Thread: Related Rates, water entering trough problem

1. Related Rates, water entering trough problem

A trough is 15ft long and 4ft across the top. Its ends are isosceles triangles with height 3 ft. Water runs into the trough at a rate of 2.5 cubic feet per minute. How fast is the water rising when it is 2 feet deep?

The answer is .0625 but I'm pretty lost on how to get there.

2. Originally Posted by Porcelain
A trough is 15ft long and 4ft across the top. Its ends are isosceles triangles with height 3 ft. Water runs into the trough at a rate of 2.5 cubic feet per minute. How fast is the water rising when it is 2 feet deep?

The answer is .0625 but I'm pretty lost on how to get there.
$\displaystyle \frac{dV}{dt} = 2.5 \, ft^3/min$

volume of water in the tank ...

$\displaystyle V = \frac{1}{2} \cdot b \cdot h \cdot 15$

using similar triangles ...

$\displaystyle \frac{b}{h} = \frac{4}{3}$

solve for $\displaystyle b$, substitute the result for $\displaystyle b$ in the volume formula to get $\displaystyle V$ as a function of $\displaystyle h$.

take the time derivative and determine $\displaystyle \frac{dh}{dt}$ when $\displaystyle h = 2$ ft

3. Skeeter you always answer my questions. Thanks SOO much