Question #f65d3
1 Answer
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
# V = pir^2h #
# \ \ \ = pi(20^2)h #
# \ \ \ = 400pi h # ..... [A]
Differentiate [A] Implicitly wrt
# 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
# [(dh)/dt]_(h=25) =50/(400pi ) #