MHB How Fast Does a Balloon's Radius Expand When Inflated?

[Monoxdifly]

Icha is going to blow a sphere-shaped rubber balloon. She uses a pump to infuse the air with the volume addition rate $$40cm^3/s$$. If the radius addition rate is 20 cm/s, the radius of the sphere after being blown is ...
A. $$\frac{1}{\sqrt\pi}$$ cm
B. $$\frac{1}{\sqrt{2\pi}}$$ cm
C. $$\frac{1}{2\sqrt\pi}$$ cm
D. $$\frac{1}{3\sqrt\pi}$$ cm
E. $$\pi$$ cm

Please give me some hints. Thanks.

 

[skeeter]

Icha is going to blow a sphere-shaped rubber balloon. She uses a pump to infuse the air with the volume addition rate $$40cm^3/s$$. If the radius addition rate is 20 cm/s, the radius of the sphere after being blown is ...
A. $$\frac{1}{\sqrt\pi}$$ cm
B. $$\frac{1}{\sqrt{2\pi}}$$ cm
C. $$\frac{1}{2\sqrt\pi}$$ cm
D. $$\frac{1}{3\sqrt\pi}$$ cm
E. $$\pi$$ cm

Please give me some hints. Thanks.

take the derivative of the sphere volume formula w/respect to time ...

 

[Monoxdifly]

$$40cm^3/s=4\pi r^2(20cm/s)$$
$$2cm^2=4\pi r^2$$
$$r^2=\frac{1cm^2}{2\pi}$$
$$r=\frac{1}{\sqrt{2\pi}}cm$$?

 

[MarkFL]

For a sphere of radius $r$, the volume is:

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

Differentiating w.r.t time $t$, we have:

$$\d{V}{t}=4\pi r^2\d{r}{t}\implies r=\frac{1}{2}\sqrt{\frac{\d{V}{t}}{\pi\d{r}{t}}}$$

Plugging in the given numbers, we have:

$$r=\frac{1}{2}\sqrt{\frac{40\frac{\text{cm}^3}{\text{s}}}{20\pi\frac{\text{cm}}{\text{s}}}}=\frac{1}{\sqrt{2\pi}}\text{ cm}\quad\checkmark$$

I will move this thread to our Calculus forum. :)

 

[Monoxdifly]

I will move this thread to our Calculus forum. :)

I thought this was pre-calculus because it was a question in high school graduation exam.

 

[MarkFL]

I thought this was pre-calculus because it was a question in high school graduation exam.

It is a related rate problem involving differential calculus (so that's why I moved it). Many places teach elementary calculus in high school. :)

 

[Monoxdifly]

What is the actual limit between pre-calculus and calculus, anyway?