Question

A water trough is 8 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 80 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.3 m3/min how fast is the water level rising when the water is 30 cm deep?

Answer 1

Answer:It is rising at a rate of $7.5cm/min$

Explanation:

We have volume of trapezoid equals

$V=Area\times Length\\\\V=\frac{1}{2}(a+b)h\times L$

Thus at any time 't' we have

$V(t)=\frac{1}{2}(a(t)+b(t))h(t)\times L\\\\\therefore V(t)=\frac{1}{2}(20+b(t))\times h(t)\times L$

Differentiating both sides with respect to time we get

$\frac{dV(t)}{dt}=\frac{1}{2}b'(t)h(t)L+\frac{1}{2}(20+b(t))\times h'(t)L$

Applying values we have

$b(t)=20+h(t)\\b'(t)=h'(t)$

Thus we have

$\frac{dV(t)}{dt}=\frac{1}{2}h'(t)h(t)L+\frac{1}{2}(20+20+h(t))\times h'(t)L\\\\2V'(t)=h'(t)L[h(t)+(40+h(t))]\\\\\therefore h'(t)=\frac{2V'(t)}{L(h(t)+(40+h(t)))}$

Applying values we get

$h'(t)=0.075m/min=7.5cm/min$