Question

A spherical balloon is being filled in such a way that the surface area is increasing at a rate of 20 cm^2/sec when the radius is 2 meters. At what rate is air being pumped in the balloon when the radius is 2 meters?

Answer 1

Surface area of a sphere?

Answer 2

A=4pi(r^2)

Answer 3

Good good. And we know \(\large \frac{dA}{dt}\) = 20 cm^2/sec

Answer 4

So would I find the derivative of the area formula?

Answer 5

Take the derivative of A = 4pi(r^2) to get \(\large \frac{dA}{dr}\)

Answer 6

Derivative with respect to r, I mean.

Answer 7

dA/dr = 4pi(2r)?

Answer 8

Right, and 4pi(2r) = 8pi*r
\(\Large \frac{dA}{dt} /div \frac{dA}{dr} = \frac{dr}[dt}\)

Answer 9

Wait, dA/dt=8pi(r)dr/dt?

Answer 10

So would I just then plug in the r and dA/dt to get my answer?

Answer 11

Okay, Ashley.
You start out knowing dA/dt. You are asked to find out dV/dt.

Take derivative of A=4pir^2 to get dA/dr
dA/dt / divided by dA/dr gives you dr/dt
Now look at your volume equation:
V = (4/3)pi*r^3
Take the derivative with respect to r to get
dV/dr

dr/dt * dV/dr = dV/dt

Answer 12

How much of that makes sense to you?

Answer 13

dV/dr=4pi(r^2)?

Answer 14

I'm so confused.

Answer 15

You are correct =)
dV/dr = 4*pi*r^2

Answer 16

I just want to chime in here. I am doing related rates right now as well and SmoothMath, you made is very clear

Answer 17

Ah, great =) Very good to hear.

Answer 18

So do I just multiply the two equations together to find dV/dt?

Answer 19

For both of you, I would say, make sure you're clear about what you are starting with and what you are trying to get to.

In this problem, we start with how quickly the surface area is changing, which is
dA/dt
And I am trying to figure out how much air is being blown into the balloon, which is how quickly the volume is changing, so
dV/dt

How do I get from dA/dt to dV/dt
\(\Large \frac{dA}{dt}*\frac{dr}{dA}*\frac{dV}{dr}\)

Answer 20

This is just like chemistry lol

Answer 21

I agree =)

Answer 22

Dimensional analysis.

Answer 23

Ask me any questions you have please, Ashley. I want you to understand.

Answer 24

How would you get dr/dA?

Answer 25

Okay, so I wrote that as \(\Large *\frac{dr}{dA}\)

but you could also do \(\Large \div \frac{dA}{dr}\)
and that would be the same thing.
Do you know how to find dA/dr?

Answer 26

Didn't we find that earlier? dA/dr = 8pi(r)

Answer 27

Right. Good.

Answer 28

So now what?

Answer 29

What have you gotten so far?

Answer 30

dA/dr=8pi(r)
dV/dr=4/3pi(r^2)

Answer 31

And you also know, from the problem, that
dA/dt = 20 cm^2/sec

Answer 32

Yes, and r=2

Answer 33

Great. Soooo you can figure it out from there.

Answer 34

\(\Large \frac{dA}{dt} \div \frac{dA}{dr}*\frac{dV}{dr}\)

Answer 35

So it'd just plug it in to get
20/16pi*16pi?

Answer 36

I think your math may be getting mildly confused.

Answer 37

\(\Large \frac{dA}{dt} \div \frac{dA}{dr} * \frac{dV}{dr}\)
\(\Large \rightarrow 20 \div (8\pi*r) * (4\pi*r^2)\)

Answer 38

Oh, I plugged in r=2

Answer 39

Great, that's fine.

Answer 40

You got the wrong result though, so check your math.

Answer 41

I still got 20 / (16pi) (16pi)

Answer 42

Oh okay. You mean \(\Large \frac{20}{16\pi}*16\pi\)

Answer 43

That simplifies.

Answer 44

Yes, so I get 20 right? I kinda think we just solved dA/dt

Answer 45

lol no don't worry. It's just a coincidence that it comes out the same.

Answer 46

20 cm^3/sec is correct.

Answer 47

Okay, thanks. :D I honestly didn't get it.