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Math Help - Weird calculus problem involving

  1. #1
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    Weird calculus problem involving

    Okay, so I have this question that has been killing me. This is it:

    If f is the focal length of a convex lens and an object is placed at a distance p from the lens, then its image will be at a distance q from the lens, where f , p , and q are related by the lens equation
    1/f=1/p+1/q
    What is the rate of change of p with respect to q if q=4 and f=1 ?

    So I've tried about a million approaches and I never get the right answer. I know you're supposed to show how you've attempted this problem, but I've tried so many ways I don't know which one to type on here. Basically, I'm hoping somebody will be able to give me a boot in the right direction because I actually haven't a clue as to how to even begin attacking this.
    Any hints?
    Thanks!
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  2. #2
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    We are given \frac{1}{f}=\frac{1}{p}+\frac{1}{q}, where

    f=1 and q=4. We must find dp/dt.

    \frac{1}{p}+\frac{1}{q}=\underbrace{1}_{\text{beca  use f=1}}

    Differentiate wrt time:

    \frac{-1}{p^{2}}\cdot \frac{dp}{dt}-\frac{1}{q^{2}}\cdot \frac{dq}{dt}=0

    Solve for dp/dt:

    \frac{dp}{dt}=\frac{-1}{q^{2}}\cdot \frac{dq}{dt}\cdot p^{2}

    We know q=4 and we can find p by 1=\frac{1}{p}+\frac{1}{4}

    p=\frac{4}{3}

    \frac{dp}{dt}=\frac{-1}{16}\cdot \frac{dq}{dt}\cdot \frac{16}{9}=\frac{-1}{9}\cdot \frac{dq}{dt}

    We do not know dq/dt. Were you given that?. Assuming I read the problem correctly.
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  3. #3
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    No, we don't have that information. I can't seem to figure out how to figure it out without anything such as dq/dt or df/dt. I can't find a way around this lack of info...
    Thanks for the help though
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  4. #4
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    So I just figured it out. In case anybody is curious, here's how it's done:
    Use the steps suggested by galactus:

    "f=1 and q=4. We must find dp/dt.



    "

    Then differentiate with respect to q.
    1/p=-(1/q)+1
    (-1/(p^2))(dp/dq)=1/(q^2)
    dp/dq=(1/q^2))/(-1/(p^2))
    Simplifies to dp/dq=-(p^2)/(q^2)

    if we solve



    for p, we end up with p=[1/(1-(1/q))]

    plug that into
    dp/dq=-(p^2)/(q^2)
    for p and plug in q=4 and we end up with (-1/9)

    Yay!
    Thanks for your help. You basis was vital in helping me figure this out. Sooo happy!!
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  5. #5
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    df/dt = 0 because it is not changing.

    As we can see, dq/dt is 9 times dp/dt.


    EDIT: yes, good approach. I just differentiated wrt time. The problem said q. My bad. Anyway, it leads to the same result.
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