Water is draining from a swimming pool in such a way that the remaining volume of water after t minutes is V = 200(50 - t)^2 cubic meters. Find :

Question

(a) the average rate at which the water leaves the pool in the first 5 minutes.

(b) Find the instantaneous rate at which the water is leaving at t = 5 minutes.

a)
V(0) = 500,000
V(5) = 405,000
(V(5)-V(0))/(5-0) = -19,000 ft^3/min

Can someone help me on question b and check if question a is correct

Answer 1

a) is correct
b)
dV/dt = 400(50-t)^1 (-1)
at t = 5
dV/dt = -400(45) = -18,000 ft^3/min

Answer 2

how did you get 400?

Answer 3

Do you not know how to find the derivative using the chain rule?
If not, this is not a forum to teach it to you from scratch.

Answer 4

400 = 200*2