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    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
    radius: r
    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.
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    (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).
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    (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.
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    (Original post by mik1a)
    seems quite simple now that it makes sense.
    Yeah, maths is always like that.
 
 
 
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