Question

The volume of a spherical balloon is increasing at the rate of 25cm^3/min, how fast is the radius increasing when the radius is equal to 20 cm?

Answer 1

Answer:

1/64pi cm/min or 0.00497 cm/min

Step-by-step explanation:

dV/dr = 4pi×r²

At r = 20, dV/dr = 1600pi

dr/dt = dr/dV × dV/dt

dr/dt = 1/1600pi × 25

dr/dt = 1/64pi cm/min

Answer 2

Answer:

So the radius is increasing at $\frac{1}{64\pi} \frac{\text{cm}}{\text{min}}$.

This is approximately 0.00497 cm/min that the radius is increasing.  

Step-by-step explanation:

$V=\frac{4}{3}\pi r^3$

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:

$V'=\frac{4}{3} \pi \cdot 3r^2r'$

I had to use constant multiple rule and chain rule.

We are given $V'=+25 \frac{\text{cm^3}}{\text{min}}$ and $r=20 \text{cm}$.

We want to find $r'$.

Let's plug in first:

$25=\frac{4}{3}\pi \cdot 3(20)^2r'$

$25=\frac{4}{3} \pi \cdot 3(400)r'$

$25=\frac{4}{3} \pi \cdot 1200r'$

Multiply both sides by 3:

$75=4 \pi \cdot 1200r'$

$75=4800 \pi r'$

Divide both sides by $4800 \pi$:

$\frac{75}{4800 \pi}=r'$

$\frac{1}{64 \pi}=r'$

So the radius is increasing at $\frac{1}{64\pi} \frac{\text{cm}}{\text{min}}$.

This is approximately 0.00497 cm/min that the radius is increasing.