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

1f=1p+1q
What is the rate of change of p with respect to q if q=4 and f=2? (Make sure you have the correct sign for the rate.)

1/f = 1/p + 1/q

0 = -1/p^2 dp/dq - 1/q^2
dp/dq = -p^2/q^2

To find the rate of change of p with respect to q, we need to differentiate the lens equation with respect to q.

Given: 1/f = 1/p + 1/q

Differentiating both sides with respect to q:

d(1/f)/dq = d(1/p)/dq + d(1/q)/dq

Since f is a constant, the derivative of 1/f with respect to q is zero.

0 = d(1/p)/dq + d(1/q)/dq

Now, substituting the given values of q=4 and f=2 into the equation, we can solve for the rate of change.

0 = d(1/p)/dq + d(1/4)/dq

Since 1/4 is a constant, the derivative of 1/4 with respect to q is also zero.

0 = d(1/p)/dq + 0

Thus, the rate of change of p with respect to q when q=4 and f=2 is zero.