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

karush

Well-known member
Jan 31, 2012
2,715
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
Jan 30, 2012
2,492
Seems good to me.
 

karush

Well-known member
Jan 31, 2012
2,715
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
Jan 26, 2012
1,403
Seems reasonable to me too :)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
$
\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.