Answer:
So the radius is increasing at [tex]\frac{1}{64\pi} \frac{\text{cm}}{\text{min}}[/tex].
This is approximately 0.00497 cm/min that the radius is increasing.
Step-by-step explanation:
[tex]V=\frac{4}{3}\pi r^3[/tex]
The volume and radius are both things that are changing with respect to time.
So their derivatives will definitely not be 0.
Let's differentiate:
[tex]V'=\frac{4}{3} \pi \cdot 3r^2r'[/tex]
I had to use constant multiple rule and chain rule.
We are given [tex]V'=+25 \frac{\text{cm^3}}{\text{min}}[/tex] and [tex]r=20 \text{cm}[/tex].
We want to find [tex]r'[/tex].
Let's plug in first:
[tex]25=\frac{4}{3}\pi \cdot 3(20)^2r'[/tex]
[tex]25=\frac{4}{3} \pi \cdot 3(400)r'[/tex]
[tex]25=\frac{4}{3} \pi \cdot 1200r'[/tex]
Multiply both sides by 3:
[tex]75=4 \pi \cdot 1200r'[/tex]
[tex]75=4800 \pi r'[/tex]
Divide both sides by [tex]4800 \pi[/tex]:
[tex]\frac{75}{4800 \pi}=r'[/tex]
[tex]\frac{1}{64 \pi}=r'[/tex]
So the radius is increasing at [tex]\frac{1}{64\pi} \frac{\text{cm}}{\text{min}}[/tex].
This is approximately 0.00497 cm/min that the radius is increasing.
