MHB Find rate of change of the depth of water when the water is 8 ft deep

[karush]

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]

Seems good to me.

 

[karush]

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]

Seems reasonable to me too :)

 

[Evgeny.Makarov]

$
\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.