Question

an oil tank is being drained for cleaning. after $t$ minutes there are $v$ litres of oil left in the tank, $v(t) = 35(25-t)^2, 0 \leq t \leq 25$.
a. determine what $v(25)$ represents. 1
b. determine the average rate of change of volume during the first 15 minutes. 4
c. determine the rate of change of volume at the time $t = 15$ minutes. 4

Explanation:

Step1: Analyze V(25) meaning

Substitute $t=25$ into $V(t)$:
$V(25)=35(25-25)^2=35\times0^2=0$
This corresponds to the volume after 25 minutes, when the tank is fully drained.

Step2: Calculate average rate of change

The average rate of change over $[0,15]$ is $\frac{V(15)-V(0)}{15-0}$.
First find $V(15)=35(25-15)^2=35\times10^2=3500$
$V(0)=35(25-0)^2=35\times625=21875$
Expression: $\frac{3500-21875}{15}=\frac{-18375}{15}$

Step3: Calculate instantaneous rate of change

First find derivative $V'(t)$ using chain rule:
$V'(t)=35\times2(25-t)\times(-1)=-70(25-t)$
Substitute $t=15$:
$V'(15)=-70(25-15)$

Answer:

a. $V(25)$ represents the volume of oil left in the tank after 25 minutes, which is 0 litres (the tank is fully drained).
b. $\frac{3500-21875}{15}=-1225$ litres per minute
c. $-70(25-15)=-700$ litres per minute