v(0) = 35(25)^2 = 21875
v(15) = 35(10)^2 = 3500
avg rate of change for those 15 minutes
= (2500 - 21875)/(15-0)
= -18375/15 L/min
= -1225 L/min
You are correct
at t=15, you are looking for the instantaneous rate of change, that is,
the derivative of v(t)
v ' (t) = 70(25-t) (-1)
v ' (15) = -70(25-15) = -700 L/min
I am re-posting this question because i wrote the incorrect function last time.
An oil tank is being drained for cleaning. After t minutes there are v litres of oil left in the tank, where v(t)=35(25-t)^2, 0<=t<=25
a) determine the average rate of change of volume during the first 15 minutes
I got -1225 as my answer, is that correct?
b) determine the rate of change of volume at the time t=15 minutes
I'm not sure how to answer this.
1 answer
2 answers