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# Rates of CHange watch

1. A circular oil slick is increasing in area at a constant rate of 3 m^2 s^-1. Find the rate of increase of the radius of the slick at the instant when the area is 1200 m^2.

Slightly disheatening to get stuck on the second question, here is what I tried:

area: A
time: t

dA/dt = 3

Also,

A = pi.r^2
dA/dr = 2pi.r

The chain rule says:

dA/dt = dA/dr * dr/dt

Here's where I get stuck. I don't understand why it tells us to find it at that specific area. I tried substituting values to get:

3 = 2pi.r (dr/dt)

But I don't know where to go now. Could you please explain clearly how do do these because the single example they give isn't enough. Thanks very much if anyone can help!

Mike.
2. (Original post by mik1a)
3 = 2pi.r (dr/dt)
You're fine up to there. Now divide both sides by 2pi.r to get

3/(2pi.r) = dr/dt

What you want is dr/dt when A = 1200

Since A = pi. r^2
when A = 1200, r = sqrt(1200/pi)

Stick that figure in to get your answer (don't have a calc handy, sorry).
3. (Original post by Squishy)
You're fine up to there. Now divide both sides by 2pi.r to get

3/(2pi.r) = dr/dt

What you want is dr/dt when A = 1200

Since A = pi. r^2
when A = 1200, r = sqrt(1200/pi)

Stick that figure in to get your answer (don't have a calc handy, sorry).
Ok thanks, I actually did this but messed up in the algebra and got convinced I'd gone wrong (which I had). Thanks for the help, seems quite simple now that it makes sense.
4. (Original post by mik1a)
seems quite simple now that it makes sense.
Yeah, maths is always like that.

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