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Math Help - Derivative Question --> Rate of Change with focal length formula

  1. #1
    s3a
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    Derivative Question --> Rate of Change with focal length formula

    Question:

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

    (My work is attached)

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    Question:

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

    (My work is attached)

    Any help would be greatly appreciated!
    Thanks in advance!
    1/p = 1/f - 1/q = (q - f)/fq, so p = fq/(q - f). So dp/dq = d/dq fq/(q - f) = f(q - f) - fq by the quotient rule. Plugging in q = 2 and f = 3 yields 3(2 - 3) - 2*3 = -3 - 6 = -9.
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  3. #3
    s3a
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    I'm lost in your explanation. I don't even get how you go from 1/p = 1/f - 1/q to (q - f)/fq .
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  4. #4
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    Quote Originally Posted by s3a View Post
    I'm lost in your explanation. I don't even get how you go from 1/p = 1/f - 1/q to (q - f)/fq .
    It's just standard subtraction of fractions.
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