QuestionDepth A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. Water is flowing into the tank at a rate of 10 cubic feet per minute. Find the rate of change of the depth of the water when the water is 8 feet deep.

Studdy Solution
Substitute the given values to find the rate of change of depth when the water is 8 feet deep.
Given $\frac{dV}{dt} = 10$ cubic feet per minute and $h = 8$ feet, substitute these values into the differentiated equation:
$10 = \frac{25\pi}{144} \cdot 8^2 \cdot \frac{dh}{dt}$
Simplify:
$10 = \frac{25\pi}{144} \cdot 64 \cdot \frac{dh}{dt}$
$10 = \frac{1600\pi}{144} \cdot \frac{dh}{dt}$
$10 = \frac{400\pi}{36} \cdot \frac{dh}{dt}$
$10 = \frac{100\pi}{9} \cdot \frac{dh}{dt}$
Solve for $\frac{dh}{dt}$:
$\frac{dh}{dt} = \frac{10 \cdot 9}{100\pi}$
$\frac{dh}{dt} = \frac{90}{100\pi}$
$\frac{dh}{dt} = \frac{9}{10\pi} \text{ feet per minute}$
The rate of change of the depth of the water when the water is 8 feet deep is:
$$\boxed{\frac{9}{10\pi} \text{ feet per minute}}$$