Question

Question #f65d3

Answer 1

`[(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 )`