{"id":867,"date":"2021-09-03T17:27:02","date_gmt":"2021-09-03T17:27:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=867"},"modified":"2022-10-29T01:52:42","modified_gmt":"2022-10-29T01:52:42","slug":"tangent-planes-and-linear-approximations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/tangent-planes-and-linear-approximations\/","title":{"raw":"Tangent Planes and Linear Approximations","rendered":"Tangent Planes and Linear Approximations"},"content":{"raw":"
\r\n

Learning Outcomes<\/h3>\r\n
\r\n
    \r\n \t
  • Determine the equation of a plane tangent to a given surface at a point.<\/li>\r\n \t
  • Use the tangent plane to approximate a function of two variables at a point.<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<\/div>\r\n

    Tangent Planes<\/h2>\r\nIntuitively, it seems clear that, in a plane, only one line can be tangent to a curve at a point. However, in three-dimensional space, many lines can be tangent to a given point. If these lines lie in the same plane, they determine the tangent plane at that point. A tangent plane at a regular point contains all of the lines tangent to that point. A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.\r\n
    \r\n

    Definition<\/h3>\r\n\r\n
    \r\n\r\nLet [latex]P_{0}=(x_0,\\ y_0,\\ z_0)[\/latex] be a point on a surface [latex]S[\/latex], and let [latex]C[\/latex] be any curve passing through [latex]P_0[\/latex] and lying entirely in [latex]S[\/latex]. If the tangent lines to all such curves [latex]C[\/latex] at [latex]P_0[\/latex] lie in the same plane, then this plane is called the\u00a0tangent plane<\/strong> to [latex]S[\/latex] at [latex]P_0[\/latex] (Figure 1).\r\n\r\n<\/div>\r\n[caption id=\"attachment_1230\" align=\"aligncenter\" width=\"697\"]\"A Figure 1. The tangent plane to a surface [latex]\\small{S}[\/latex] at a point [latex]\\small{P_{0}}[\/latex] contains all the tangent lines to curves in [latex]\\small{S}[\/latex] that pass through [latex]\\small{P_{0}}[\/latex].[\/caption]For a tangent plane to a surface to exist at a point on that surface, it is sufficient for the function that defines the surface to be differentiable at that point, defined later in this section. We define the term tangent plane here and then explore the idea intuitively.\r\n
    \r\n

    Definition<\/h3>\r\n\r\n
    \r\n\r\nLet [latex]S[\/latex] be a surface defined by a differentiable function [latex]z=f\\,(x,\\ y)[\/latex], and let [latex]P_0=(x_0,\\ y_0)[\/latex] be a point in the domain of [latex]f[\/latex]. Then, the equation of the tangent plane to [latex]S[\/latex] at [latex]P_0[\/latex] is given by\r\n
    [latex]\\large{z=f\\,(x_0,\\ y_0)+f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)}[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\nTo see why this formula is correct, let\u2019s first find two tangent lines to the surface [latex]S[\/latex].\u00a0 First, we can recall the notion of a tangent line from single-variable calculus.\r\n
    \r\n

    Recall: Definition of a Tangent Line<\/h3>\r\n\r\n
    \r\n\r\nIf [latex] y = f(x) [\/latex] is differentiable at [latex] x = x_0 [\/latex], then the equation of the tangent line to the curve at [latex] x=x_0 [\/latex] is given by:\r\n\r\n[latex] y = f(x_0)+f'(x_0)(x-x_0) [\/latex]\r\n\r\n<\/div>\r\nThe equation of the tangent line to the curve that is represented by the intersection of [latex]S[\/latex] with the vertical trace given by [latex]x=x_0[\/latex] is [latex]z=f\\,(x_0,\\ y_0)+f_y\\,(x_0,\\ y_0)(y-y_0)[\/latex]. Similarly, the equation of the tangent line to the curve that is represented by the intersection of [latex]S[\/latex] with the vertical trace given by [latex]y=y_0[\/latex] is [latex]z=f\\,(x_0,\\ y_0)+f_x\\,(x_0,\\ y_0)(x-x_0)[\/latex]. A parallel vector to the first tangent line is [latex]{\\bf{a}}={\\bf{j}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{k}}[\/latex]; a parallel vector to the second tangent line is [latex]{\\bf{b}}={\\bf{i}}+f_x\\,(x_0,\\ y_0)\\,{\\bf{k}}[\/latex]. We can take the cross product of these two vectors:\r\n
    [latex]\\begin{array}{ccc}\\hfill {{\\bf{a}}\\times{\\bf{b}}} & =\\hfill & {({\\bf{j}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{k}})\\times({\\bf{i}}+f_x\\,(x_0,\\ y_0)\\,{\\bf{k}})} \\hfill \\\\ \\hfill & =\\hfill & {\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}}\\\\0&1&f_y\\,(x_0,\\ y_0)\\\\1&0&f_x(x_0,\\ y_0)\\end{vmatrix}} \\hfill \\\\ \\hfill & =\\hfill & {f_x\\,(x_0,\\ y_0)\\,{\\bf{i}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{j}}-{\\bf{k}}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nThis vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. We can use this vector as a normal vector to the tangent plane, along with the point [latex]P_0=(x_0,\\ y_0,\\ f\\,(x_0,\\ y_0))[\/latex] in the equation for a plane:\r\n
    [latex]\\begin{array}{ccc}\\hfill {{\\bf{n}}\\cdot ((x-x_0)\\,{\\bf{i}}+(y-y_0)\\,{\\bf{j}}+(z-f\\,(x_0,\\ y_0))\\,{\\bf{k}})} & =\\hfill & {0} \\hfill \\\\ \\hfill {(f_x\\,(x_0,\\ y_0)\\,{\\bf{i}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{j}}-{\\bf{k}})\\cdot ((x-x_0)\\,{\\bf{i}}+(y-y_0)\\,{\\bf{j}}+(z-f\\,(x_0,\\ y_0))\\,{\\bf{k}})} & =\\hfill & {0} \\hfill \\\\ \\hfill {f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)-(z-f\\,(x_0,\\ y_0))} & =\\hfill & {0} \\hfill \\\\ \\hfill \\end{array} [\/latex]<\/div>\r\n \r\n\r\nSolving this equation for [latex]z[\/latex] gives us the above definition.\r\n
    \r\n

    Example: Finding a Tangent Plane<\/h3>\r\nFind the equation of the tangent plane to the surface defined by the function\u00a0[latex]f\\,(x,\\ y)=2x^{2}-3xy+8y^{2}+2x-4y+4[\/latex] at point [latex](2,-1)[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1367793933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367793933124\"]\r\n

    First, we must calculate [latex]f_x\\,(x,\\ y)[\/latex] and [latex]f_y\\,(x,\\ y)[\/latex], then use our definition with [latex]x_0=2[\/latex] and [latex]y_0=-1[\/latex]:<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill {f_x\\,(x,\\ y)} & =\\hfill & {4x-3y+2}\\hfill \\\\ \\hfill {f_y\\,(x,\\ y)} & =\\hfill & {-3x+16y-4} \\hfill \\\\ \\hfill {f\\,(2,\\ -1)} & =\\hfill & {2(2)^{2}-3(2)(-1)+8(-1)^{2}+2(2)-4(-1)+4=34.}\\hfill \\\\ \\hfill {f_x\\,(2,\\ -1)} & =\\hfill & {4(2)-3(-1)+2=13} \\hfill \\\\ \\hfill {f_y\\,(2,\\ -1)} & =\\hfill & {-3(2)+16(-1)-4=-26.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nThen our equation becomes\r\n
    [latex]\\begin{array}{ccc}\\hfill {z} & =\\hfill & {f\\,(x_0,\\ y_0)+f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)}\\hfill \\\\ \\hfill {z} & =\\hfill & {34+13(x-2)-26(y-(-1))}\\hfill \\\\ \\hfill {z} & =\\hfill & {34+13x-26-26y-26}\\hfill \\\\ \\hfill {z} & =\\hfill & {13x-26y-18.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n(See the following figure).\r\n
    \r\n\r\n[caption id=\"attachment_1232\" align=\"aligncenter\" width=\"764\"]\"A Figure 2.\u00a0Calculating the equation of a tangent plane to a given surface at a given point.[\/caption]\r\n\r\n<\/div>\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    TRY IT<\/h3>\r\nFind the equation of the tangent plane to the surface defined by the function [latex]f\\,(x,\\ y)=x^{3}+x^{2}y+y^{2}-2x+3y-2[\/latex] at point [latex](-1,\\ 3)[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1467793933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793933124\"]\r\n
    [latex]z=7x+8y-3[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Example: Finding Another Tangent Plane<\/h3>\r\nFind the equation of the tangent plane to the surface defined by the function [latex]f\\,(x,y)=\\sin{(2x)}\\cos{(3y)}[\/latex] at point [latex](\\pi \/3,\\pi \/4)[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1377793933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1377793933124\"]\r\n

    First, calculate [latex]f_x\\,(x,\\ y)[\/latex] and [latex]f_y\\,(x,\\ y)[\/latex], then use our definition with [latex]x_0=\\pi \/3[\/latex] and [latex]y_0=\\pi \/4[\/latex]:<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill {f_x\\,(x,\\ y)} & =\\hfill & {2\\cos{(2x)}\\cos{(3y)}}\\hfill \\\\ \\hfill {f_y\\,(x,y)} & =\\hfill & {-3\\sin{(2x)}\\sin{(3y)}}\\hfill \\\\ \\hfill {f,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} & =\\hfill & {\\sin{(2(\\frac{\\pi}{3}))}\\cos{(3\\,(\\frac{\\pi}{4}))}=(\\frac{\\sqrt{3}}{2})(-\\frac{\\sqrt{2}}{2})=-\\frac{\\sqrt{6}}{4}}\\hfill \\\\ \\hfill {f_x\\,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} & =\\hfill & {2\\cos{(2(\\frac{\\pi}{3}))}\\cos{(3,(\\frac{\\pi}{4}))}=2\\,(-\\frac{1}{2})(-\\frac{\\sqrt{2}}{2})=\\frac{\\sqrt{2}}{2}}\\hfill \\\\ \\hfill {f_y\\,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} & =\\hfill & {-3\\sin{(2\\,(\\frac{\\pi}{3}))}\\sin{(3\\,(\\frac{\\pi}{4}))}=-3\\,(\\frac{\\sqrt{3}}{2})(\\frac{\\sqrt{2}}{2})=-\\frac{3\\sqrt{6}}{4}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n
    \r\n\r\nThen our definition becomes\r\n\r\n[latex]\\begin{alignat}{2} \\hspace{6cm}z&=f(x_0,y_0)+f_x(x_0,y_0) (x-x_0)+f_y(x_0,y_0) (y-y_0)\\\\\r\n\r\nz&=-\\frac{\\sqrt6}{4}+\\frac{\\sqrt2}{2}\\left(x-\\frac{\\pi}3\\right)-\\frac{3\\sqrt6}4\\left(y-\\frac{\\pi}4\\right)&\\quad\\\\\r\n\r\nz&=\\frac{\\sqrt2}{2}x-\\frac{3\\sqrt6}{4}y-\\frac{\\sqrt6}4-\\frac{\\pi\\sqrt2}6+\\frac{3\\pi\\sqrt6}{16}.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n<\/div>\r\n
    [\/hidden-answer]<\/span><\/div>\r\n<\/div>\r\nA tangent plane to a surface does not always exist at every point on the surface. Consider the function\r\n
    [latex]f\\,(x,\\ y)=\\begin{cases}\\frac{xy}{\\sqrt{x^{2}+y^{2}}}\\ \\ (x,\\ y)\\neq(0,\\ 0) \\\\ 0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (x,\\ y)=(0,\\ 0).\\end{cases}[\/latex]<\/div>\r\n \r\n\r\nThe graph of this function follows.\r\n\r\n[caption id=\"attachment_1234\" align=\"aligncenter\" width=\"581\"]\"A Figure 3.\u00a0Graph of a function that does not have a tangent plane at the origin.[\/caption]\r\n\r\nIf either [latex]x=0[\/latex] or [latex]y=0[\/latex], then [latex]f\\,(x,\\ y)=0[\/latex], so the value of the function does not change on either the [latex]x-[\/latex] or [latex]y-[\/latex]axis. Therefore, [latex]f_x\\,(x,\\ 0)=f_y\\,(0,\\ y)=0[\/latex], so as either [latex]x[\/latex] or [latex]y[\/latex] approach zero, these partial derivatives stay equal to zero. Substituting them into our definition gives [latex]z=0[\/latex] as the equation of the tangent line. However, if we approach the origin from the different direction, we get a different story. For example, suppose we approach the origin along the line [latex]y=x[\/latex]. If we put [latex]y=x[\/latex] into the original function, it becomes\r\n
    [latex]\\LARGE{f\\,(x,\\ x)=\\frac{x(x)}{\\sqrt{x^{2}+(x)^{2}}}=\\frac{x^{2}}{\\sqrt{2x^{2}}}=\\frac{|x|}{\\sqrt{2}}}.[\/latex]<\/div>\r\n \r\n\r\nWhen [latex]x>0[\/latex], the slope of this curve is equal to [latex]\\sqrt{2}\/2[\/latex]; when [latex]x<0[\/latex], the slope of this curve is equal to [latex]-(\\sqrt{2}\/2)[\/latex]. This presents a problem. In the definition of\u00a0tangent plane<\/em>, we presumed that all tangent lines through point [latex]P[\/latex] (in this case, the origin) lay in the same plane. This is clearly not the case here. When we study differentiable functions, we will see that this function is not differentiable at the origin.\r\n

    Linear Approximations<\/h2>\r\nRecall from Linear Approximations and Differentials that the formula for the linear approximation of a function [latex]f\\,(x)[\/latex] at the point [latex]x=a[\/latex] is given by\r\n
    [latex]y\\approx f\\,(a)+f'\\,(a)(x-a).[\/latex]<\/div>\r\n \r\n\r\nThe diagram for the linear approximation of a function of one variables appears in the following graph.\r\n\r\n[caption id=\"attachment_1236\" align=\"aligncenter\" width=\"342\"]\"A Figure 4. Linear approximation of a function in one variable.[\/caption]\r\n\r\nThe tangent line can be used as an approximation to the function [latex]f\\,(x)[\/latex] for values of [latex]x[\/latex] reasonably close to [latex]x=a[\/latex]. When working with a function of two variables, the tangent line is replaced by a tangent plane, but the approximation idea is much the same.\r\n
    \r\n

    Definition<\/h3>\r\n\r\n
    \r\n\r\nGiven a \u00a0function [latex]z=f\\,(x,\\ y)[\/latex] with continuous partial derivatives that exist at the point [latex](x_0,\\ y_0)[\/latex], the\u00a0linear approximation<\/strong> of [latex]f[\/latex] at the point [latex](x_0,\\ y_0)[\/latex] is given by the equation\r\n
    [latex]\\large{L\\,(x,\\ y)=f\\,(x_0,\\ y_0)+f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)}.[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\nNotice that this equation also represents the tangent plane to the surfaced defined by [latex]z=f\\,(x,\\ y)[\/latex] at the point [latex](x_0,\\ y_0)[\/latex]. The idea behind using a linear approximation is that, if there is a point [latex](x_0,\\ y_0)[\/latex] at which the precise value of [latex]f\\,(x,\\ y)[\/latex] is known, then for values of [latex](x,\\ y)[\/latex] reasonably close to [latex](x_0,\\ y_0)[\/latex], the linear approximation (i.e., tangent plane) yields a value that is also reasonably close to the exact value of [latex]f\\,(x,\\ y)[\/latex] (Figure 5). Furthermore the plane that is used to find the linear approximation is also the tangent plane to the surface at the point [latex](x_0,\\ y_0)[\/latex].\r\n\r\n[caption id=\"attachment_1238\" align=\"aligncenter\" width=\"650\"]\"A Figure 5. Using a tangent plane for linear approximation at a point.[\/caption]\r\n\r\n
    \r\n

    Example: Using a Tangent Plane Approximation<\/h3>\r\nGiven the function [latex]f\\,(x,\\ y)=\\sqrt{41-4x^{2}-y^{2}}[\/latex], approximate [latex]f\\,(2.1,\\ 2.9)[\/latex] using point [latex](2,\\ 3)[\/latex] for [latex](x_0,\\ y_0)[\/latex]. What is the approximate value of [latex]f\\,(2.1,\\ 2.9)[\/latex] to four decimal places?\r\n\r\n[reveal-answer q=\"fs-id1467793993124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793993124\"]\r\n

    To apply our definition, we first must calculate [latex]f\\,(x_0,\\ y_0)[\/latex], [latex]f_x\\,(x_0,\\ y_0)[\/latex], and [latex]f_y\\,(x_0,\\ y_0)[\/latex] using [latex]x_0=2[\/latex] and [latex]y_0=3[\/latex]:<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill {f\\,(x_0,\\ y_0)} & =\\hfill & {f\\,(2,\\ 3)=\\sqrt{41-4(2)^{2}-(3)^{2}}=\\sqrt{41-16-9}=\\sqrt{16}=4}\\hfill \\\\ \\hfill {f_x\\,(x,\\ y)} & =\\hfill & {-\\frac{4x}{\\sqrt{41-4x^{2}-y^{2}}}\\ so\\ f_x\\,(x_0,\\ y_0)=-\\frac{4(2)}{\\sqrt{41-4(2)^{2}-(3)^{2}}}=-2}\\hfill \\\\ \\hfill {f_y\\,(x,\\ y)} & =\\hfill & {-\\frac{y}{\\sqrt{41-4x^{2}-y^{2}}}\\ so\\ f_y\\,(x_0,\\ y_0)=-\\frac{3}{\\sqrt{41-4(2)^{2}-(3)^{2}}}=-\\frac{3}{4}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nNow we substitute these values into our definition equation:\r\n
    [latex]\\begin{array}{ccc}\\hfill {L\\,(x,\\ y)} & =\\hfill & {f\\,(x_0,\\ y_0)+f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)}\\hfill \\\\ \\hfill & =\\hfill & {4-2(x-2)-\\frac{3}{4}(y-3)} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{41}{4}-2x-\\frac{3}{4}y.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nLast, we substitute [latex]x=2.1[\/latex] and [latex]y=2.9[\/latex] into [latex]L\\,(x,\\ y)[\/latex]:\r\n
    [latex]L\\,(2.1,\\ 2.9)=\\frac{41}{4}-2(2.1)-\\frac{3}{4}(2.9)=10.25-4.2-2.175=3.875.[\/latex]<\/div>\r\n \r\n\r\nThe approximate value of [latex]f\\,(2.1,\\ 2.9([\/latex] to four decimal places is\r\n
    [latex]f\\,(2.1,\\ 2.9)=\\sqrt{41-4(2.1)^{2}-(2.9)^{2}}=\\sqrt{14.95}\\approx 3.8665[\/latex],<\/div>\r\n \r\n\r\nwhich corresponds to a [latex]0.2\\%[\/latex] error in approximation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    TRY IT<\/h3>\r\nGiven the function [latex]f\\,(x,\\ y)=e^{5-2x+3y}[\/latex] approximate [latex]f\\,(4.1,\\ 0.9)[\/latex] using point [latex](4,\\ 1)[\/latex] for [latex](x_0,\\ y_0)[\/latex]. What is the approximate value of [latex]f\\,(4.1,\\ 0.9)[\/latex] to four decimal places?\r\n\r\n[reveal-answer q=\"fs-id1467793733124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793733124\"]\r\n
    [latex]L\\,(x,\\ y)=6-2x+3y[\/latex], so [latex]L\\,(4.1,\\ 0.9)=6-2(4.1)+3(0.9)=0.5[\/latex]<\/div>\r\n \r\n
    [latex]f\\,(4.1,\\ 0.9)=e^{5-2(4.1)+3(0.9)}=e^{-0.5}\\approx 0.6065.[\/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