# [SOLVED]find rate of change of the depth of water when the water is 8 ft deep

#### karush

##### Well-known member
A conical tank (with the vertex down)

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

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Seems good to me.

#### karush

##### Well-known member
so moving along..

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

$\displaystyle\frac{dV}{dt}=\frac{dh}{dt}\frac{75}{432}\pi h^2$

$\displaystyle\frac{dh}{dt}=\frac{dV}{dt}\frac{432}{75\pi h^2} =\frac{10 \text{ ft}^3}{\text{min}}\cdot\frac{432}{75\pi\cdot {(8\text{ft})^2}} =\frac{9}{10}\pi \frac{\text{ft}}{\text{min}}$

sorta seems reasonable...

#### Prove It

##### Well-known member
MHB Math Helper
Seems reasonable to me too

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
$\displaystyle\frac{dh}{dt}=\frac{dV}{dt}\frac{432}{75\pi h^2} =\frac{10 \text{ ft}^3}{\text{min}}\cdot\frac{432}{75\pi\cdot {(8\text{ft})^2}} =\frac{9}{10}\pi \frac{\text{ft}}{\text{min}}$
In the last expression, $\pi$ should be in the denominator.