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Math Help - Rate of change problem

  1. #1
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    Rate of change problem

    So the whole problem is:

    If f is the focal length of a convex lens and an object is placed at a distance q from the lens, then its image will be at a distance p from the lens, where f,g, and
    p are related by the lens equation

    \frac{1}{f}=\frac{1}{q}+\frac{1}{p}

    Suppose the focal length of a particular lens is 20cm. What is the rate of change of q with respect to p when p=3?

    So to start out I plug in f=20cm and p=3 into the equation above but then I can't figure what to do from here. I just having trouble comprehending the problem and figuring out how to start. Any help is greatly appreciated!
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  2. #2
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    Quote Originally Posted by ascendancy523 View Post
    So the whole problem is:

    If f is the focal length of a convex lens and an object is placed at a distance q from the lens, then its image will be at a distance p from the lens, where f,g, and
    p are related by the lens equation

    \frac{1}{f}=\frac{1}{q}+\frac{1}{p}

    Suppose the focal length of a particular lens is 20cm. What is the rate of change of q with respect to p when p=3?

    So to start out I plug in f=20cm and p=3 into the equation above but then I can't figure what to do from here. I just having trouble comprehending the problem and figuring out how to start. Any help is greatly appreciated!
    Once put f= 20 cm into the formula you have
    \frac{1}{20}= \frac{1}{q]+ \frac{1}{p}.

    Now the problem asks yuo to find "the rate of change of q with respect to p". That means, of course, that you are asked to differentiate q with respect to p.

    One way to do that would be to first solve for q as a function of p and differentiate.

    Another would be to use "implicit differentiation".

    Perhaps it would help to write the equation as \frac{1}{20}= q^{-1}+ p^{-1}. Can you differentiate that with respect to p?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Once put f= 20 cm into the formula you have
    \frac{1}{20}= \frac{1}{q]+ \frac{1}{p}.

    Now the problem asks yuo to find "the rate of change of q with respect to p". That means, of course, that you are asked to differentiate q with respect to p.

    One way to do that would be to first solve for q as a function of p and differentiate.

    Another would be to use "implicit differentiation".

    Perhaps it would help to write the equation as \frac{1}{20}= q^{-1}+ p^{-1}. Can you differentiate that with respect to p?
    So how would you go about solving that implicitly? My whole chapter deals with implicit differentiation so I'm assuming it want us to practice using that method.

    My guess would be to find a relationship between the two and create an equation which would be \frac{dq}{dp}?
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