# calculus

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 = 2 and f = 6? (Make sure you have the correct sign for the rate.)

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1. 1/6 = 1/p + 1/2

1/p = 1/6 - 3/6 = - 2/6 = -1/3
so
p = -3

now the calculus, assume f is a constant of the lens so df/d anything = 0.
0 = (-dp/dq)/p^2 - (-1)/q^2

dp/dq = p^2/q^2 = 9/2 = 4.5

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posted by Damon
2. chinga tu madre no jala wa

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posted by diego

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