Answer:
V'(t) = [tex]-250(1 - \frac{1}{40}t)[/tex]
If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.
Step-by-step explanation:
Given:
V = [tex]5000(1 - \frac{1}{40}t )^2[/tex] , where 0≤t≤40.
Here we have to find the derivative with respect to "t"
We have to use the chain rule to find the derivative.
V'(t) = [tex]2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )[/tex]
V'(t) = [tex]2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )[/tex]
When we simplify the above, we get
V'(t) = [tex]-250(1 - \frac{1}{40}t)[/tex]
If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.