# Thread: related rate unlike others

1. ## related rate unlike others

Hi. I had pretty well gotten down the concept of related rates and then this problem came along. Three hours later I still don't have a correct answer.

the following equation is given and I am supposed to find dq/dp when f = 20 and p = 10. No proportional relationship is given to relate f and p so I am not sure if it is just coincidence that at this point in time f =2p.

1/f = 1/q + 1/p

so what is dq/dp?

I've tried getting rid of f by making it 2p, also I tried replacing p with 1/2 f. I have tried everything I could come up with to make this thing work but I never get an answer that is correct (I submit answers online).

Any help would be greatly appreciated!!

2. 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?

by littlejodo on 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!?!?

3. Originally Posted by littlejodo
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?

by littlejodo on 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).
You say "the focal length doesn't change" so that can't possibly be true! In your original problem you were told to find dq/dp when q= 10 and p= 20. That means that 1/f= 1/10+ 1/20= 3/20 so f= 20/3.

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!?!?
1/p+ 1/q= 1/f, a constant. $p^{-1}+ q^{-1}= 1/f$. Differentiating $-p^{-2}- q^{-2}dq/dp= 0$ so $q^{-2} dq/dp= -p^{-2}$. Finally, $dq/dp= q^2p^{-2}= (q/p)^2$.

Yes, that involves "q"- there's nothing wrong with that! You are asked for the vaue when q= 10 and p= 20. Put those in.

4. what is given is p=10 and f=20

5. Now I see... thank you so much!!