160\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1177795055160\"]\r\n<div class=\"exercise\">\r\n<p class=\"exercise\">It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]27.3\\text{ days}\\times\\frac{24\\text{ hr}}{1\\text{ day}}\\times\\frac{3600\\text{ sec}}{1\\text{ hour}}=2,358,720\\text{ sec}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nSubstitute all the data into the equation given in the problem and solve for [latex]a[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {(2,358,720\\text{ sec})^{2}} &amp; =\\hfill &amp; {\\frac{4\\pi^{2}a^{3}}{\\big(6.67\\,\\times\\,10^{-11}\\frac{m^{3}}{\\text{kg}\\cdot{\\text{sec}}^{2}}\\big)(7.35\\,\\times\\,10^{22}\\text{ kg}+5.97\\,\\times\\,10^{24}\\text{ kg})}} \\hfill \\\\ \\hfill {5.563\\,\\times\\,10^{12}}&amp; =\\hfill &amp; {\\frac{4\\pi^{2}a^{3}}{(6.67\\,\\times\\,10^{-11}\\,m^{3})(6.04\\,\\times\\,10^{24})}}\\hfill \\\\ \\hfill {(5.563\\,\\times\\,10^{12})(6.67\\,\\times\\,10^{-11}\\text{ m}^{3})(6.04\\,\\times\\,10^{24})}&amp; =\\hfill &amp; {4\\pi^{2}a^{3}}\\hfill \\\\ \\hfill {a^{3}}&amp; =\\hfill &amp; {\\frac{2.241\\,\\times\\,10^{27}}{4\\pi^{2}}\\text{m}^{3}}\\hfill \\\\ \\hfill {a}&amp; =\\hfill &amp; {3.84\\,\\times\\,10^{8}\\text{ m}}\\hfill \\\\ \\hfill &amp; \\approx\\hfill &amp; {384,000\\text{ km}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n[\/hidden-answer]\r\n<strong>Analysis<\/strong>\r\n\r\nAccording to solarsystem.nasa.gov, the actual average distance from the Moon to Earth is [latex]384,400\\text{ km}[\/latex]. This is calculated using reflectors left on the Moon by Apollo astronauts back in the 1960s.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nTitan is the largest moon of Saturn. The mass of Titan is approximately [latex]1.35\\,\\times\\,10^{23}\\text{ kg}[\/latex]. The mass of Saturn is approximately [latex]5.68\\,\\times\\,10^{26}\\text{ kg}[\/latex]. Titan takes approximately [latex]16[\/latex] days to orbit Saturn. \u00a0Use this information, along with the universal gravitation constant [latex]G=6.67\\,\\times\\,10^{-11}\\text{ m}^{3}\/\\text{kg}\\cdot{\\text{sec}}^{2}[\/latex] to estimate the distance from Titan to Saturn.\r\n\r\n[reveal-answer q=\"fs-id1168894933165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1168894933165\"]\r\n<div style=\"text-align: center;\">[latex]\\large{a=1.224\\,\\times\\,10^{9}\\approx1,224,000\\text{ km}}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949632&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Mk28qcuDB6A&amp;video_target=tpm-plugin-k8n0m4oy-Mk28qcuDB6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.17_transcript.html\">transcript for \u201cCP 3.17\u201d here (opens in new window).<\/a><\/center>\r\n<div data-type=\"title\">\r\n<div class=\"textbox tryit\">\r\n<h3>Activity: Navigating a Banked Turn<\/h3>\r\n<p id=\"fs-id1169737919206\" class=\" \">How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could depend on several factors:<\/p>\r\n\r\n<ul>\r\n \t<li>The weight of the car;<\/li>\r\n \t<li>The friction between the tires and the road;<\/li>\r\n \t<li>The radius of the circle;<\/li>\r\n \t<li>The \"steepness\" of the turn;<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1169737919231\" class=\" \">In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.<\/p>\r\n<p id=\"fs-id1169737919236\" class=\" \">A car of mass\u00a0[latex]m[\/latex] moves with constant angular speed [latex]\\omega[\/latex]\u00a0around a circular curve of radius\u00a0[latex]R[\/latex] (Figure 10). The curve is banked at an angle [latex]\\theta[\/latex].\u00a0If the height of the car off the ground is\u00a0[latex]h[\/latex], then the position of the car at time\u00a0[latex]t[\/latex] is given by the function [latex]r(t)=\\langle{\\bf{R}}\\cos{(\\omega{t})},\\ {\\bf{R}}\\sin{(\\omega{t})},\\ h\\rangle[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_935\" align=\"aligncenter\" width=\"706\"]<img class=\"size-full wp-image-935\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163437\/3-4-10.jpeg\" alt=\"This figure has two graphics. The first is a circle with a car on the circle. The circle is labeled \u201coverhead view\u201d. From the car there is a vector labeled \u201cv\u201d tangent to the circle. There is also a vector towards the center from the car labeled \u201ca\u201d. The second graphic is labeled \u201cfront view\u201d. It is the car at an angle. The angle is labeled \u201ctheta\u201d. The height of the cars tilt is labeled \u201ch\u201d.\" width=\"706\" height=\"276\" \/> Figure 6.\u00a0Views of a race car moving around a track.[\/caption]\r\n<ol>\r\n \t<li>Find the velocity function [latex]{\\bf{v}}\\,(t)[\/latex] of the car. Show that [latex]{\\bf{v}}[\/latex] is\u00a0tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it.<\/li>\r\n \t<li>Show that the speed of the car is [latex]\\omega{R}[\/latex].\u00a0Use this to show that [latex](2\\pi{r})\\,\/\\,{\\bf{v}}=(2\\pi)\\,\/\\,\\omega[\/latex].<\/li>\r\n \t<li>Find the acceleration [latex]{\\bf{a}}[\/latex]. Show that\u00a0this vector points toward the center of the circle and that [latex]|{\\bf{a}}|=R\\omega^{2}[\/latex].<\/li>\r\n \t<li>The force required to produce this circular motion is called the <em data-effect=\"italics\">centripetal force<\/em>, and it is denoted [latex]{\\bf{F}}_{\\text{cent}}[\/latex].\u00a0This force points toward the center of the circle (not toward the ground). Show that [latex]|{\\bf{F}}_{\\text{cent}}|=(m|{\\bf{v}}|^{2})\\,\/\\,R[\/latex]. As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure 11). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that [latex]|{\\bf{f}}|=\\mu|{\\bf{N}}|[\/latex]for some positive constant [latex]\\mu[\/latex]. The constant\u00a0[latex]\\mu[\/latex] is called the <em>coefficient of friction<\/em>.[caption id=\"attachment_936\" align=\"aligncenter\" width=\"334\"]<img class=\"size-full wp-image-936\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163551\/3-4-11.jpeg\" alt=\"This figure is the front of a car tilted to the left. The angle of the tilt is theta. From the center of the car are three vectors. The first vector is labeled \u201cN\u201d and is coming out of the top of the car perpendicular to the car. The second vector is coming out of the bottom of the car labeled \u201cmg\u201d. The third vector is labeled \u201cf\u201d and is coming out of the side of the car, orthogonal to \u201cN\u201d.\" width=\"334\" height=\"272\" \/> Figure 7.\u00a0The car has three forces acting on it: gravity (denoted by\u00a0[latex]m{\\bf{g}}[\/latex]), the friction force\u00a0[latex]\\bf{f}[\/latex], and the force exerted by the road\u00a0[latex]\\bf{N}[\/latex].[\/caption]Let [latex]v_{\\text{max}}[\/latex]\u00a0denote the maximum speed the car can attain through the curve without skidding. In other words,\u00a0[latex]v_{\\text{max}}[\/latex]\u00a0is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the magnitude of the centripetal force is\r\n<div style=\"text-align: center;\">[latex]|\\large{{\\bf{F}}_{\\text{cent}}|=\\frac{mv^{2}_{\\text{max}}}{R}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThe next three questions deal with developing a formula that relates the speed [latex]v_{\\text{max}}[\/latex] to the banking angle [latex]\\theta[\/latex].<\/li>\r\n \t<li>Show that [latex]|{\\bf{N}}|\\cos{\\theta}=mg+|{\\bf{f}}|\\sin{\\theta}[\/latex]. Conclude that [latex]|{bf{N}}|=(mg)\\,\/\\,(\\cos{\\theta}-\\mu\\sin{\\theta})[\/latex].<\/li>\r\n \t<li>The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force points toward the center of the circular curve. Show that\r\n<div style=\"text-align: center;\">[latex]|{\\bf{F}}_{\\text{cent}}|=|{\\bf{N}}|\\sin{\\theta}+|{\\bf{f}}|\\cos{\\theta}[\/latex].<\/div>\r\nConclude that\r\n<div style=\"text-align: center;\">[latex]|\\large{{\\bf{F}}_{\\text{cent}}|=\\frac{\\sin{\\theta}+\\mu\\cos{\\theta}}{\\cos{\\theta}-\\mu\\sin{\\theta}}mg}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Show that [latex]\\text{v}^{2}_{\\text{max}}=((\\sin{\\theta}+\\mu\\cos{\\theta}) \\ \/ \\ (\\cos{\\theta}-\\mu\\sin{\\theta}))gR[\/latex].\u00a0Conclude that the maximum speed does not actually depend on the mass of the car.<span data-type=\"newline\">\r\n<\/span>Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.<span data-type=\"newline\">\r\n<\/span>The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure 12. Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately [latex]211\\ ft[\/latex] with a banking angle of [latex]24^{\\circ}[\/latex]. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of [latex]28^{\\circ}[\/latex]. (The track has variable angle banking.)\r\n<div>\r\n\r\n[caption id=\"attachment_937\" align=\"aligncenter\" width=\"900\"]<img class=\"size-full wp-image-937\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163833\/3-4-12.jpeg\" alt=\"This figure has two graphics. The first is a picture of a raceway. There are cars on the track and fans in the stands. The second graphic is an oval drawing of a raceway. The inner radius of a curve is labeled \u201c211 ft\u201d and the width of the radius is labeled \u201c40 ft\u201d.\" width=\"900\" height=\"304\" \/> Figure 8.\u00a0At the Bristol Motor Speedway, Bristol, Tennessee (a), the turns have an inner radius of about 211 ft and a width of 40 ft (b). (credit: part (a) photo by Raniel Diaz, Flickr)[\/caption]\r\n\r\n<\/div>\r\nThe coefficient of friction for a normal tire in dry conditions is approximately [latex]0.7[\/latex]. Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately [latex]0.98[\/latex].\r\n\r\nBefore answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.<\/li>\r\n \t<li data-mark-suffix=\".\">In dry conditions, how fast can the car travel through the bottom of the turn without skidding?<\/li>\r\n \t<li data-mark-suffix=\".\">In dry conditions, how fast can the car travel through the top of the turn without skidding?<\/li>\r\n \t<li data-mark-suffix=\".\">In wet conditions, the coefficient of friction can become as low as [latex]0.1[\/latex]. If this is the case, how fast can the car travel through the bottom of the turn without skidding?<\/li>\r\n \t<li data-mark-suffix=\".\">Suppose the measured speed of a car going along the outside edge of the turn is [latex]105\\text{ mph}[\/latex]. Estimate the coefficient of friction for the car\u2019s tires.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>State Kepler\u2019s laws of planetary motion.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169737393059\" class=\"\">Now let\u2019s look at an application of vector functions. In particular, let\u2019s consider the effect of gravity on the motion of an object as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignore the effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as <span id=\"term145\" data-type=\"term\">projectile motion<\/span>. It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs.<\/p>\n<p id=\"fs-id1169737818838\" class=\"\">First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose the positive <em data-effect=\"italics\">y<\/em>-axis to be up, the negative <em data-effect=\"italics\">y-<\/em>axis to be down, and the positive <em data-effect=\"italics\">x-<\/em>axis to be forward (i.e., away from the thrower of the object). The effect of gravity is in a downward direction, so Newton\u2019s second law tells us that the force on the object resulting from gravity is equal to the mass of the object times the acceleration resulting from to gravity, or [latex]F_{g}=mg[\/latex], where [latex]F_{g}[\/latex]\u00a0represents the force from gravity and\u00a0[latex]g[\/latex] represents the acceleration resulting from gravity at Earth\u2019s surface. The value of\u00a0[latex]g[\/latex] in the English system of measurement is approximately [latex]32\\ ft\/sec^{2}[\/latex] and it is approximately [latex]9.8\\ m\/sec^{2}[\/latex] in the metric system.\u00a0This is the only force acting on the object. Since gravity acts in a downward direction, we can write the force resulting from gravity in the form [latex]F_{g}=-mg\\,{\\bf{j}}[\/latex], as shown in the following figure.<\/p>\n<div id=\"attachment_929\" style=\"width: 449px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-929\" class=\"size-full wp-image-929\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21162913\/3-4-5.jpeg\" alt=\"This figure is a ball falling in a vertical path. The ball is at the top at the initial position. From the ball a vector is drawn vertically downward labeled \u201cacceleration\u201d. The vertical line is labeled \u201cdistance\u201d. At the bottom of the line it is labeled \u201cfinal position\u201d. There is also grass at the bottom of the figure.\" width=\"439\" height=\"401\" \/><\/p>\n<p id=\"caption-attachment-929\" class=\"wp-caption-text\">Figure 1.\u00a0An object is falling under the influence of gravity.<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Interactive<\/h3>\n<p>Visit this <a href=\"http:\/\/www.openstax.org\/l\/20_projectile\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a> for a video showing projectile motion.<\/p>\n<\/div>\n<p>Newton\u2019s second law also tells us that [latex]F=m{\\bf{a}}[\/latex], where [latex]\\bf{a}[\/latex] represents the acceleration vector of the object. This force must be equal to the force of gravity at all times, so we therefore know that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {F} & =\\hfill & {F_{g}} \\hfill \\\\ \\hfill {m{\\bf{a}}}& =\\hfill & {-mg\\,{\\bf{j}}} \\hfill \\\\ \\hfill {{\\bf{a}}}& =\\hfill & {-g\\,{\\bf{j}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>Now we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite the last equation in the form<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{v}}'\\,(t)=-g\\,{\\bf{j}}.[\/latex]<\/p>\n<p>By taking the antiderivative of each side of this equation we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}\\,(t)} & =\\hfill & {\\displaystyle\\int_{}\\ -g\\,{\\bf{j}}\\,dt} \\hfill \\\\ \\hfill & =\\hfill & {-gt\\,{\\bf{j}}+{\\bf{C}}_{1}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>for some constant vector [latex]{\\bf{C}}_{1}[\/latex]. To determine the value of this vector, we can use the velocity of the object at a fixed time, say at time [latex]t=0[\/latex]. We call this velocity the <em>initial velocity<\/em>: [latex]{\\bf{v}}\\,(0)={\\bf{v}}_{0}[\/latex]. Therefore, [latex]{\\bf{v}}\\,(0)=-g\\,(0)\\,{\\bf{j}}+{\\bf{C}}_{1}={\\bf{v}}_{0}[\/latex] and [latex]{\\bf{C}}_{1}={\\bf{v}}_{0}[\/latex]. This gives the velocity vector as [latex]{\\bf{v}}\\,(t)=-gt\\,{\\bf{j}}+{\\bf{v}}_{0}[\/latex].<\/p>\n<p>Next we use the fact that velocity [latex]{\\bf{v}}\\,(t)[\/latex] is the derivative of position [latex]{\\bf{s}}\\,(t)[\/latex]. This gives the equation<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{s}}'\\,(t)=-gt\\,{\\bf{j}}+{\\bf{v}}_{0}[\/latex].<\/p>\n<p>Taking the antiderivative of both sides of this equation leads to<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(t)} & =\\hfill & {\\displaystyle\\int_{}\\ -gt\\,{\\bf{j}}+{\\bf{v}}_{0}\\,dt} \\hfill \\\\ \\hfill & =\\hfill & {-\\frac{1}{2}\\,gt^{2}\\,{\\bf{j}}+{\\bf{v}}_{0}t+{\\bf{C}}_{2},}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>with another unknown constant vector [latex]{\\bf{C}}_{2}[\/latex]. To determine the value of [latex]{\\bf{C}}_{2}[\/latex], we can use the position of the object at a given time, say at time [latex]t=0[\/latex]. We call this position the\u00a0<em>initial position<\/em>: [latex]{\\bf{s}}\\,(0)={\\bf{s}}_{0}[\/latex]. Therefore, [latex]{\\bf{s}}\\,(0)=-(1\/2)\\,g(0)^{2}\\,{\\bf{j}}+{\\bf{v}}_{0}\\,(0)+{\\bf{C}}_{2}={\\bf{s}}_{0}[\/latex] and [latex]{\\bf{C}}_{2}={\\bf{s}}_{0}[\/latex]. This gives the position of the object at any time as<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{s}}\\,(t)=-\\frac{1}{2}\\,gt^{2}\\,{\\bf{j}}+{\\bf{v}}_{0}\\,t+{\\bf{s}}_{0}.[\/latex]<\/p>\n<p>Let\u2019s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from the origin at an angle [latex]\\theta[\/latex] to the horizontal, with initial speed [latex]v_{0}[\/latex]. How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. Therefore, [latex]{\\bf{s}}_{0}={\\bf{0}}[\/latex], as shown in the following figure.<\/p>\n<div id=\"attachment_930\" style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-930\" class=\"size-full wp-image-930\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163018\/3-4-6.jpeg\" alt=\"This figure has an upside down parabolic curve representing projectile motion. The figure is labeled \u201cconstant horizontal velocity; constant vertical acceleration\u201d. The curve is in the first quadrant beginning and ending on the x-axis. The height of the curve is labeled \u201cH\u201d. The distance on the x-axis is labeled \u201cR\u201d. Angle theta represents the direction of the projectile at the origin. Five points are labeled on the graph with vectors. The vectors are labeled \u201cv\u201d with subscripts representing directions.\" width=\"650\" height=\"352\" \/><\/p>\n<p id=\"caption-attachment-930\" class=\"wp-caption-text\">Figure 2.\u00a0Projectile motion when the object is thrown upward at an angle [latex]\\theta[\/latex].\u00a0The horizontal motion is at constant velocity and the vertical motion is at constant acceleration.<\/p>\n<\/div>\n<p>We can rewrite the initial velocity vector in the form [latex]{\\bf{v}}_{0}=v_{0}\\cos{\\theta}{\\bf{i}}+v_{0}\\sin{\\theta}{\\bf{j}}[\/latex]. Then the equation for the position functions [latex]{\\bf{s}}(t)[\/latex] becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(t)} & =\\hfill & {-\\frac{1}{2}gt^{2}{\\bf{j}}+v_{0}t\\cos{\\theta{\\bf{i}}}+v_{0}t\\sin{\\theta{\\bf{j}}}} \\hfill \\\\ \\hfill & =\\hfill & {v_{0}t\\cos{\\theta{\\bf{i}}}+v_{0}t\\sin{\\theta{\\bf{j}}}-\\frac{1}{2}gt^{2}\\,{\\bf{j}}}\\hfill \\\\ \\hfill & =\\hfill & {v_{0}t\\cos{\\theta{\\bf{i}}}+(v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2})\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>The coefficient of [latex]{\\bf{i}}[\/latex] represents the horizontal component of [latex]{\\bf{s}}(t)[\/latex] and is the horizontal distance of the object from the origin at time [latex]t[\/latex]. The maximum value of the horizontal distance (measured at the same initial and final attitude) is called the range [latex]R[\/latex]. The coefficient of [latex]{\\bf{j}}[\/latex] represents the vertical component of [latex]{\\bf{s}}(t)[\/latex] and is the altitude of the object at time [latex]t[\/latex]. The maximum value of the vertical distance is the height [latex]H[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Motion of a Cannonball<\/h3>\n<p>During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of [latex]30^\\circ[\/latex] above horizontal and the initial speed of the cannonball is [latex]600\\ ft\/sec[\/latex]. The cliff is [latex]100\\ ft[\/latex] above the water (Figure 7).<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Find the maximum height of the cannonball.<\/li>\n<li>How long will it take for the cannonball to splash into the sea?<\/li>\n<li>How far out to sea will the cannonball hit the water?\n<div id=\"attachment_931\" style=\"width: 702px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-931\" class=\"size-full wp-image-931\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163109\/3-4-7.jpeg\" alt=\"This figure has a cannon at the edge of a cliff aimed upwards. There is a cannonball coming out of the cannon. The path of the cannonball is an upside down parabola represented with a broken line. The maximum height is labeled with \u201c?\u201d. The height of the cliff is 100 feet.\" width=\"692\" height=\"412\" \/><\/p>\n<p id=\"caption-attachment-931\" class=\"wp-caption-text\">Figure 3.\u00a0The flight of a cannonball (ignoring air resistance) is projectile motion.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1267795055160\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1267795055160\" class=\"hidden-answer\" style=\"display: none\">We use the equation<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{s}}\\,(t)=v_{0}t\\cos{\\theta{\\bf{i}}}+\\big(v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2}\\big){\\bf{j}}[\/latex]<\/div>\n<div><\/div>\n<p>with [latex]\\theta={30^\\circ}[\/latex],<span style=\"font-size: 1rem; text-align: initial;\">\u00a0[latex]g=32\\ ft\/sec^{2}[\/latex], and [latex]v_{0}=600\\ ft\/sec[\/latex]. Then the position equation becomes<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(t)} & =\\hfill & {600t(\\cos{30}){\\bf{i}}+(600t\\sin{30}-\\frac{1}{2}(32)t^{2}){\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {300t\\sqrt{3}\\,{\\bf{i}}+(300t-16t^{2})\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The cannonball reaches its maximum height when the vertical component of its velocity is zero, because the cannonball is neither rising nor falling at that point. The velocity vector is\n<div><\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}\\,(t)} & =\\hfill & {{\\bf{s}}'\\,(t)} \\hfill \\\\ \\hfill & =\\hfill & {300\\sqrt{3}\\,{\\bf{i}}+(300-32t)\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, the vertical component of velocity is given by the expression [latex]300-32t[\/latex]. Setting this expression equal to zero and solving [latex]t[\/latex] gives [latex]t=9.375\\ sec[\/latex]. The height of the cannonball at this time is given by the vertical component of the position vector, evaluated at [latex]t=9.375[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(9.375)} & =\\hfill & {300\\,(9.375)\\,\\sqrt{3}\\,{\\bf{i}}+(300\\,(9.375)-16\\,(9.375)^{2}){\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {4871.39\\,{\\bf{i}}+1406.25\\,{\\bf{j}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, the maximum height of the cannonball is [latex]1406.39\\text{ }ft[\/latex] above the cannon, or [latex]1506.39\\text{ }ft[\/latex] above sea level.<\/li>\n<li>When the cannonball lands in the water, is is [latex]100\\text{ }ft[\/latex] below the cannon. Therefore, the vertical component of the position vector is equal to [latex]-100[\/latex]. Setting the vertical component of [latex]{\\bf{s}}\\,(t)[\/latex] equal to [latex]-100[\/latex] and solving, we obtain\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {300t-16t^{2}} & =\\hfill & {-100} \\hfill \\\\ \\hfill {16t^{2}-300t-100}& =\\hfill & {0} \\hfill \\\\ \\hfill {4t^{2}-75t-25}& =\\hfill & {0} \\hfill \\\\ \\hfill {t}& =\\hfill & {\\frac{75\\pm\\sqrt{(-75)^{2}-4(4)(-25)}}{2(4)}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{75\\pm\\sqrt{6025}}{8}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{75\\pm5\\sqrt{241}}{8}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The positive value of [latex]t[\/latex] that solves this equation is approximately [latex]19.08[\/latex]. Therefore, the cannonball hits the water after approximately [latex]19.08\\text{ }sec[\/latex].<\/li>\n<li>To find the distance out to sea, we simply substitute the answer from part (b) into [latex]{\\bf{s}}\\,(t)[\/latex]:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}\\,(19.08)} & =\\hfill & {300\\,(19.08)\\,\\sqrt{3}\\,{\\bf{i}}+(300\\,(19.08)-16\\,(19.08)^{2})\\,{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {9914.26\\,{\\bf{i}}-100.7424\\,{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, the ball hits the water about [latex]9914.26\\text{ }ft[\/latex] away from the base of the cliff. Notice that the vertical component of the position vector is very close to [latex]-100[\/latex], which tells us that the ball just hit the water. Note that [latex]9914.26[\/latex] feet is not the true range of the cannon since the cannonball lands in the ocean at a location below the cannon. The range of the cannon would be determined by finding how far out the cannonball is when its height is [latex]100\\text{ }ft[\/latex] above the water (the same as the altitude of the cannon).<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>An archer fires an arrow at an angle of [latex]40^\\circ[\/latex] above the horizontal with an initial speed of [latex]98\\text{ }m\/sec[\/latex]. The height of the archer is [latex]171.5\\text{ }cm[\/latex]. Find the horizontal distance the arrow travels before it hits the ground.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168794933165\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168794933165\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]967.15\\ m[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949633&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=oFRnQ2QtMeY&amp;video_target=tpm-plugin-05gnio59-oFRnQ2QtMeY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.16_transcript.html\">transcript for \u201cCP 3.16\u201d here (opens in new window).<\/a><\/div>\n<p>One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? To determine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level. In other words, we want to determine an equation for the range. In this case, the equation of projectile motion is<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{s}}(t)=v_{0}t\\cos{\\theta{\\bf{i}}}+(v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2})\\,{\\bf{j}}.[\/latex]<\/p>\n<p>Setting the second component equal to zero and solving for [latex]t[\/latex] yields<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2}} & =\\hfill & {0} \\hfill \\\\ \\hfill {t(v_{0}\\sin{\\theta}-\\frac{1}{2}gt)}& =\\hfill & {0}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, either [latex]t=0[\/latex] or [latex]t=\\frac{2v_{0}\\sin{\\theta}}{g}[\/latex]. We are interested in the second value of [latex]t[\/latex], so we substitute this into [latex]{\\bf{s}}(t)[\/latex], which gives<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{{\\bf{s}}\\,\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)} & =\\hfill & {v_{0}\\,\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)\\cos{\\theta{\\bf{i}}}+\\Big(v_{0}\\,\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)\\sin{\\theta}-\\frac{1}{2}g\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)^{2}\\Big)\\,{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {\\bigg(\\frac{2v^2_0\\sin{\\theta}\\cos{\\theta}}{g}\\bigg)\\,{\\bf{i}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{v^2_0\\sin{2\\theta}}{g}\\,{\\bf{i}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>Thus, the expression for the range of a projectile fired at an angle [latex]\\theta[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{R=\\frac{v^2_0\\sin{2\\theta}}{g}\\,{\\bf{i}}}.[\/latex]<\/p>\n<p>The only variable in this expression is [latex]\\theta[\/latex]. To maximize the distance traveled, take the derivative of the coefficient of [latex]{\\bf{i}}[\/latex] with respect to [latex]\\theta[\/latex] and set it equal to zero:<\/p>\n<p>[latex]\\begin{align}    \\hspace{10cm} \\frac{d}{d\\theta}\\left(\\frac{v_{0}^2t\\sin{2\\theta}}{g}\\right) & =0 \\\\    \\frac{2v_{0}\\cos{2\\theta}}{g} & = 0 \\\\    \\theta & = 45^\\circ.    \\end{align}[\/latex]<\/p>\n<div>\n<p>This value of [latex]\\theta[\/latex] is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air resistance, the best angle to fire a projectile (to maximize the range) is at a [latex]45^\\circ[\/latex] angle. The distance it travels is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\bf{s}}\\bigg(\\frac{2v_{0}\\sin{45}}{g}\\bigg)=\\frac{v^2_0\\sin{90}}{g}\\,{\\bf{i}}=\\frac{v^2_0}{g}\\,{\\bf{j}}}.[\/latex]<\/p>\n<p>Therefore, the range for an angle of [latex]45^\\circ[\/latex] is [latex]v^2_0\/g[\/latex].<\/p>\n<h2>Kepler&#8217;s Laws<\/h2>\n<p>During the early 1600s, Johannes <span id=\"term146\" class=\"no-emphasis\" data-type=\"term\">Kepler<\/span> was able to use the amazingly accurate data from his mentor Tycho <span id=\"term147\" class=\"no-emphasis\" data-type=\"term\">Brahe<\/span>to formulate his three laws of planetary motion, now known as <strong><span id=\"term148\" data-type=\"term\">Kepler\u2019s laws of planetary motion<\/span><\/strong>. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley\u2019s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Kepler&#8217;s Laws of Planetary Motion Theorem<\/h3>\n<hr \/>\n<ol>\n<li>The path of any planet about the Sun is elliptical in shape, with the center of the Sun located at one focus of the ellipse (the law of ellipses).<\/li>\n<li>A line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal time intervals (the law of equal areas) (Figure 8).<\/li>\n<li>The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of the lengths of their semi-major orbital axes (the law of harmonies).<\/li>\n<\/ol>\n<\/div>\n<div id=\"attachment_932\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-932\" class=\"size-full wp-image-932\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163209\/3-4-8.jpeg\" alt=\"This figure is an elliptical curve labeled \u201cplanets orbit\u201d. The sun is represented towards the left inside the ellipse, at a focal point. Along the ellipse there are points A,B,C,D,E,F. There are line segments from the sun to each point.\" width=\"417\" height=\"345\" \/><\/p>\n<p id=\"caption-attachment-932\" class=\"wp-caption-text\">Figure 4. Kepler\u2019s first and second laws are pictured here. The Sun is located at a focus of the elliptical orbit of any planet. Furthermore, the shaded areas are all equal, assuming that the amount of time measured as the planet moves is the same for each region.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Kepler\u2019s third law is especially useful when using appropriate units. In particular, [latex]1\\text{ }\\textit{astronomical unit}[\/latex] is defined to be the average distance from Earth to the Sun, and is now recognized to be [latex]149,597,870,700\\text{ }m[\/latex] or, approximately [latex]93,000,000[\/latex]mi. We therefore write [latex]1\\text{ }A.U. = 93,000,000[\/latex]mi. Since the time it takes for Earth to orbit the Sun is [latex]1[\/latex] year, we use Earth years for units of time. Then, substituting [latex]1[\/latex] year for the period of Earth and [latex]1[\/latex]A.U. for the average distance to the Sun, Kepler\u2019s third law can be written as<\/p>\n<p style=\"text-align: center;\">[latex]T^2_p=D^3_p[\/latex]<\/p>\n<p>for any planet in the solar system, where [latex]T_p[\/latex] is the period of that planet measured in Earth years and [latex]D_p[\/latex]\u00a0is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.<\/p>\n<p id=\"fs-id1169738059916\" class=\"\">Kepler\u2019s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler\u2019s third law to other orbital systems, such as a moon orbiting around a planet. Kepler\u2019s original third law only applies to objects orbiting the Sun.<\/p>\n<h3>Proof<\/h3>\n<p>Let\u2019s now prove Kepler\u2019s first law using the calculus of vector-valued functions. First we need a coordinate system. Let\u2019s place the Sun at the origin of the coordinate system and let the vector-valued function [latex]{\\bf{r}}\\,(t)[\/latex] represent the location of a planet as a function of time. Newton proved Kepler&#8217;s law using his second law of\u00a0motion and his law of universal gravitation. Newton\u2019s second law of motion can be written as [latex]{\\bf{F}}=m{\\bf{a}}[\/latex], where [latex]{\\bf{F}}[\/latex] represents the net force acting on the planet. His law of universal gravitation can be written in the form [latex]{\\bf{F}}=-\\frac{GmM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\cdot\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}[\/latex], which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude [latex]\\frac{GmM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}[\/latex] (Figure 9).<\/p>\n<div id=\"attachment_933\" style=\"width: 388px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-933\" class=\"size-full wp-image-933\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163250\/3-4-9.jpeg\" alt=\"This figure is an ellipse with a circle to the left on the inside at a focal point. The circle represents the sun. On the ellipse is a smaller circle representing Earth. The line segment drawn between the circles is labeled \u201cgravitational force\u201d.\" width=\"378\" height=\"286\" \/><\/p>\n<p id=\"caption-attachment-933\" class=\"wp-caption-text\">Figure 5.\u00a0The gravitational force between Earth and the Sun is equal to the mass of the earth times its acceleration.<\/p>\n<\/div>\n<p>Setting these two forces equal to each other, and using the fact that [latex]{\\bf{a}}\\,(t)={\\bf{v}}'\\,(t)[\/latex], we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\large{m{\\bf{v}}'\\,(t)=-\\frac{GmM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\cdot\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}}[\/latex],<\/p>\n<p>which can be rewritten as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{d{\\bf{v}}}{dt}=-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}{\\bf{r}}}.[\/latex]<\/p>\n<p>This equation shows that the vectors [latex]d{\\bf{v}}\/dt[\/latex] and [latex]{\\bf{r}}[\/latex] are parallel to each other, so [latex]d{\\bf{v}}\/dt\\,\\times\\,{\\bf{r}}=0[\/latex]. Next, let&#8217;s differentiate [latex]{\\bf{r}}\\,\\times\\,{\\bf{v}}[\/latex] with respect to time:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{d}{dt}({\\bf{r}}\\,\\times\\,{\\bf{v}})=\\frac{d{\\bf{r}}}{dt}\\,\\times\\,{\\bf{v}}+{\\bf{r}}\\,\\times\\frac{d{\\bf{v}}}{dt}={\\bf{v}}\\,\\times\\,{\\bf{v}}+{\\bf{0}}={\\bf{0}}}.[\/latex]<\/p>\n<p>This proves that [latex]{\\bf{r}}\\,\\times\\,{\\bf{v}}[\/latex] is a constant vector, which we call [latex]{\\bf{C}}[\/latex]. Since [latex]{\\bf{r}}[\/latex] and [latex]{\\bf{v}}[\/latex] are both perpendicular to\u00a0[latex]{\\bf{C}}[\/latex] for all values of [latex]t[\/latex], they must lie in a plane perpendicular to [latex]{\\bf{C}}[\/latex]. Therefore, the motion of the planet lies in a plane.<\/p>\n<p>Next we calculate the expression [latex]d{\\bf{v}}\/dt\\,\\times\\,{\\bf{C}}[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}=-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}{\\bf{r}}\\,\\times\\,({\\bf{r}}\\,\\times\\,{\\bf{v}})=-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}[({\\bf{r}}\\cdot{\\bf{v}})\\,{\\bf{r}}-({\\bf{r}}\\cdot{\\bf{r}})\\,{\\bf{v}}].[\/latex]<\/p>\n<p>The last equality in the above equation is from the triple cross product formula (Introduction to Vectors in Space). We need an expression for [latex]{\\bf{r}}\\cdot{\\bf{v}}[\/latex]. To calculate this, we differentiate [latex]{\\bf{r}}\\cdot{\\bf{r}}[\/latex] with respect to time:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}({\\bf{r}}\\,\\cdot\\,{\\bf{r}})=\\frac{d{\\bf{r}}}{dt}\\cdot{\\bf{r}}+{\\bf{r}}\\cdot\\frac{d{\\bf{r}}}{dt}=2{\\bf{r}}\\cdot\\frac{d{\\bf{r}}}{dt}=2{\\bf{r}}\\cdot{\\bf{v}}.[\/latex]<\/p>\n<p>Since [latex]{\\bf{r}}\\,\\cdot\\,{\\bf{r}}=\\left\\Vert{\\bf{r}}\\right\\Vert^{2}[\/latex], we also have<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dt}({\\bf{r}}\\cdot{\\bf{r}})=\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert^{2}=2\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert[\/latex].<\/p>\n<p>Combining the last two equations, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {2{\\bf{r}}\\cdot{\\bf{v}}} & =\\hfill & {2\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert} \\hfill \\\\ \\hfill {{\\bf{r}}\\cdot{\\bf{v}}}& =\\hfill & {\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert{.}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>Substituting into our expression for [latex]\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}[\/latex] gives us<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}} & =\\hfill & {-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}[({\\bf{r}}\\cdot{\\bf{r}})\\,{\\bf{r}}-({\\bf{r}}\\cdot{\\bf{r}})\\,{\\bf{v}}]} \\hfill \\\\ \\hfill & =\\hfill & {-\\frac{GM}{\\left\\Vert{\\bf{r}}\\right\\Vert^{3}}\\big[\\left\\Vert{\\bf{r}}\\right\\Vert{\\big(}\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert\\big)\\,{\\bf{r}}-\\left\\Vert{\\bf{r}}\\right\\Vert^{2}{\\bf{v}}\\big]}\\hfill \\\\ \\hfill & =\\hfill & {-GM\\Big[\\frac{1}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\big(\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert\\big)\\,{\\bf{r}}-\\frac{1}{\\left\\Vert{\\bf{r}}\\right\\Vert}{\\bf{v}}\\Big]}\\hfill \\\\ \\hfill & =\\hfill & {GM\\Big[\\frac{\\bf{v}}{\\left\\Vert{\\bf{r}}\\right\\Vert}-\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\big(\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert\\big)\\Big].}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>However,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{d}{dt}\\,\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}} & =\\hfill & {\\frac{\\frac{d}{dt}({\\bf{r}})\\left\\Vert{\\bf{r}}\\right\\Vert-{\\bf{r}}\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert}{\\left\\Vert{\\bf{r}}\\right\\Vert}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\frac{d{\\bf{r}}}{dt}}{\\left\\Vert{\\bf{r}}\\right\\Vert}-\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\bf{v}}{\\left\\Vert{\\bf{r}}\\right\\Vert}-\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{v}}\\right\\Vert^{2}}\\,\\frac{d}{dt}\\left\\Vert{\\bf{r}}\\right\\Vert.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{d{\\bf{v}}}{dt}\\,\\times\\,{\\bf{C}}=GM\\Big(\\frac{d}{dt}\\,\\frac{\\bf{r}}{\\left\\Vert{\\bf{r}}\\right\\Vert}\\Big)}.[\/latex]<\/p>\n<p>Since [latex]\\bf{C}[\/latex] is a constant vector, we can integrate both sides and obtain<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{v}}\\,\\times\\,{\\bf{C}}=GM\\frac{{\\bf{r}}}{\\left\\Vert{\\bf{r}}\\right\\Vert}+{\\bf{D}},[\/latex]<\/p>\n<p>where [latex]\\bf{D}[\/latex] is a constant vector. Our goal is to solve for [latex]\\left\\Vert{\\bf{r}}\\right\\Vert[\/latex]. Let&#8217;s start by calculating [latex]{\\bf{r}}\\cdot({\\bf{v}}\\,\\times\\,{\\bf{C}})[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}\\cdot({\\bf{v}}\\,\\times\\,{\\bf{C}})={\\bf{r}}\\cdot\\Big(GM\\frac{\\bf{r}}{\\left\\Vert{\\bf{r}}\\right\\Vert}+{\\bf{D}}\\Big)=GM\\frac{\\left\\Vert{\\bf{r}}\\right\\Vert^{2}}{\\left\\Vert{\\bf{r}}\\right\\Vert}+{\\bf{r}}\\cdot{\\bf{D}}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+{\\bf{r}}\\cdot{\\bf{D}}.[\/latex]<\/p>\n<p>However, [latex]{\\bf{r}}\\cdot({\\bf{v}}\\,\\times\\,{\\bf{C}})=({\\bf{r}}\\,\\times\\,{\\bf{v}})\\,\\times\\,{\\bf{C}}[\/latex], so<\/p>\n<p style=\"text-align: center;\">[latex]({\\bf{r}}\\,\\times\\,{\\bf{v}})\\,\\times\\,{\\bf{C}}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+{\\bf{r}}\\cdot{\\bf{D}}[\/latex].<\/p>\n<p>Since [latex]{\\bf{r}}\\,\\times\\,{\\bf{v}}={\\bf{C}}[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{C}}\\right\\Vert^{2}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+{\\bf{r}}\\cdot{\\bf{D}}[\/latex].<\/p>\n<p>Note that [latex]{\\bf{r}}\\cdot{\\bf{D}}=\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\left\\Vert{\\bf{D}}\\right\\Vert\\cos{\\theta}[\/latex], where [latex]\\theta[\/latex] is the angle between [latex]\\bf{r}[\/latex] and [latex]\\bf{D}[\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{C}}\\right\\Vert^{2}=GM\\left\\Vert{\\bf{r}}\\right\\Vert+\\left\\Vert{\\bf{r}}\\right\\Vert\\,\\left\\Vert{\\bf{D}}\\right\\Vert\\cos{\\theta}[\/latex].<\/p>\n<p>Solving for [latex]\\left\\Vert{\\bf{r}}\\right\\Vert[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\Vert{\\bf{r}}\\right\\Vert=\\frac{\\left\\Vert{\\bf{C}}\\right\\Vert^{2}}{GM+\\left\\Vert{\\bf{D}}\\right\\Vert\\cos{\\theta}}=\\frac{\\left\\Vert{\\bf{C}}\\right\\Vert^{2}}{GM}\\bigg(\\frac{1}{1+e\\cos{\\theta}}\\bigg)[\/latex],<\/p>\n<p>where [latex]e=\\left\\Vert{\\bf{D}}\\right\\Vert\/GM[\/latex].\u00a0This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if [latex]e\\,>\\,1[\/latex], a parabola if [latex]e=1[\/latex], or an ellipse if [latex]e\\,<\\,1[\/latex]. Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.\n\n[latex]_\\blacksquare[\/latex]\n\n\n<div class=\"textbox exercises\">\n<h3>Example: Using Kepler&#8217;s Third Law for Nonheliocentric Orbits<\/h3>\n<p>Kepler\u2019s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler\u2019s third law becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\large{P^{2}=\\frac{4\\pi^{2}a^{3}}{G(m+M)}},[\/latex]<\/p>\n<p>where [latex]m[\/latex] is the mass of the Moon and [latex]M[\/latex] is the mass of Earth,\u00a0[latex]a[\/latex] represents the length of the major axis of the elliptical orbit, and\u00a0[latex]P[\/latex] represents the period.<\/p>\n<p>Given that the mass of the Moon is [latex]7.35\\ \\times\\ 10^{22}\\text{ kg}[\/latex], the mass of the Earth is [latex]5.97\\ \\times\\ 10^{24}\\text{ kg}[\/latex],<\/p>\n<p>[latex]G=6.67\\ \\times\\ 10^{-11}\\text{m}_{\\small{3}}\/\\text{kg}\\cdot\\text{sec}^{2}[\/latex], and the period of the moon is [latex]27.3[\/latex] days, let&#8217;s find the length of the major axis of the orbit of the Moon around Earth.<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1177795055160\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1177795055160\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"exercise\">\n<p class=\"exercise\">It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well:<\/p>\n<div style=\"text-align: center;\">[latex]27.3\\text{ days}\\times\\frac{24\\text{ hr}}{1\\text{ day}}\\times\\frac{3600\\text{ sec}}{1\\text{ hour}}=2,358,720\\text{ sec}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Substitute all the data into the equation given in the problem and solve for [latex]a[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {(2,358,720\\text{ sec})^{2}} & =\\hfill & {\\frac{4\\pi^{2}a^{3}}{\\big(6.67\\,\\times\\,10^{-11}\\frac{m^{3}}{\\text{kg}\\cdot{\\text{sec}}^{2}}\\big)(7.35\\,\\times\\,10^{22}\\text{ kg}+5.97\\,\\times\\,10^{24}\\text{ kg})}} \\hfill \\\\ \\hfill {5.563\\,\\times\\,10^{12}}& =\\hfill & {\\frac{4\\pi^{2}a^{3}}{(6.67\\,\\times\\,10^{-11}\\,m^{3})(6.04\\,\\times\\,10^{24})}}\\hfill \\\\ \\hfill {(5.563\\,\\times\\,10^{12})(6.67\\,\\times\\,10^{-11}\\text{ m}^{3})(6.04\\,\\times\\,10^{24})}& =\\hfill & {4\\pi^{2}a^{3}}\\hfill \\\\ \\hfill {a^{3}}& =\\hfill & {\\frac{2.241\\,\\times\\,10^{27}}{4\\pi^{2}}\\text{m}^{3}}\\hfill \\\\ \\hfill {a}& =\\hfill & {3.84\\,\\times\\,10^{8}\\text{ m}}\\hfill \\\\ \\hfill & \\approx\\hfill & {384,000\\text{ km}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<p><strong>Analysis<\/strong><\/p>\n<p>According to solarsystem.nasa.gov, the actual average distance from the Moon to Earth is [latex]384,400\\text{ km}[\/latex]. This is calculated using reflectors left on the Moon by Apollo astronauts back in the 1960s.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Titan is the largest moon of Saturn. The mass of Titan is approximately [latex]1.35\\,\\times\\,10^{23}\\text{ kg}[\/latex]. The mass of Saturn is approximately [latex]5.68\\,\\times\\,10^{26}\\text{ kg}[\/latex]. Titan takes approximately [latex]16[\/latex] days to orbit Saturn. \u00a0Use this information, along with the universal gravitation constant [latex]G=6.67\\,\\times\\,10^{-11}\\text{ m}^{3}\/\\text{kg}\\cdot{\\text{sec}}^{2}[\/latex] to estimate the distance from Titan to Saturn.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168894933165\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168894933165\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\large{a=1.224\\,\\times\\,10^{9}\\approx1,224,000\\text{ km}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949632&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Mk28qcuDB6A&amp;video_target=tpm-plugin-k8n0m4oy-Mk28qcuDB6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.17_transcript.html\">transcript for \u201cCP 3.17\u201d here (opens in new window).<\/a><\/div>\n<div data-type=\"title\">\n<div class=\"textbox tryit\">\n<h3>Activity: Navigating a Banked Turn<\/h3>\n<p id=\"fs-id1169737919206\" class=\"\">How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could depend on several factors:<\/p>\n<ul>\n<li>The weight of the car;<\/li>\n<li>The friction between the tires and the road;<\/li>\n<li>The radius of the circle;<\/li>\n<li>The &#8220;steepness&#8221; of the turn;<\/li>\n<\/ul>\n<p id=\"fs-id1169737919231\" class=\"\">In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.<\/p>\n<p id=\"fs-id1169737919236\" class=\"\">A car of mass\u00a0[latex]m[\/latex] moves with constant angular speed [latex]\\omega[\/latex]\u00a0around a circular curve of radius\u00a0[latex]R[\/latex] (Figure 10). The curve is banked at an angle [latex]\\theta[\/latex].\u00a0If the height of the car off the ground is\u00a0[latex]h[\/latex], then the position of the car at time\u00a0[latex]t[\/latex] is given by the function [latex]r(t)=\\langle{\\bf{R}}\\cos{(\\omega{t})},\\ {\\bf{R}}\\sin{(\\omega{t})},\\ h\\rangle[\/latex].<\/p>\n<div id=\"attachment_935\" style=\"width: 716px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-935\" class=\"size-full wp-image-935\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163437\/3-4-10.jpeg\" alt=\"This figure has two graphics. The first is a circle with a car on the circle. The circle is labeled \u201coverhead view\u201d. From the car there is a vector labeled \u201cv\u201d tangent to the circle. There is also a vector towards the center from the car labeled \u201ca\u201d. The second graphic is labeled \u201cfront view\u201d. It is the car at an angle. The angle is labeled \u201ctheta\u201d. The height of the cars tilt is labeled \u201ch\u201d.\" width=\"706\" height=\"276\" \/><\/p>\n<p id=\"caption-attachment-935\" class=\"wp-caption-text\">Figure 6.\u00a0Views of a race car moving around a track.<\/p>\n<\/div>\n<ol>\n<li>Find the velocity function [latex]{\\bf{v}}\\,(t)[\/latex] of the car. Show that [latex]{\\bf{v}}[\/latex] is\u00a0tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it.<\/li>\n<li>Show that the speed of the car is [latex]\\omega{R}[\/latex].\u00a0Use this to show that [latex](2\\pi{r})\\,\/\\,{\\bf{v}}=(2\\pi)\\,\/\\,\\omega[\/latex].<\/li>\n<li>Find the acceleration [latex]{\\bf{a}}[\/latex]. Show that\u00a0this vector points toward the center of the circle and that [latex]|{\\bf{a}}|=R\\omega^{2}[\/latex].<\/li>\n<li>The force required to produce this circular motion is called the <em data-effect=\"italics\">centripetal force<\/em>, and it is denoted [latex]{\\bf{F}}_{\\text{cent}}[\/latex].\u00a0This force points toward the center of the circle (not toward the ground). Show that [latex]|{\\bf{F}}_{\\text{cent}}|=(m|{\\bf{v}}|^{2})\\,\/\\,R[\/latex]. As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure 11). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that [latex]|{\\bf{f}}|=\\mu|{\\bf{N}}|[\/latex]for some positive constant [latex]\\mu[\/latex]. The constant\u00a0[latex]\\mu[\/latex] is called the <em>coefficient of friction<\/em>.\n<div id=\"attachment_936\" style=\"width: 344px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-936\" class=\"size-full wp-image-936\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163551\/3-4-11.jpeg\" alt=\"This figure is the front of a car tilted to the left. The angle of the tilt is theta. From the center of the car are three vectors. The first vector is labeled \u201cN\u201d and is coming out of the top of the car perpendicular to the car. The second vector is coming out of the bottom of the car labeled \u201cmg\u201d. The third vector is labeled \u201cf\u201d and is coming out of the side of the car, orthogonal to \u201cN\u201d.\" width=\"334\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-936\" class=\"wp-caption-text\">Figure 7.\u00a0The car has three forces acting on it: gravity (denoted by\u00a0[latex]m{\\bf{g}}[\/latex]), the friction force\u00a0[latex]\\bf{f}[\/latex], and the force exerted by the road\u00a0[latex]\\bf{N}[\/latex].<\/p>\n<\/div>\n<p>Let [latex]v_{\\text{max}}[\/latex]\u00a0denote the maximum speed the car can attain through the curve without skidding. In other words,\u00a0[latex]v_{\\text{max}}[\/latex]\u00a0is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the magnitude of the centripetal force is<\/p>\n<div style=\"text-align: center;\">[latex]|\\large{{\\bf{F}}_{\\text{cent}}|=\\frac{mv^{2}_{\\text{max}}}{R}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>The next three questions deal with developing a formula that relates the speed [latex]v_{\\text{max}}[\/latex] to the banking angle [latex]\\theta[\/latex].<\/li>\n<li>Show that [latex]|{\\bf{N}}|\\cos{\\theta}=mg+|{\\bf{f}}|\\sin{\\theta}[\/latex]. Conclude that [latex]|{bf{N}}|=(mg)\\,\/\\,(\\cos{\\theta}-\\mu\\sin{\\theta})[\/latex].<\/li>\n<li>The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force points toward the center of the circular curve. Show that\n<div style=\"text-align: center;\">[latex]|{\\bf{F}}_{\\text{cent}}|=|{\\bf{N}}|\\sin{\\theta}+|{\\bf{f}}|\\cos{\\theta}[\/latex].<\/div>\n<p>Conclude that<\/p>\n<div style=\"text-align: center;\">[latex]|\\large{{\\bf{F}}_{\\text{cent}}|=\\frac{\\sin{\\theta}+\\mu\\cos{\\theta}}{\\cos{\\theta}-\\mu\\sin{\\theta}}mg}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Show that [latex]\\text{v}^{2}_{\\text{max}}=((\\sin{\\theta}+\\mu\\cos{\\theta}) \\ \/ \\ (\\cos{\\theta}-\\mu\\sin{\\theta}))gR[\/latex].\u00a0Conclude that the maximum speed does not actually depend on the mass of the car.<span data-type=\"newline\"><br \/>\n<\/span>Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.<span data-type=\"newline\"><br \/>\n<\/span>The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure 12. Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately [latex]211\\ ft[\/latex] with a banking angle of [latex]24^{\\circ}[\/latex]. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of [latex]28^{\\circ}[\/latex]. (The track has variable angle banking.)<\/p>\n<div>\n<div id=\"attachment_937\" style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-937\" class=\"size-full wp-image-937\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/21163833\/3-4-12.jpeg\" alt=\"This figure has two graphics. The first is a picture of a raceway. There are cars on the track and fans in the stands. The second graphic is an oval drawing of a raceway. The inner radius of a curve is labeled \u201c211 ft\u201d and the width of the radius is labeled \u201c40 ft\u201d.\" width=\"900\" height=\"304\" \/><\/p>\n<p id=\"caption-attachment-937\" class=\"wp-caption-text\">Figure 8.\u00a0At the Bristol Motor Speedway, Bristol, Tennessee (a), the turns have an inner radius of about 211 ft and a width of 40 ft (b). (credit: part (a) photo by Raniel Diaz, Flickr)<\/p>\n<\/div>\n<\/div>\n<p>The coefficient of friction for a normal tire in dry conditions is approximately [latex]0.7[\/latex]. Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately [latex]0.98[\/latex].<\/p>\n<p>Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.<\/li>\n<li data-mark-suffix=\".\">In dry conditions, how fast can the car travel through the bottom of the turn without skidding?<\/li>\n<li data-mark-suffix=\".\">In dry conditions, how fast can the car travel through the top of the turn without skidding?<\/li>\n<li data-mark-suffix=\".\">In wet conditions, the coefficient of friction can become as low as [latex]0.1[\/latex]. If this is the case, how fast can the car travel through the bottom of the turn without skidding?<\/li>\n<li data-mark-suffix=\".\">Suppose the measured speed of a car going along the outside edge of the turn is [latex]105\\text{ mph}[\/latex]. Estimate the coefficient of friction for the car\u2019s tires.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-825\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 3.16. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 3.17. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":19,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 3.16\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 3.17\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-825","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/825","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":130,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/825\/revisions"}],"predecessor-version":[{"id":6445,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/825\/revisions\/6445"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/825\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=825"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=825"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=825"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=825"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}