Question

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 7 cubic feet per minute. if the pool has radius 5 feet and height 8 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 5 feet?

Answer 1

v = (pi)r² h
v = (pi)5² h
v = 25(pi) h
take derivative with respect to h
dv/dh = 25(pi)

(dv/dt)*(dt/dh) = dv/dh

dv/dt is given to be 5 cm³
solved for dv/dh = 25(pi)

5(dt/dh) = 25(pi)
dt/dh = 25(pi)/5
dt/dh = 5(pi)

then the recipricol of dt/dh; we want to find dh/dt = 1/(5(pi))

Answer 2

The rate of change of the height of the water  in the pool when the depth of the water in the pool is 5 feet which is $\rm \bold{ \dfrac{dt}{dh}}$ will  $\rm \bold{ \dfrac{1}{5 \pi}}$

The volume of the cylindrical tank,

$\rm \bold{ V = \pi r^2.h}$

r- radius = 5 ft

$\rm \bold{ V = \pi 5^2.h}\\\\\rm \bold{ V = 25 \pi h}$

Derivative  respect to Height h

$\rm \bold{ \dfrac{dv}{dt} = 25\pi }$

$\rm \bold{ \dfrac{dv}{dh} = \dfrac{dv}{dt}\times \dfrac{dt}{dh} }$

$\rm \bold{ \dfrac{dv}{dt} }$ is given to be $\rm \bold{ 5cm^3}$

$\rm \bold{ \dfrac{dt}{dh} = \dfrac{25\pi}{5} }\\\\ \rm \bold{ \dfrac{dt}{dh} = 5\pi}$

Hence, we can conclude that the rate of change of the height of the water which is $\rm \bold{ \dfrac{dt}{dh}}$ will  $\rm \bold{ \dfrac{1}{5 \pi}}$

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