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Find how fast the radius of a circle get smaller as its area gets smaller

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data
    (Translated, sorry my English) Find how fast does the radius of a circle get smaller when its area is 75*pi cm^2 and it (the area) gets smaller at a rate of 2*pi cm^2/s?

    2. Relevant equations
    Not really any given equations, but I used these:
    A = r^2 * pi
    s = v*t + s0 (If v is a constant)
    t=A/Va (where t is the time when A=0)

    3. The attempt at a solution
    From now on I'll call area, A = 75*pi cm^2 and the rate v_a = 2*pi cm^2/s.
    I gave myself that the speed (v) was constant, that shouldn't really change the answer.
    So s = dv/dt = v*t + s0

    By using A = r^2 * pi we know that:
    r^2 = A/pi
    r = sqrt(A/pi) and since at t = 0 then A = 75*pi so
    r = sqrt(75) at t = 0, lets call r from now on s.

    Let's find the time until A = 0, and therefore s=0
    t = A/v_a = 75*pi/(2*pi) = 37,5 sec

    So in 37,5 sec the circle area will be zero with is radius. Therefore we can see that:
    s = v*t + s0
    0 = v*37,5 + sqrt(75)
    -sqrt(75) = v*37,5
    v = -sqrt(75)/37,5
    v = -2*sqrt(75)/75
    v = -2/sqrt(75)

    So my final answer was v = -2/sqrt(75) (or v = -2/(5*sqrt(3))


    But according to my teacher the correct answer is v = -1/sqrt(75).


    --
    Here's how my teacher did it:

    We know that F(t) = pi * r(t)^2 and
    (1) F'(t0) = 2*pi*r(t0)*r'(t0) = -2*pi
    (2) F(t0) = pi*r(t)^2 = 75*pi
    for a specific t0.

    According to (2) we know that r(t0) = sqrt(75) = 5*sqrt(3) and according to (1) we get r'(t0) = -1/r(t0) = -1/sqrt(75)

    --
    He called my solution "a waste of time" although he said it should work too, but couldn't explain to me why I didn't get the same answer.

    Where did I go wrong (or even better, where did he go wrong?)
     
  2. jcsd
  3. Oct 15, 2009 #2

    Mark44

    Staff: Mentor

    The area of the circle is A(r) = pi r2. Assuming that both A and r are differentiable functions of t, and differentiating with respect to t, you get
    dA/dt = 2pi r dr/dt

    Solve this equation for dr/dt. That will give you dr/dt for any arbitrary t.
    Now evaluate dr/dt at the time when A is 75 pi cm2 and when dA/dt is -2 pi cm2/sec. You will need to find the radius r for which the area A is 75 pi cm2 to do this.
     
  4. Oct 15, 2009 #3
    Thank you for a quick reply!

    dr/dt = 1/(2*pi*r) dA/dt and dA/dt = -2*pi so
    dr/dt = -1/r

    And like I showed before r = sqrt(75) so the final answer is indeed:
    dr/dt = -1/sqrt(75)


    **SOLVED**
    Thank you again. I understand your solution a lot better than my teacher's.
     
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