Everytime I do this problem I end up with a q left and I am not allowed to use a q in my answer.

If anyone knows how to set this problem up it would be helpful.

Re: Strange related rates problem?

bylittlejodoon Wed Oct 29, 2008 3:24 pm

I'm not sure how to set it up.

Normally with related rates I would identify what changes and then differentiate and then plug in the values for f and p.

The focal length doesn't change. I guess the distance of the object to the lens (q) does as well as the distance from the lens to the image(p).

Here is what I have been trying. Since the question is asking for dq/dp I tried putting f in terms of p. Since f = 20 and p = 10, f = 2p

1/2p = 1/q + 1/p

or

2p^-1 = q^-1 + p^-1

I differentiate to get:

-2p^-2 = -q^-2 dq/dp + -p^-2

I try to isolate dq/dp to get:

(-2p^2 + p^2 )/ -q^-2 = dq/dp

I plug in the values

(-2(10)^-2 + 10^-2)/ q = dq/dp

but again, I can't use q in my answer. help!?!?