Answer to Question #329558 in Calculus for Sync

Question

A trough is 75cm long and its ends in the form of an isosceles triangle have an altitude of 20cm 

and a base of 30cm. Water is flowing into the trough at the rate of 100cm3/sec. Find the rate at 

which the water level is rising when the water is 10cm deep?

Answer 1

The volume of the water when the water is $h$ cm deep:

$V=\frac{b\cdot L\cdot h}{2}$

From the similarity of triangles we can write:

$\frac{B}{H}=\frac{b}{h}$ $\implies$ $b=\frac{Bh}{H}$

Now the volume of the water:

$V=\frac{B\cdot L\cdot h^2}{2H}$

Let's find $\frac {dV}{dt}$ :

$\frac {dV}{dt}=\frac{B\cdot L\cdot h}{H}\frac{dh}{dt}$

The rate at which the water level is rising:

$\frac{dh}{dt}=\frac{H}{B\cdot L\cdot h}\frac {dV}{dt}=\frac{20}{30\cdot 75\cdot 10}\cdot 100\approx0.089$ cm/sec