Question #329558

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?


1
Expert's answer
2022-04-19T03:37:26-0400


The volume of the water when the water is hh cm deep:

V=bLh2V=\frac{b\cdot L\cdot h}{2}

From the similarity of triangles we can write:

BH=bh\frac{B}{H}=\frac{b}{h}     \implies b=BhHb=\frac{Bh}{H}

Now the volume of the water:

V=BLh22HV=\frac{B\cdot L\cdot h^2}{2H}

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

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

The rate at which the water level is rising:

dhdt=HBLhdVdt=203075101000.089\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


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