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alt="" src="../../files/differentiation/related-rates/main-1.png" style="float: left; width: 100px; height: 100px;"> This page is about finding related rates of change, otherwise know as connected rates of change. In questions on this topic, you will be given a rate of change of some quantity and be required to find the rate of change of another, that is related to the first by some relationship. This is an application of the Chain Rule and a knowlegde of implicit differentiation is helpful.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <p>On this page, you should learn how to</p> <ul> <li>Connect rates of change using the <a href="../742/chain-rule.html" title="Chain Rule">Chain Rule</a></li> <li>Solve problems with Related rates of Change using <a href="../741/implicit-differentiation.html" title="Implicit Differentiation">Implicit Differentiation</a></li> </ul> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-yellow panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div>Example - Ferris Wheel</div> </div> <div class="panel-body"> <div> <div>The London Eye is a tall observation wheel that carries passengers around the rim of the wheel.</div> <div style="text-align: center;"><img alt="" src="../../files/differentiation/related-rates/example2q-ferris-wheel.png" style="width: 350px; height: 320px;"></div> <div> </div> <div>The circular wheel has radius 70 metres and revolves at a rate of one revolution per 30 minutes. Determine how fast a passenger on the wheel is moving vertically upwards when the passenger is at a point P, 10 metres higher than the centre of the wheel, and is rising. <hr class="hidden-separator"></div> <div> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/767268626"></iframe></div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Example - Volume of a Cone</p> </div> </div> <div class="panel-body"> <div> <p>A right circular cone is being filled with water. Initially the cone is empty, then water is added at <span class="math-tex">\(\large 6\pi\)</span> cm<sup>3</sup>s<sup>-1</sup>. At the time when the radius is 3cm, the volume of the solid is <span class="math-tex">\(\large 15\pi\)</span> cm3, the radius is changing at a rate of 0.5 cms<sup>-1</sup>.</p> <p>Find the rate of change of the depth of water at this time.</p> <hr class="hidden-separator"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/767044844"></iframe></div> <p> </p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <p> </p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-violet"> <div class="panel-heading"><a class="expander pull-right" href="#"><span class="fa fa-plus"></span></a> <div> <p>Summary</p> </div> </div> <div class="panel-body"> <div> <p><iframe frameborder="0" height="480" scrolling="no" src="../../files/differentiation/related-rates/related-rates-revision-notes.pdf" width="100%"></iframe></p> <p> </p> </div> </div> <div class="panel-footer"> <div> <p>text</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander pull-right" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <p>Try this quiz to practise the skills from this page</p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#eded9a18"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="eded9a18"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Related Rates of Change <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="11-954-2872" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>A = 4x²</p><p>Find <span class="math-tex">\(\large \frac{dA}{dt}\)</span> when x = 4 and <span class="math-tex">\(\large \frac{dx}{dt}=1.5\)</span></p></div><div class="q-answer"><p><span class="math-tex">\(\large \frac{dA}{dt}\)</span>= <input type="text" style="height: auto;" data-c="48"> <span class="review"></span></p></div><div class="q-explanation"><p><span class="math-tex">\(\large A = 4x²\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dA}{dt}=8x\cdot\frac{dx}{dt}\)</span></p><p><span class="math-tex">\(\large \frac{dA}{dt}=8\times4\times1.5=48\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>E = 5v²</p><p>Find <span class="math-tex">\(\large \frac{dE}{dt}\)</span> when v = 3 and <span class="math-tex">\(\large \frac{dv}{dt}=0.5\)</span></p></div><div class="q-answer"><p><span class="math-tex">\(\large \frac{dE}{dt}\)</span>= <input type="text" style="height: auto;" data-c="15"> <span class="review"></span></p></div><div class="q-explanation"><p><span class="math-tex">\(\large E=5v^2\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dE}{dt}=10v\cdot\frac{dv}{dt}\)</span></p><p><span class="math-tex">\(\large \frac{dE}{dt}=10\times3\times0.5=15\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><span class="math-tex">\(\large V=\frac{4}{3}\pi r^3\)</span></p><p>Find <span class="math-tex">\(\large \frac{dV}{dt}\)</span> when r = 5 and <span class="math-tex">\(\large \frac{dr}{dt}=\frac{3}{\pi}\)</span></p></div><div class="q-answer"><p><span class="math-tex">\(\large \frac{dV}{dt}\)</span>= <input type="text" style="height: auto;" data-c="300"> <span class="review"></span></p></div><div class="q-explanation"><p><span class="math-tex">\(\large V=\frac{4}{3}\pi r^3\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dV}{dt}=4\pi r^2\cdot\frac{dr}{dt}\)</span></p><p><span class="math-tex">\(\large \frac{dV}{dt}=4\pi \times 5^2\times\frac{3}{\pi}=300\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><span class="math-tex">\(\large A = 4\pi r²\)</span></p><p>When r = 4 and <span class="math-tex">\(\large \frac{dr}{dt}=0.25\)</span>, then <span class="math-tex">\(\large \frac{dA}{dt}=k\pi\)</span></p><p>Find the value of <strong><em>k</em></strong></p></div><div class="q-answer"><p><strong><em>k </em></strong>= <input type="text" style="height: auto;" data-c="8"> <span class="review"></span></p></div><div class="q-explanation"><p><span class="math-tex">\(\large A = 4\pi r²\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dA}{dt}=8\pi r\cdot\frac{dr}{dt}\)</span></p><p><span class="math-tex">\(\large \frac{dA}{dt}=8\pi \times4\times0.25=8\pi\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><span class="math-tex">\(\large x=2\sin \theta\)</span></p><p>Find <span class="math-tex">\(\large \frac{dx}{dt}\)</span> when <span class="math-tex">\(\large \theta=\frac{\pi}{6}\)</span> and <span class="math-tex">\(\large \frac{d\theta}{dt}=\sqrt{3}\)</span></p></div><div class="q-answer"><p><span class="math-tex">\(\large \frac{dx}{dt}\)</span>= <input type="text" style="height: auto;" data-c="3"> <span class="review"></span></p></div><div class="q-explanation"><p><span class="math-tex">\(\large x=2\sin \theta\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dx}{dt}=2\cos \theta\cdot\frac{d\theta}{dt}\)</span></p><p><span class="math-tex">\(\large \frac{dx}{dt}=2 \cos (\frac{\pi}{6})\times \sqrt{3}\)</span></p><p><span class="math-tex">\(\large \frac{dx}{dt}=2 \times \frac{\sqrt{3}}{2}\times \sqrt{3}\)</span></p><p><span class="math-tex">\(\large \frac{dx}{dt}=3\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The sides of a cube are increasing at 0.5 cms<sup>-1</sup></p><p>When the sides of the cube are 5cm, what is the rate at which the <strong>surface area </strong>is increasing?</p></div><div class="q-answer"><p>Answer = <input type="text" style="height: auto;" data-c="30"> <span class="review"></span> cm²s<sup>-1</sup></p></div><div class="q-explanation"><p>The surface area, <span class="math-tex">\(\large A=6x^2\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dA}{dt}=12x\cdot\frac{dx}{dt}\\ \large \frac{dA}{dt}=12\times5\times 0.5=30\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The radius of a circle is increasing at a rate of 2cm per min. When the radius is 10cm, the rate the area is increasing is <span class="math-tex">\(\large k\pi\)</span>.</p><p>Find the value of <strong><em>k</em></strong></p></div><div class="q-answer"><p><strong><em>k</em></strong> = <input type="text" style="height: auto;" data-c="40"> <span class="review"></span></p></div><div class="q-explanation"><p>The area, <span class="math-tex">\(\large A=\pi r^2\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dA}{dt}=2\pi r\cdot\frac{dr}{dt}\\ \large \frac{dA}{dt}=2\pi \times10\times 2=40\pi\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The edges of a cube are increasing at 2 cm per minute.</p><p>When the edges of the cube are 6 cm, what is the rate at which the volume is increasing?</p></div><div class="q-answer"><p>Answer = <input type="text" style="height: auto;" data-c="216"> <span class="review"></span></p></div><div class="q-explanation"><p>The volume of the cube, <span class="math-tex">\(\large V=x^3\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dV}{dt}=3x^2\cdot\frac{dx}{dt}\\ \large \frac{dV}{dt}=3\times6^2\times 2=216\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A hemispherical bowl with radius 10 cm is being filled with water.</p><p>The volume of water can be found using the formula <span class="math-tex">\(\large V=10\pi h^2-\frac{1}{3}\pi h^3\)</span></p><p>The rate at which the volume is increasing is <span class="math-tex">\(\large 15\pi\)</span> cm<sup>3</sup> s<sup>-1 </sup></p><p>Find the rate at which the height of water is increasing when <strong><em>h</em></strong> = 5cm</p></div><div class="q-answer"><p>Answer = <input type="text" style="height: auto;" data-c="0.2"> <span class="review"></span></p></div><div class="q-explanation"><p><span class="math-tex">\(\large V=10\pi h^2-\frac{1}{3}\pi h^3\)</span></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><span class="math-tex">\(\large \frac{dV}{dt}=20 \pi h\cdot\frac{dh}{dt}- \pi h^2\cdot\frac{dh}{dt}\\ \large 15\pi=20 \pi \times 5\times\frac{dh}{dt}- \pi \times 5^2\cdot\frac{dh}{dt}\\ \large 15\pi=100 \pi \frac{dh}{dt}- 25\pi \frac{dh}{dt}\\ \large 15\pi=75 \pi \frac{dh}{dt}\\ \large \frac{dh}{dt}=0.2\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The radius of a cone is 6cm. The radius is increasing at 0.5 cms<sup>-1</sup>.</p><p>The height of the same cone is 5cm. The height is increasing at a rate of 1cms<sup>-1</sup>.</p><p>The rate at which the volume of the cone is increasing is <span class="math-tex">\(\large k\pi\)</span> cm<sup>3</sup>s<sup>-1</sup>.</p><p>Find <strong><em>k</em></strong></p><p style="text-align: right;"><em>The formula for the volume of a cone, <span class="math-tex">\(\large V=\frac{1}{3}\pi r^2h\)</span></em></p></div><div class="q-answer"><p><em><strong>k</strong></em> = <input type="text" style="height: auto;" data-c="22"> <span class="review"></span></p></div><div class="q-explanation"><p><em><span class="math-tex">\(\large V=\frac{1}{3}\pi r^2h\)</span></em></p><p>Use Implicit differentiation to differentiate with respect to <strong><em>t</em></strong></p><p><strong><em>r</em></strong> and <strong><em>h</em></strong> are both variables, and so we should consider that <em><span class="math-tex">\(\large r^2\cdot h\)</span></em> is a product and we need to use the Product Rule:</p><p><span class="math-tex">\(\large \frac{dV}{dt}=\frac{1}{3}\pi(r^2\cdot h) \\ \large \frac{dV}{dt}=\frac{1}{3}\pi (2r\frac{dr}{dt}\cdot h+r^2\cdot \frac{dh}{dt})\\ \large \frac{dV}{dt}=\frac{1}{3}\pi (2\times6\times 0.5\times 5+6^2\times1)\\ \large \frac{dV}{dt}=\frac{1}{3}\pi(66)\\ \large \frac{dV}{dt}=22\pi\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next <i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> </div> <div class="panel-footer"> <div> <p>text</p> </div> </div> </div> <div class="panel panel-default panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Exam-style Questions</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-default panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 1</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="951"> <p><img class="sibico" src="../../../img/sibico/hl-green.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL easy"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>The radius of a sphere is increasing at 2.5 cms<sup>-1</sup></p> <p>Find the rate at which the volume of the sphere is increasing when the radius is 8 cm.</p> <p>Give your answer in terms of <span class="math-tex">\(\large \pi\)</span></p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>The rate at which the volume of the sphere is increasing is <span class="math-tex">\(\large \frac{dV}{dt}\)</span><content> </content></p> <p>To find it, use the chain rule</p> <p style="margin-left: 80px;"><span class="math-tex">\(\large \frac{dV}{dt}=\)</span><span style="color:#B22222;"><span class="math-tex">\(\large \frac{dV}{dr}\)</span></span><span class="math-tex">\(\large \times\)</span><span style="color:#0000FF;"><span class="math-tex">\(\large \frac{dr}{dt}\)</span></span></p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>The volume of a sphere, <span class="math-tex">\(\large V = \frac{4}{3}\pi r^3\)</span></p> <p>Differentiate with respect to <strong><em>r</em></strong> , <span class="math-tex">\(\large \frac{dV}{dr}=4\pi r^2\)</span></p> <p>When <strong><em>r</em></strong> = 8 cm , <span style="color:#B22222;"><span class="math-tex">\(\large \frac{dV}{dr}=4\pi \times 8^2=256 \pi\)</span></span></p> <hr class="hidden-separator"> <p>The radius of a sphere is increasing at 2.5 cms<sup>-1</sup>, <span style="color:#0000FF;"><span class="math-tex">\(\large \frac{dr}{dt}=2.5\)</span></span></p> <hr class="hidden-separator"> <p>Using the chain rule</p> <p style="margin-left: 80px;"><span class="math-tex">\(\large \frac{dV}{dt}=\)</span><span style="color:#B22222;"><span class="math-tex">\(\large \frac{dV}{dr}\)</span></span><span class="math-tex">\(\large \times\)</span><span style="color:#0000FF;"><span class="math-tex">\(\large \frac{dr}{dt}\)</span></span></p> <p style="margin-left: 80px;"><span class="math-tex">\(\large \frac{dV}{dt}=\)</span><span style="color:#B22222;"><span style="color:#B22222;"><span class="math-tex">\(\large 256 \pi\)</span></span></span><span class="math-tex">\(\large \times\)</span><span style="color:#0000FF;"><span style="color:#0000FF;"><span class="math-tex">\(\large 2.5\)</span></span></span></p> <p style="margin-left: 80px;"><span class="math-tex">\(\large \frac{dV}{dt}=640 \pi\)</span></p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 2</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="952"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p><img alt="" src="../../files/differentiation/related-rates/esq2-cone.png" style="float: right; width: 250px; height: 248px;">The diagram shows a container in the form of a right circular cone. The height of the cone is equal to its diameter. Initially the cone is empty, then water is added at a rate of <span class="math-tex">\(\large 18\pi\)</span> cm<sup>3 </sup>per minute. the depth of water in the container at the time is given by <strong><em>h</em></strong> cm.</p> <p>a) Show that the volume, <strong><em>V</em></strong> cm<sup>3</sup> , of water in the container when the depth is <strong><em>x</em></strong> cm is given by</p> <p style="margin-left: 80px;"><span class="math-tex">\(\large V=\frac{1}{12} \pi x^3\)</span></p> <p>b) Find the rate at which the depth of the water is increasing at the instant when the depth is 12 cm</p> <hr class="hidden-separator"> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>a) height = diameter</p> <p>Therefore, <span class="math-tex">\(\large r=\frac{h}{2}\)</span></p> <p>In this case, the height of water = <strong><em>x</em></strong></p> <p>b) The rate at which the depth of the water is increasing is <span class="math-tex">\(\large \frac{dx}{dt}\)</span><content> </content></p> <p>To find it, use the chain rule</p> <p style="margin-left: 80px;"><span style="color:#0000FF;"><span class="math-tex">\(\large \frac{dV}{dt}\)</span>=</span><span style="color:#B22222;"><span class="math-tex">\(\large \frac{dV}{dx}\)</span></span><span class="math-tex">\(\large \times\)</span><span class="math-tex">\(\large \frac{dx}{dt}\)</span></p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>a) The volume of a cone, <span class="math-tex">\(\large V = \frac{1}{3}\pi r^2h\)</span></p> <p>In this case the height = diameter</p> <p>Therefore, <span class="math-tex">\(\large r=\frac{h}{2}\)</span></p> <p>The volume of a cone, <span class="math-tex">\(\large V = \frac{1}{3}\pi (\frac{h}{2})^2h\\ \large V = \frac{1}{3}\pi (\frac{h^2}{4})h\\ \large V = \pi \frac{h^3}{12}\)</span></p> <p>When the depth (height) of water= <strong><em>x</em></strong>, then</p> <p><span class="math-tex">\(\large V=\frac{1}{12} \pi x^3\)</span></p> <hr class="hidden-separator"> <p>b) <span class="math-tex">\(\large V=\frac{1}{12} \pi x^3\)</span></p> <p>Differentiate with respect to <strong><em>x</em></strong> , <span class="math-tex">\(\large \frac{dV}{dx}=\frac{1}{4}\pi r^2\)</span></p> <p>When <strong><em>x</em></strong> = 12 cm , <span style="color:#B22222;"><span class="math-tex">\(\large \frac{dV}{dx}=\frac{1}{4}\pi \times12^2=36\pi\)</span></span></p> <hr class="hidden-separator"> <p>Water is added at a rate of <span class="math-tex">\(\large 18\pi\)</span> cm<sup>3 </sup>per minute, <span style="color:#0000FF;"><span class="math-tex">\(\large \frac{dV}{dt}=18 \pi\)</span></span></p> <hr class="hidden-separator"> <p>Using the chain rule</p> <p style="margin-left: 80px;"><span style="color:#0000FF;"><span class="math-tex">\(\large \frac{dV}{dt}\)</span>=</span><span style="color:#B22222;"><span class="math-tex">\(\large \frac{dV}{dx}\)</span></span><span class="math-tex">\(\large \times\)</span><span class="math-tex">\(\large \frac{dx}{dt}\)</span></p> <p style="margin-left: 80px;"><span style="color:#0000FF;"><span style="color:#0000FF;"><span class="math-tex">\(\large 18 \pi\)</span> </span>=</span><span style="color:#B22222;"><span style="color:#B22222;"><span class="math-tex">\(\large36\pi\)</span></span></span><span class="math-tex">\(\large \times\)</span><span class="math-tex">\(\large \frac{dx}{dt}\)</span></p> <p style="margin-left: 80px;"><span style="color:#B22222;"></span><span style="color:#0000FF;"></span><span class="math-tex">\(\large \frac{dx}{dt}=\frac{18\pi}{36\pi}\)</span></p> <p style="margin-left: 80px;"><span style="color:#B22222;"><span style="color:#B22222;"></span></span><span style="color:#0000FF;"><span style="color:#0000FF;"></span></span><span class="math-tex">\(\large \frac{dx}{dt}=0.5\)</span> cm per minute</p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 3</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="953"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="Calculator"></p> <p>A searchlight rotates at 2 revolutions per minute. The beam hits a wall 30m away and produces a spot of light that moves horizontally along the wall. How fast is the spot moving along the wall when the angle, <span class="math-tex">\(\large \theta\)</span> between the beam and the line through the spotlight perpendicular to the wall is 45°?</p> <p style="text-align: center;"><img alt="" src="../../files/differentiation/related-rates/esq3_1-searchlight.png" style="width: 300px; height: 253px;"></p> <hr class="hidden-separator"> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>If we let <strong><em>x </em></strong>be the distance along the wall, then the speed of the spotlight is<span class="math-tex">\(\large \frac{dx}{dt}\)</span><content> </content></p> <p>To find it, use the chain rule</p> <p style="margin-left: 80px;"><span style="color:#0000FF;"></span><span class="math-tex">\(\large \frac{dx}{dt}\)</span><span style="color:#0000FF;">=</span><span style="color:#B22222;"><span class="math-tex">\(\large \frac{dx}{d\theta}\)</span></span><span class="math-tex">\(\large \times\)</span><span style="color:#0000FF;"><span class="math-tex">\(\large \frac{d\theta}{dt}\)</span></span></p> <p>To find <span style="color:#B22222;"><span class="math-tex">\(\large \frac{dx}{d\theta}\)</span></span> use right-angled triangle trigonometry to find <strong><em>x</em></strong> in terms of <span class="math-tex">\(\large \theta\)</span></p> <p><img alt="" src="../../files/differentiation/related-rates/esq3_2-searchlight.png" style="width: 200px; height: 134px;"></p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>The searchlight is rotating at 2 revolutions per minute</p> <p>1 revolution is <span class="math-tex">\(\large 2\pi \)</span> radians</p> <p>Therefore, <span style="color:#0000FF;"><span class="math-tex">\(\large \frac{d\theta}{dt}=4\pi\)</span></span> radians per minute</p> <hr class="hidden-separator"> <p>Let the distance fom from the perpendicular be <strong><em>x</em></strong></p> <p><img alt="" src="../../files/differentiation/related-rates/esq3_2-searchlight.png" style="width: 200px; height: 134px;"></p> <p><span class="math-tex">\(\large \sin \theta=\frac{x}{30}\)</span></p> <p><span class="math-tex">\(\large x=30 \sin\theta\)</span></p> <p>Differentiate with respect to <span class="math-tex">\(\large \theta\)</span></p> <p><span class="math-tex">\(\large \frac{dx}{d\theta}=30 \cos \theta\)</span></p> <p>When <span class="math-tex">\(\theta\)</span>=45°, <span style="color:#B22222;"><span class="math-tex">\(\large \frac{dx}{d\theta}=30 \cos45°=15\sqrt{2}\)</span></span></p> <hr class="hidden-separator"> <p>Use the chain rule</p> <p style="margin-left: 80px;"><span style="color:#0000FF;"></span><span class="math-tex">\(\large \frac{dx}{dt}\)</span><span style="color:#0000FF;">=</span><span style="color:#B22222;"><span class="math-tex">\(\large \frac{dx}{d\theta}\)</span></span><span class="math-tex">\(\large \times\)</span><span style="color:#0000FF;"><span class="math-tex">\(\large \frac{d\theta}{dt}\)</span></span></p> <p style="margin-left: 80px;"><span class="math-tex">\(\large \frac{dx}{dt}\)</span><span style="color:#0000FF;">=</span><span style="color:#B22222;"><span style="color:#B22222;"><span class="math-tex">\(\large 15\sqrt{2}\)</span></span></span><span class="math-tex">\(\large \times\)</span><span style="color:#0000FF;"><span style="color:#0000FF;"><span class="math-tex">\(\large 4\pi\)</span></span></span></p> <p style="margin-left: 80px;"><span class="math-tex">\(\large \frac{dx}{dt}=60\sqrt{2}\pi\approx267\)</span>m per minute</p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 4</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="954"> <p><img class="sibico" src="../../../img/sibico/hl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL difficult"> <img class="sibico" src="../../../img/sibico/calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="Calculator"></p> <p>A ladder AB of length 8m has one endA on horizontal ground and the other end B resting against a vertical wall.</p> <p>The end A slips away from the wall at a constant speed of 0.5 ms<sup>-1</sup> and the end B slips down the wall.</p> <p>Determine the speed of the end B is slipping down the wall when the top of the ladder is 5 m above the ground.</p> <p style="text-align: center;"><img alt="" src="../../files/differentiation/related-rates/esq4_1-ladder.png" style="width: 257px; height: 326px;"></p> <hr class="hidden-separator"> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>If we let <strong><em>x </em></strong>be the distance along the ground, then the speed end A slips away from the wall is <span class="math-tex">\(\large \frac{dx}{dt}\)</span><content> </content></p> <p>If w let <strong><em>y</em></strong> be the distance up the wall, then the speed end B slips down the wall is <span class="math-tex">\(\large \frac{dy}{dt}\)</span></p> <p>Use Pythagoras' Theorem</p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>The speed end A slips away from the wall is <span class="math-tex">\(\large \frac{dx}{dt}\)</span><content> </content></p> <p><span class="math-tex">\(\large \frac{dx}{dt}=0.5\)</span> ms<sup>-1</sup></p> <hr class="hidden-separator"> <p>The speed end B slips down the wall is <span class="math-tex">\(\large \frac{dy}{dt}\)</span></p> <hr class="hidden-separator"> <p>Using Pythagoras' Theorem</p> <p style="margin-left: 40px;"><span class="math-tex">\(\large x^2+y^2=8^2\)</span></p> <p style="margin-left: 40px;"><span class="math-tex">\(\large x^2+y^2=64\)</span></p> <p>Differentiate with respect to <strong><em>t </em></strong>(implicit differentiation)</p> <p style="margin-left: 40px;"><span class="math-tex">\(\large 2x \frac{dx}{dt}+2y\frac{dy}{dt}=0\)</span></p> <hr class="hidden-separator"> <p>When <strong><em>y</em></strong> =5 ,</p> <p style="margin-left: 40px;"><span class="math-tex">\(\large x^2+5^2=64\\ \large x^2=39\\ \large x=\sqrt {39}\)</span></p> <hr class="hidden-separator"> <p>Substitute these values in the differential equation</p> <p style="margin-left: 40px;"><span class="math-tex">\(\large 2x \frac{dx}{dt}+2y\frac{dy}{dt}=0\)</span></p> <p style="margin-left: 40px;"><span class="math-tex">\(\large 2\sqrt{39}\times 0.5+2\times 5\frac{dy}{dt}=0\\ \large 10\frac{dy}{dt}=-\sqrt{39}\\ \large \frac{dy}{dt}=-\frac{\sqrt{39}}{10}\\ \large \frac{dy}{dt}=-0.624\)</span></p> <p>Notice that the value is negative. This represents the <strong>velocity </strong>and shows that it is slipping <strong>down</strong><strong>.</strong></p> <p><strong>The speed is 0.624 ms<sup>-1</sup></strong></p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 5</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="955"> <p><img class="sibico" src="../../../img/sibico/hl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL difficult"> <img class="sibico" src="../../../img/sibico/calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="Calculator"></p> <p> </p> <p style="text-align: center;"><img alt="" src="../../files/differentiation/related-rates/esq5-solid.png" style="width: 350px; height: 394px;"></p> <p> </p> <p>A solid is made up of a cylinder and a hemisphere as shown in the diagram.</p> <p>a) Write down a formula for the volume of the solid.</p> <p>b) At the time when the radius is 6cm, the volume of the solid is <span class="math-tex">\(\large 684\pi\)</span> cm<sup>3</sup> , the radius is changing at a rate of 1.5 cm/minute and the volume is changing at a rate of <span class="math-tex">\(\large 1800\pi\)</span> cm<sup>3</sup> /min. Find the rate of change of the height at this time.</p> <hr class="hidden-separator"> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>We can find the volume of the solid to be <span class="math-tex">\(\large V=\frac{2}{3}\pi r^3+\pi r^2h\)</span></p> <p>We know <span class="math-tex">\(\large \frac{dr}{dt}\)</span> and <span class="math-tex">\(\large \frac{dV}{dt}\)</span>. We are required to find <span class="math-tex">\(\large \frac{dh}{dt}\)</span>.</p> <p>Use implicit differentiation and differentiate the formula for <strong><em>V</em></strong> with respect to <strong><em>t</em></strong></p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>a) The volume of a hemisphere is <span class="math-tex">\(\large \frac{2}{3}\pi r^3\)</span></p> <p>The volume of a cylinder is <span class="math-tex">\(\large \pi r^2h\)</span></p> <p>The volume of the solid, <span class="math-tex">\(\large V=\frac{2}{3}\pi r^3+\pi r^2h\)</span></p> <hr class="hidden-separator"> <p>b) We can find the value of h when r = 6 and V = <span class="math-tex">\(\large 684\pi\)</span></p> <p style="margin-left: 40px;"><span class="math-tex">\(\large 684\pi=\frac{2}{3}\pi \times 6^3+\pi \times 6^2\times h\\ \large 684\pi=144\pi+36\pi h\\ \large 540\pi=36\pi h\\ \large h = 15\)</span></p> <hr class="hidden-separator"> <p>The radius is changing at a rate of 1.5 cm/minute, <span class="math-tex">\(\large \frac{dr}{dt}=1.5\)</span></p> <p>The volume is changing at a rate of <span class="math-tex">\(\large 1800\pi\)</span> cm<sup>3</sup> /min, <span class="math-tex">\(\large \frac{dV}{dt}=1800\pi\)</span></p> <p>We are required to find <span class="math-tex">\(\large \frac{dh}{dt}\)</span>.</p> <p>Use implicit differentiation and differentiate the formula for <strong><em>V</em></strong> with respect to <strong><em>t</em></strong></p> <p><span class="math-tex">\(\large V=\frac{2}{3}\pi r^3+\pi r^2h\)</span></p> <p>Notice that the second part of the expression <img alt="" src="../../files/differentiation/related-rates/esq5-ans-solid.png" style="width: 190px; height: 50px;"> is a product (both <strong><em>r</em></strong> and <strong><em>h</em></strong> are variables). Therefore we need to use the Product Rule for differentiation</p> <p style="margin-left: 40px;"><span class="math-tex">\(\large \frac{dV}{dt}=2\pi r^2\frac{dr}{dt}+\pi r^2\frac{dh}{dt}+2\pi r^2\frac{dr}{dt}\cdot h\)</span></p> <p style="margin-left: 40px;"><span class="math-tex">\(\large 1800\pi=2\pi \times 6^2\times 1.5+\pi \times 6^2\frac{dh}{dt}+2\pi \times 6^2\times 1.5\times 15\\ \large 1800\pi=108\pi+36\pi \cdot \frac{dh}{dt}+1620\pi\\ \large 72\pi=36\pi \cdot \frac{dh}{dt}\\ \large \frac{dh}{dt}=2\)</span></p> <p style="margin-left: 40px;"><span class="math-tex">\(\large \frac{dh}{dt}=2\)</span> cm/min</p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="page-container panel-self-assessment" data-id="2872"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label 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