Question

"A tank holds 1000L of water, which takes an hour to drain from the bottom of the tank, then the volume V of water remaining in the tank after t minutes is: V = 1000(1-(t/60))^2 where t is between 0 and 60 inclusive.

Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t) after 10 min."

I found the volume at 10 min and at 0 min, then did the slope formula, but I'm not getting answer I'm supposed to. Am I reading the question incorrectly?

Answer 1

This is clearly a Calculus question.
You have taken the average rate of change from t=0 to t=10

It asked for the "instantaneous rate of change", which is the derivative dV/dt

So find dV/dt, then sub in t=10

Answer 2

Oh! I haven't been taught derivatives yet. Thank you for pointing that out. I'm getting ahead of myself here. :)