{"id":200,"date":"2021-07-30T17:31:05","date_gmt":"2021-07-30T17:31:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=200"},"modified":"2022-10-29T00:30:31","modified_gmt":"2022-10-29T00:30:31","slug":"putting-it-together-vectors-in-space","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/putting-it-together-vectors-in-space\/","title":{"raw":"Putting It Together: Vectors in Space","rendered":"Putting It Together: Vectors in Space"},"content":{"raw":"

Finding the Focus of a Parabolic Reflector<\/h2>\r\nEnergy hitting the surface of a parabolic reflector is concentrated at the focal point of the reflector (Figure 1).\u00a0If the surface of a parabolic reflector is described by equation [latex]\\frac{x^2}{100}+\\frac{y^2}{100}=\\frac{z}{4}[\/latex],\u00a0where is the focal point of the reflector?\r\n\r\n[caption id=\"attachment_320\" align=\"aligncenter\" width=\"650\"]\"This Figure 1. Energy reflects off of the parabolic reflector and is collected at the focal point. (credit: modification of CGP Grey, Wikimedia Commons)[\/caption]\r\n

Solution<\/span><\/h4>\r\nSince [latex]z[\/latex]\u00a0is the first-power variable, the axis of the reflector corresponds to the [latex]z[\/latex]-axis. The coefficients of [latex]x^2[\/latex]\u00a0and [latex]y^2[\/latex]\u00a0are equal, so the cross-section of the paraboloid perpendicular to the\u00a0[latex]z[\/latex]-axis is a circle. We can consider a trace in the\u00a0[latex]xz[\/latex]-plane or the\u00a0[latex]yz[\/latex]-plane; the result is the same. Setting [latex]y=0[\/latex],\u00a0the trace is a parabola opening up along the\u00a0[latex]z[\/latex]-axis, with standard equation [latex]x^2=4pz[\/latex],\u00a0where [latex]p[\/latex]\u00a0is the focal length of the parabola.\u00a0In this case, this equation becomes [latex]x^2=100\\cdot\\frac{z}{4}=4pz[\/latex] or [latex]25=4p[\/latex].\u00a0So\u00a0[latex]p[\/latex]\u00a0is\u00a0[latex]6.25[\/latex] m, which tells us that the focus of the paraboloid is [latex]6.25[\/latex]\u00a0m up the axis from the vertex. Because the vertex of this surface is the origin, the focal point is\u00a0[latex](0,0,6.25)[\/latex].","rendered":"

Finding the Focus of a Parabolic Reflector<\/h2>\n

Energy hitting the surface of a parabolic reflector is concentrated at the focal point of the reflector (Figure 1).\u00a0If the surface of a parabolic reflector is described by equation [latex]\\frac{x^2}{100}+\\frac{y^2}{100}=\\frac{z}{4}[\/latex],\u00a0where is the focal point of the reflector?<\/p>\n

\"This<\/p>\n

Figure 1. Energy reflects off of the parabolic reflector and is collected at the focal point. (credit: modification of CGP Grey, Wikimedia Commons)<\/p>\n<\/div>\n

Solution<\/span><\/h4>\n

Since [latex]z[\/latex]\u00a0is the first-power variable, the axis of the reflector corresponds to the [latex]z[\/latex]-axis. The coefficients of [latex]x^2[\/latex]\u00a0and [latex]y^2[\/latex]\u00a0are equal, so the cross-section of the paraboloid perpendicular to the\u00a0[latex]z[\/latex]-axis is a circle. We can consider a trace in the\u00a0[latex]xz[\/latex]-plane or the\u00a0[latex]yz[\/latex]-plane; the result is the same. Setting [latex]y=0[\/latex],\u00a0the trace is a parabola opening up along the\u00a0[latex]z[\/latex]-axis, with standard equation [latex]x^2=4pz[\/latex],\u00a0where [latex]p[\/latex]\u00a0is the focal length of the parabola.\u00a0In this case, this equation becomes [latex]x^2=100\\cdot\\frac{z}{4}=4pz[\/latex] or [latex]25=4p[\/latex].\u00a0So\u00a0[latex]p[\/latex]\u00a0is\u00a0[latex]6.25[\/latex] m, which tells us that the focus of the paraboloid is [latex]6.25[\/latex]\u00a0m up the axis from the vertex. Because the vertex of this surface is the origin, the focal point is\u00a0[latex](0,0,6.25)[\/latex].<\/p>\n\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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