calculus
posted by Sam .
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.)

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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