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Expert-verified Found in: Page 656 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Determine the rate of change of volume when $$r = 120in$$and $$h = 140in$$.

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

See the step by step solution

## 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$$\frac{{dV}}{{dt}} = 8160\pi {\rm{i}}{{\rm{n}}^3}/{\rm{s}}$$. ### Want to see more solutions like these? 