Question #f65d3

1 Answer
Steve M
Sep 16, 2017

[(dh)/dt]_(h=25) = 1/(8pi) ~~ 0.0398 \ cms^-1

Explanation:

Let us summarise the variables:

{(r=20, "Radius", (cm)), (h(t), "Height of water at time t", (cm)), (V(t), "Volume of the water at time t", (cm^3)), (t, "time", (s)) :}

We are given that:

[(dV)/dt]_(h=25) = 50 \ cm^3s^-1

and, we seek the value of:

[ (dh)/dt ]_(h=25)

The volume of the water in the cylinder, at time t:

V = pir^2h
\ \ \ = pi(20^2)h
\ \ \ = 400pi h ..... [A]

Differentiate [A] Implicitly wrt t, and applying the Chain Rule:

d/dt V = 400pi d/dt h
:. (dV)/dt = 400pi (dh)/dt d/(dh) h
:. (dV)/dt = 400pi (dh)/dt 1

:. (dV)/dt = 400pi (dh)/dt

:. (dh)/dt =1/(400pi ) ( (dV)/dt )

So, when h=25, we have:

[(dh)/dt]_(h=25) =50/(400pi )

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