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#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

is $10 \text{ ft}$ across the top and $12 \text{ ft}$ deep.

If water is flowing into the tank at rate of $\displaystyle\frac{10 \text{ ft}^3}{\text{min}}$

Find the rate of change of the depth of the water when the water is $8 \text{ ft}$ deep.

$\displaystyle V=\frac{1}{3} \pi r^2 h$

$\displaystyle r=\frac{5}{12} h$

$\displaystyle\frac{dV}{dt}=\frac{10 \text{ ft}^3}{\text{min}}$

$

\displaystyle V=\frac{1}{3}\pi\left(\frac{5}{12} h\right)^2 h

=\frac{25}{432} \pi h^3

$

$\displaystyle\frac{d}{dt}V=\frac{d}{dt}\frac{25}{432} \pi h^3$

want to see if this set up right so far before continue??? h=depth