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Math Help - Implicit Differentiation word problem

  1. #1
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    Implicit Differentiation word problem

    "A stone is dropped into a still pond. Concentric circular ripples spread out and the radius of the disturbed area increases at a rate of 16 cm/sec. At what rate does the area of the disturbed region increase when its radius is 4 cm?"

    don't really know where to start.

    I said A = πr^2 --> r = sqrt(A/π) --> d/dt(r) = sqrt(A/π) . d/dt

    Now I'm lost...
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by satx View Post
    "A stone is dropped into a still pond. Concentric circular ripples spread out and the radius of the disturbed area increases at a rate of 16 cm/sec. At what rate does the area of the disturbed region increase when its radius is 4 cm?"

    don't really know where to start.

    I said A = πr^2 --> r = sqrt(A/π) --> d/dt(r) = sqrt(A/π) . d/dt
    Why did you do that? (and what is it you think you have done for that matter?)

    You had:

    A=\pi r^2

    Now just differentiate this with respect to time.

    CB
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    So, d/dt (A) = πr^2(d/dt) = π4^2/.25s = 64π cm^2 per second?
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    Quote Originally Posted by satx View Post
    So, d/dt (A) = (d/dt)πr^2 = π4^2/.25s = 64π cm^2 per second?
    first note what is in red above

    Now, A = \pi r^2

    \Rightarrow \frac {dA}{dt} = 2 \pi r \cdot \frac {dr}{dt}

    Now continue
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    So (2π4^2)/.25s = 128π sq cm per sec?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by satx View Post
    So (2π4^2)/.25s = 128π sq cm per sec?
    are you looking at what's posted? where is 4^2 coming from? There is nothing squared in what I posted. And what's that 0.25?

    \frac {dA}{dt} = 2 \pi (4) (16) ~\frac {\text {cm}^2}{\text{sec}}= 128 \pi ~\frac {\text {cm}^2}{\text{sec}}
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    Quote Originally Posted by Jhevon View Post
    first note what is in red above

    Now, A = \pi r^2

    \Rightarrow \frac {dA}{dt} = 2 \pi r \cdot \frac {dr}{dt}

    Now continue
    Quote Originally Posted by Jhevon View Post
    are you looking at what's posted? where is 4^2 coming from? There is nothing squared in what I posted. And what's that 0.25?

    \frac {dA}{dt} = 2 \pi (4) (16) ~\frac {\text {cm}^2}{\text{sec}}= 128 \pi ~\frac {\text {cm}^2}{\text{sec}}
    I was multiplying the r by the r in dr/dt. And the .25 sec is the time it takes the radius to reach 4 cm. I told you I didn't really know where to start...

    Anyway, so the end result is right then?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by satx View Post
    I was multiplying the r by the r in dr/dt. And the .25 sec is the time it takes the radius to reach 4 cm. I told you I didn't really know where to start...

    Anyway, so the end result is right then?
    dr/dt is notation. The r in that is not available to be multiplied.

    Do not complicate things. You do exactly as I did, and just plug in. dr/dt is the rate at which the radius is changing. That was given in the problem, it is 16 cm/sec, so you just plug in dr/dt = 16 the form I gave you. Then, r = 4, and compute.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by satx View Post
    I was multiplying the r by the r in dr/dt. And the .25 sec is the time it takes the radius to reach 4 cm. I told you I didn't really know where to start...

    Anyway, so the end result is right then?
    You are not yet ready for implicit differentiation, you need to go back and revise the basics of differentiation.

    CB
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