Thread: 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!

Originally Posted by s3a

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.

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 .

Originally Posted by s3a

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.