Strange related rates problem?

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littlejodo

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Oct 29, 2008
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I've done several related rate problems successfully but this one has me stumped!! I keep getting wrong answers.

the equation for the focal length of a lens is f. The equation relating focal length to the distance of the object to the lens, q, and the distance to the image, p is given with:

1/f = 1/q + 1/p

suppose that the focal length is 20cm. What is the rate of change of q with respect to p when p = 10?

dq/dp = ?


Even if someone could help me set this up it would amazing. I've tried putting f in terms of p and p in terms of f... then differentiate and put in values ... but I get the wrong answer.
 
littlejodo said:
I've done several related rate problems successfully but this one has me stumped!! I keep getting wrong answers.

the equation for the focal length of a lens is f. The equation relating focal length to the distance of the object to the lens, q, and the distance to the image, p is given with:

1/f = 1/q + 1/p

suppose that the focal length is 20cm. What is the rate of change of q with respect to p when p = 10?

dq/dp = ?


Even if someone could help me set this up it would amazing. I've tried putting f in terms of p and p in terms of f... then differentiate and put in values ... but I get the wrong answer.
Click to expand...

Please show your work - indicating exactly where you are stuck - so that we know where to begin to help you.
 
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 I can't use "q" in my answer. It is supposed to be numeric only. This is why I am lost. I figured that I must not even be setting this up correctly.
 
littlejodo said:
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 <<<< Your problem is here

1/p = 1/f - 1/q

-1/p^2 * (dp/dq) = 1/q^2

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

when p = 10, f = 20 ---> q = -20

So at p =10 (& f = 20)

dp/dq = ????

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 I can't use "q" in my answer. It is supposed to be numeric only. This is why I am lost. I figured that I must not even be setting this up correctly.
Click to expand...
 
unfortunately the answer is not -20, I just submitted it online and it said "incorrect".

Is that what you meant? the answer is -20?

The question is asking for dq/dp not dp/dq. I know it is confusing... at least, I'm completely confused.
 
littlejodo said:
unfortunately the answer is not -20, I just submitted it online and it said "incorrect".

Is that what you meant? the answer is -20? <<< No I left that for you to calculate!! READ carefully.

The question is asking for dq/dp not dp/dq. <<<< That really does not matter - it is the same procedure.


I know it is confusing... at least, I'm completely confused.
Click to expand...
 
littlejodo said:
I've done several related rate problems successfully but this one has me stumped!! I keep getting wrong answers.

the equation for the focal length of a lens is f. The equation relating focal length to the distance of the object to the lens, q, and the distance to the image, p is given with:

1/f = 1/q + 1/p

suppose that the focal length is 20cm. What is the rate of change of q with respect to p when p = 10?

dq/dp = ?


Even if someone could help me set this up it would amazing. I've tried putting f in terms of p and p in terms of f... then differentiate and put in values ... but I get the wrong answer.
Click to expand...

1/20 = 1/q + 1/p < use this two ways:

First: Differentiate implicitly

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

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

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

Now put p = 10.

Since

1/20 = 1/q + 1/p,

you can solve for q in terms of p=10:

1/20 = 1/q + 1/10,

and put that in.
 
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