# Rates of CHange

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#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.
0
15 years ago
#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).
0
#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.
0
15 years ago
#4
(Original post by mik1a)
seems quite simple now that it makes sense.
Yeah, maths is always like that.
0
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