A conical tank (with vertex down) is feet across the top and feet deep. If water is flowing into the tank at a rate of cubic feet per minute, find the rate of change of the Depth of the water when the water is feet deep.
A) $\frac { 9 } { 40 \pi } \mathrm { t } / \mathrm { min }$
B) $\frac { 9 } { 100 \pi } \mathrm { f } / \mathrm { min }$
C) $\frac { 81 } { 20 \pi } \mathrm { tt } / \mathrm { min }$
D) $\frac { 81 } { 50 \pi } \mathrm { ft } / \mathrm { min }$
E) $\frac { 81 } { 200 \pi } \mathrm { ft } / \mathrm { min }$

Question

A conical tank (with vertex down) is feet across the top and feet deep. If water is flowing into the tank at a rate of cubic feet per minute, find the rate of change of the Depth of the water when the water is feet deep. 
A) $\frac { 9 } { 40 \pi } \mathrm { t } / \mathrm { min }$
B) $\frac { 9 } { 100 \pi } \mathrm { f } / \mathrm { min }$
C) $\frac { 81 } { 20 \pi } \mathrm { tt } / \mathrm { min }$
D) $\frac { 81 } { 50 \pi } \mathrm { ft } / \mathrm { min }$
E) $\frac { 81 } { 200 \pi } \mathrm { ft } / \mathrm { min }$

Quizplus · Accepted Answer

Let the height of the conical tank be $x$ at time $t$. Let the radius be $r$. Then the volume of the water in the tank is $\frac{1}{3}\pi r^2 x$.

Differentiating with respect to time, we get:

$$\frac{dV}{dt}=\frac{1}{3}\pi r^2\frac{dx}{dt}$$

We also know that $\frac{dV}{dt}=r^2\frac{dx}{dt}\frac{d}{dt}(x\pi)$. Setting these two expressions equal, we get:

$$\frac{dx}{dt}=\frac{1}{3}\pi r^2\cdot\frac{1}{r^2\frac{d}{dt}(x\pi)}\cdot\frac{dV}{dt}=\frac{1}{3}\cdot\frac{1}{x\pi}\cdot\frac{dV}{dt}$$

Plugging in the numbers given in the problem, we get:

$$\frac{dx}{dt}=\frac{1}{3}\cdot\frac{1}{9\pi}\cdot6\pi\cdot\frac{9^2}{50}\cdot\frac{1}{25}=\frac{81}{50\pi}\mathrm{ft/min}$$

Therefore, the answer is $\boxed{	ext{(D)}}$.

A Conical Tank (With Vertex Down) Is Feet Across the Top