Answer to Question #329559 in Calculus for Keila

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

Let $u=100~cm^3/sec,~l=75\space cm,$

$H=20~cm,~B=30~cm,~h=10~cm,$

$x$ - the current depth of the water layer.

The current volume of the water:

$V=\frac{1}{2}bxl,$

where $b$ is the base of the triangular cross-section of the water layer; this triangle is similar to the cross-section of the trough with similarity factor of $k=\frac{x}{H}$, so $b=kB=x\frac{B}{H}$.

$V=\frac{1}{2}x\frac{B}{H}xl=\frac{Bl}{2H}x^2\\ u=\frac{dV}{dt}=\frac{Bl}{H}x\frac{dx}{dt}\Rarr\frac{dx}{dt}=\frac{uH}{Blx}\\ \frac{dx}{dt}(x=h)=\frac{uH}{Blh}=\frac{100\cdot20}{30\cdot75\cdot10}\approx\\ \approx0.089~(cm/sec)=0.89~(mm/sec)$