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

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

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

So the radius is increasing at
`(1)/(64\pi) \frac{\text{cm}}{\text{min}}`.

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

Explanation:

`V=(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'=(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=(4)/(3)\pi \cdot 3(20)^2r'`

`25=(4)/(3) \pi \cdot 3(400)r'`

`25=(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`:

`(75)/(4800 \pi)=r'`

`(1)/(64 \pi)=r'`

So the radius is increasing at
`(1)/(64\pi) \frac{\text{cm}}{\text{min}}`.

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