Question

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

Short answer

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

Step-by-step solution

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