Short Answer

Determine the rate of change of volume when \(r = 120in\)and \(h = 140in\).

The rate of change of volume is\(\frac{{dV}}{{dt}} = 8160\pi {\rm{i}}{{\rm{n}}^3}/{\rm{s}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Chain rule

"Suppose that \(z = f(x,y)\) is a differentiable function of \(x\) and \(y\), where \(x = g(t)\) and \(y = h(t)\) are both differentiable functions of \(t\). Then, \(z\) is differentiable function of \(t\) and\(\frac{{dz}}{{dt}} = \frac{{dz}}{{dx}} \cdot \frac{{dx}}{{dt}} + \frac{{dz}}{{dy}} \cdot \frac{{dy}}{{dt}}\)”

Step 2: Estimate the rate of change of volume

The volume of a cone is \(V = \frac{1}{3}\pi {r^2}h\) where, \(r\) is the radius and \(h\) is height.

By using chain rule, obtain the value of \(\frac{{dV}}{{dt}}\)

\(\begin{aligned}{l}\frac{{dV}}{{dt}} = \frac{\partial }{{\partial r}}\left( {\frac{1}{3}\pi {r^2}h} \right)\frac{{dr}}{{dt}} + \frac{\partial }{{\partial h}}\left( {\frac{1}{3}\pi {r^2}h} \right)\frac{{dh}}{{dt}}\\\frac{{dV}}{{dt}} = \frac{1}{3}\pi h\frac{\partial }{{\partial r}}\left( {{r^2}} \right)\frac{{dr}}{{dt}} + \frac{1}{3}\pi {r^2}\frac{\partial }{{\partial h}}(h)\frac{{dh}}{{dt}}\\\frac{{dV}}{{dt}} = \frac{1}{3}\pi h(2r)\frac{{dr}}{{dt}} + \frac{1}{3}\pi {r^2}(1)\frac{{dh}}{{dt}}\\\frac{{dV}}{{dt}} = \frac{2}{3}\pi rh\frac{{dr}}{{dt}} + \frac{1}{3}\pi {r^2}\frac{{dh}}{{dt}}\end{aligned}\)

Substitute \(r = 120{\rm{in}},h = 140{\rm{in}},\frac{{dr}}{{dt}} = 1.8{\rm{in}}/{\rm{s}}\) and \(\frac{{dh}}{{dt}} = - 2.5{\rm{in}}/{\rm{s}}\) to the above equation,

\(\begin{aligned}{l}\frac{{dV}}{{dt}} = \frac{2}{3}\pi rh\frac{{dr}}{{dt}} + \frac{1}{3}\pi {r^2}\frac{{dh}}{{dt}}\\\frac{{dV}}{{dt}} = \frac{2}{3}\pi (120)(140)(1.8) + \frac{1}{3}\pi {(120)^2}( - 2.5)\\\frac{{dV}}{{dt}} = \frac{{60480}}{3}\pi - \frac{{36000}}{3}\pi \\\frac{{dV}}{{dt}} = 20160\pi - 12000\pi \end{aligned}\)

Simplify further as,\(\frac{{dV}}{{dt}} = 8160\pi \)

Hence,\(\frac{{dV}}{{dt}} = 8160\pi {\rm{i}}{{\rm{n}}^3}/{\rm{s}}\)

Therefore, the rate of change of volume is\(\frac{{dV}}{{dt}} = 8160\pi {\rm{i}}{{\rm{n}}^3}/{\rm{s}}\).

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Essential Calculus: Early Transcendentals

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Short Answer

Determine the rate of change of volume when \(r = 120in\)and \(h = 140in\).

Step by Step Solution

Step 1: Chain rule

Step 2: Estimate the rate of change of volume

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Determine the rate of change of volume when \(r = 120in\)and \(h = 140in\).

Step by Step Solution

Step 1: Chain rule

Step 2: Estimate the rate of change of volume

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