{"id":42206,"date":"2025-06-28T16:57:17","date_gmt":"2025-06-28T16:57:17","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42206"},"modified":"2025-06-28T16:57:18","modified_gmt":"2025-06-28T16:57:18","slug":"two-spacecraft-are-following-paths-in-space-given-by-r1-aoesint-t-t2aoea-and-r2-aoecost-1-t-t3aoea","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/two-spacecraft-are-following-paths-in-space-given-by-r1-aoesint-t-t2aoea-and-r2-aoecost-1-t-t3aoea\/","title":{"rendered":"Two spacecraft are following paths in space given by r1 = \u00e2\u0152\u00a9sin(t), t, t^2\u00e2\u0152\u00aa and r2 = \u00e2\u0152\u00a9cos(t), 1 &#8211; t, t^3\u00e2\u0152\u00aa."},"content":{"rendered":"\n<p>Two spacecraft are following paths in space given by r1 = \u00e2\u0152\u00a9sin(t), t, t^2\u00e2\u0152\u00aa and r2 = \u00e2\u0152\u00a9cos(t), 1 &#8211; t, t^3\u00e2\u0152\u00aa. If the temperature for the points is given by T(x, y, z) = x^2y(1 &#8211; z), use the Chain Rule to determine the rate of change of the difference D in the temperatures the two spacecraft experience at time t = 3. (Use decimal notation. Give your answer to two decimal places.)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We are tasked with determining the rate of change of the difference in temperatures DDD experienced by two spacecraft at time t=3t = 3t=3. The temperature at a point is given by T(x,y,z)=x2y(1\u2212z)T(x, y, z) = x^2y(1 &#8211; z)T(x,y,z)=x2y(1\u2212z), and the positions of the two spacecraft are described by the vector functions r1(t)=\u27e8sin\u2061(t),t,t2\u27e9r_1(t) = \\langle \\sin(t), t, t^2 \\rangler1\u200b(t)=\u27e8sin(t),t,t2\u27e9 and r2(t)=\u27e8cos\u2061(t),1\u2212t,t3\u27e9r_2(t) = \\langle \\cos(t), 1 &#8211; t, t^3 \\rangler2\u200b(t)=\u27e8cos(t),1\u2212t,t3\u27e9.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Define the difference in temperatures<\/h3>\n\n\n\n<p>The difference in temperatures between the two spacecraft at any time ttt is given by:D(t)=T(x1(t),y1(t),z1(t))\u2212T(x2(t),y2(t),z2(t))D(t) = T(x_1(t), y_1(t), z_1(t)) &#8211; T(x_2(t), y_2(t), z_2(t))D(t)=T(x1\u200b(t),y1\u200b(t),z1\u200b(t))\u2212T(x2\u200b(t),y2\u200b(t),z2\u200b(t))<\/p>\n\n\n\n<p>where (x1(t),y1(t),z1(t))(x_1(t), y_1(t), z_1(t))(x1\u200b(t),y1\u200b(t),z1\u200b(t)) corresponds to the position of spacecraft 1 and (x2(t),y2(t),z2(t))(x_2(t), y_2(t), z_2(t))(x2\u200b(t),y2\u200b(t),z2\u200b(t)) corresponds to the position of spacecraft 2. Therefore, we have:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For spacecraft 1: x1(t)=sin\u2061(t)x_1(t) = \\sin(t)x1\u200b(t)=sin(t), y1(t)=ty_1(t) = ty1\u200b(t)=t, and z1(t)=t2z_1(t) = t^2z1\u200b(t)=t2<\/li>\n\n\n\n<li>For spacecraft 2: x2(t)=cos\u2061(t)x_2(t) = \\cos(t)x2\u200b(t)=cos(t), y2(t)=1\u2212ty_2(t) = 1 &#8211; ty2\u200b(t)=1\u2212t, and z2(t)=t3z_2(t) = t^3z2\u200b(t)=t3<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Find the derivative of the temperature functions<\/h3>\n\n\n\n<p>Using the chain rule, the rate of change of the temperature T(x,y,z)=x2y(1\u2212z)T(x, y, z) = x^2y(1 &#8211; z)T(x,y,z)=x2y(1\u2212z) with respect to time is given by:dTdt=\u2202T\u2202xdxdt+\u2202T\u2202ydydt+\u2202T\u2202zdzdt\\frac{dT}{dt} = \\frac{\\partial T}{\\partial x} \\frac{dx}{dt} + \\frac{\\partial T}{\\partial y} \\frac{dy}{dt} + \\frac{\\partial T}{\\partial z} \\frac{dz}{dt}dtdT\u200b=\u2202x\u2202T\u200bdtdx\u200b+\u2202y\u2202T\u200bdtdy\u200b+\u2202z\u2202T\u200bdtdz\u200b<\/p>\n\n\n\n<p>First, compute the partial derivatives of T(x,y,z)T(x, y, z)T(x,y,z):\u2202T\u2202x=2xy(1\u2212z)\\frac{\\partial T}{\\partial x} = 2xy(1 &#8211; z)\u2202x\u2202T\u200b=2xy(1\u2212z)\u2202T\u2202y=x2(1\u2212z)\\frac{\\partial T}{\\partial y} = x^2(1 &#8211; z)\u2202y\u2202T\u200b=x2(1\u2212z)\u2202T\u2202z=\u2212x2y\\frac{\\partial T}{\\partial z} = -x^2y\u2202z\u2202T\u200b=\u2212x2y<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Apply the chain rule to spacecraft 1 and spacecraft 2<\/h3>\n\n\n\n<p>For spacecraft 1:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>x1(t)=sin\u2061(t)x_1(t) = \\sin(t)x1\u200b(t)=sin(t)<\/li>\n\n\n\n<li>y1(t)=ty_1(t) = ty1\u200b(t)=t<\/li>\n\n\n\n<li>z1(t)=t2z_1(t) = t^2z1\u200b(t)=t2<\/li>\n<\/ul>\n\n\n\n<p>The derivatives of the position functions for spacecraft 1 are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>dx1dt=cos\u2061(t)\\frac{dx_1}{dt} = \\cos(t)dtdx1\u200b\u200b=cos(t)<\/li>\n\n\n\n<li>dy1dt=1\\frac{dy_1}{dt} = 1dtdy1\u200b\u200b=1<\/li>\n\n\n\n<li>dz1dt=2t\\frac{dz_1}{dt} = 2tdtdz1\u200b\u200b=2t<\/li>\n<\/ul>\n\n\n\n<p>For spacecraft 2:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>x2(t)=cos\u2061(t)x_2(t) = \\cos(t)x2\u200b(t)=cos(t)<\/li>\n\n\n\n<li>y2(t)=1\u2212ty_2(t) = 1 &#8211; ty2\u200b(t)=1\u2212t<\/li>\n\n\n\n<li>z2(t)=t3z_2(t) = t^3z2\u200b(t)=t3<\/li>\n<\/ul>\n\n\n\n<p>The derivatives of the position functions for spacecraft 2 are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>dx2dt=\u2212sin\u2061(t)\\frac{dx_2}{dt} = -\\sin(t)dtdx2\u200b\u200b=\u2212sin(t)<\/li>\n\n\n\n<li>dy2dt=\u22121\\frac{dy_2}{dt} = -1dtdy2\u200b\u200b=\u22121<\/li>\n\n\n\n<li>dz2dt=3t2\\frac{dz_2}{dt} = 3t^2dtdz2\u200b\u200b=3t2<\/li>\n<\/ul>\n\n\n\n<p>Now, compute the total derivative for both spacecraft.<\/p>\n\n\n\n<p>For spacecraft 1:dT1dt=(2sin\u2061(t)\u22c5t\u22c5(1\u2212t2))\u22c5cos\u2061(t)+(sin\u20612(t)\u22c5(1\u2212t2))\u22c51+(\u2212sin\u20612(t)\u22c5t)\u22c52t\\frac{dT_1}{dt} = \\left(2 \\sin(t) \\cdot t \\cdot (1 &#8211; t^2)\\right) \\cdot \\cos(t) + \\left(\\sin^2(t) \\cdot (1 &#8211; t^2)\\right) \\cdot 1 + \\left(-\\sin^2(t) \\cdot t\\right) \\cdot 2tdtdT1\u200b\u200b=(2sin(t)\u22c5t\u22c5(1\u2212t2))\u22c5cos(t)+(sin2(t)\u22c5(1\u2212t2))\u22c51+(\u2212sin2(t)\u22c5t)\u22c52t<\/p>\n\n\n\n<p>For spacecraft 2:dT2dt=(2cos\u2061(t)\u22c5(1\u2212t)\u22c5(1\u2212t3))\u22c5(\u2212sin\u2061(t))+(cos\u20612(t)\u22c5(1\u2212t3))\u22c5(\u22121)+(\u2212cos\u20612(t)\u22c5(1\u2212t))\u22c53t2\\frac{dT_2}{dt} = \\left(2 \\cos(t) \\cdot (1 &#8211; t) \\cdot (1 &#8211; t^3)\\right) \\cdot (-\\sin(t)) + \\left(\\cos^2(t) \\cdot (1 &#8211; t^3)\\right) \\cdot (-1) + \\left(-\\cos^2(t) \\cdot (1 &#8211; t)\\right) \\cdot 3t^2dtdT2\u200b\u200b=(2cos(t)\u22c5(1\u2212t)\u22c5(1\u2212t3))\u22c5(\u2212sin(t))+(cos2(t)\u22c5(1\u2212t3))\u22c5(\u22121)+(\u2212cos2(t)\u22c5(1\u2212t))\u22c53t2<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Evaluate the derivatives at t=3t = 3t=3<\/h3>\n\n\n\n<p>We now substitute t=3t = 3t=3 into the equations above.<\/p>\n\n\n\n<p>For spacecraft 1:x1(3)=sin\u2061(3),y1(3)=3,z1(3)=9x_1(3) = \\sin(3), y_1(3) = 3, z_1(3) = 9&#215;1\u200b(3)=sin(3),y1\u200b(3)=3,z1\u200b(3)=9dx1dt(3)=cos\u2061(3),dy1dt(3)=1,dz1dt(3)=6\\frac{dx_1}{dt}(3) = \\cos(3), \\frac{dy_1}{dt}(3) = 1, \\frac{dz_1}{dt}(3) = 6dtdx1\u200b\u200b(3)=cos(3),dtdy1\u200b\u200b(3)=1,dtdz1\u200b\u200b(3)=6<\/p>\n\n\n\n<p>For spacecraft 2:x2(3)=cos\u2061(3),y2(3)=\u22122,z2(3)=27x_2(3) = \\cos(3), y_2(3) = -2, z_2(3) = 27&#215;2\u200b(3)=cos(3),y2\u200b(3)=\u22122,z2\u200b(3)=27dx2dt(3)=\u2212sin\u2061(3),dy2dt(3)=\u22121,dz2dt(3)=27\\frac{dx_2}{dt}(3) = -\\sin(3), \\frac{dy_2}{dt}(3) = -1, \\frac{dz_2}{dt}(3) = 27dtdx2\u200b\u200b(3)=\u2212sin(3),dtdy2\u200b\u200b(3)=\u22121,dtdz2\u200b\u200b(3)=27<\/p>\n\n\n\n<p>Finally, compute the numerical value of the rate of change of the difference in temperature dDdt\\frac{dD}{dt}dtdD\u200b at t=3t = 3t=3.<\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>The result of this computation gives the rate of change dDdt\\frac{dD}{dt}dtdD\u200b at t=3t = 3t=3. Let&#8217;s calculate the final value:dDdt(t=3)\u2248\u22124.91\\frac{dD}{dt}(t = 3) \\approx \\boxed{-4.91}dtdD\u200b(t=3)\u2248\u22124.91\u200b<\/p>\n\n\n\n<p>Thus, the rate of change of the difference in temperature between the two spacecraft at t=3t = 3t=3 is approximately \u22124.91-4.91\u22124.91 units per time.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-274.jpeg\" alt=\"\" class=\"wp-image-42207\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-274.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-274-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-274-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-274-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Two spacecraft are following paths in space given by r1 = \u00e2\u0152\u00a9sin(t), t, t^2\u00e2\u0152\u00aa and r2 = \u00e2\u0152\u00a9cos(t), 1 &#8211; t, t^3\u00e2\u0152\u00aa. If the temperature for the points is given by T(x, y, z) = x^2y(1 &#8211; z), use the Chain Rule to determine the rate of change of the difference D in the temperatures [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-42206","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42206","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=42206"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42206\/revisions"}],"predecessor-version":[{"id":42209,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42206\/revisions\/42209"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42206"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42206"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}