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Homework Help: Rate of increase

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    The radius of a circle is increasing at a constant rate of 0.2 meter per second. What is the rate of increase in the area of the circle at the instant when the circumference of the circle is 20pi meters?

    2. Relevant equations

    3. The attempt at a solution
    I have no clue how to approach the problem
  2. jcsd
  3. Oct 2, 2008 #2


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    Homework Helper

    The usual way to approach such problems is to find a relationship between the two quantities and then apply the chain rule to find the quantity you're interested in.

    What is the relationship between the radius of a circle and its area?
    Once you have that, try differentiating both sides with respect to time.
  4. Oct 2, 2008 #3
    how could i differentiate a?.. dA/dt?
  5. Oct 2, 2008 #4
    It's quite simple when you look at it really.
    You have dr/dt (rate of increase of radius) = 0.2 where r = radius and t = time (seconds).
    you are looking for dA/dt where A = area of circle

    [tex] A = \pi r^2 [/tex]

    [tex] \frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} [/tex] <-- chain rule.

    [tex] \frac{dA}{dr} = \frac{d}{dr}[ \pi r^2 ]= 2 \pi r [/tex]

    Therefore the rule for rate of area increase = [tex] 0.4 \pi r [/tex]

    Now you just have to find it at the instant when circumference = [tex] 20 \pi [/tex]

    I'm sure you can take it from here easily. I hope I didn't give away too much.
    Last edited: Oct 2, 2008
  6. Oct 2, 2008 #5
    Thank you :)

    C = 2(pi)r
    r = C/2(pi)
    r = 20pi/2pi
    r = 10
    dA/dt = 2(pi)r * dr/dt
    dA/dt = 2(pi)r * 0.2
    dA/dt = 2(pi)(10) * 0.2
    dA/dt = 4pi
    is that correct?
  7. Oct 2, 2008 #6
    Looks fine to me. You may get marked down if you don't include units, ie. 4pi m^2 per second.
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