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Derivatives in Action, Change in radius per time of circle.

  1. Jul 17, 2015 #1
    Hi, I'm trying to understand how Differentiation and

    integral works in practice, and would really appreciate

    some help interpreting this calculation-



    If we have a circle with Area A=pi*r^2


    1) If i want to find the change in Area with respect to

    radius then

    dA/dr= 2pi*r


    2) If I'm told that the area of the circle changes with

    time

    dA/dt = 0.03 (in)^2/sec

    And i wanted to find how the radius changes with time,

    would this then be the right conclusion

    dr/dt = dr/dA*dA/dt = (1/dA/dr)*dA/dt = 1/(2pi*r)*0.03


    So the change in radius with respect to time decreases as

    the radius increases. So is it correct to leave the

    expression like this, or should i express r in the equation

    as time, t, since it says dr/dt?

    If i want to find the radius at a specific time when the

    area increases as
    dA/dt = 0.03 (in)^2/sec would it the be correct to

    integrate

    dr/dt= 0.03/(2pi*r)

    to give me a function r which depends on t?

    If i want to express r as t in the eqation
    dr/dt= 1/(2pi*r)*0.03 , how exactly would i do it:


    I could isolate r in A=pi*r^2 --> sqrt(A/pi) and

    substitute sqrt(A/pi) into the equation and then to

    substitute t into A, i guess i would integrate dA/dt = 0.03

    (in)^2/sec to get A=A(t)=0.03*t and then substitute A=0.03t

    like this

    dr/dt= 0.03/(2pi*(sqrt(0.03*t/pi)


    I don't know if I'm breaking any rules here? OR if my

    reasoning is wrong?
     
  2. jcsd
  3. Jul 17, 2015 #2
    Is this a homework problem or just general interest?
     
  4. Jul 17, 2015 #3
    The first formula is wrong:
    A=2 Pi r(t)^2
    dA/dt= 2 Pi 2 r(t) (dr/dt)
     
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