Derivative Question --> Rate of Change with focal length formula

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October 16th 2009, 04:04 PM
s3a

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Derivative Question --> Rate of Change with focal length formula

Question:

If http://gauss.vaniercollege.qc.ca/web...faee426371.png is the focal length of a convex lens and an object is placed at a distance http://gauss.vaniercollege.qc.ca/web...da050b4dd1.png from the lens, then its image will be at a distance http://gauss.vaniercollege.qc.ca/web...12a8ea5921.png from the lens, where http://gauss.vaniercollege.qc.ca/web...faee426371.png, http://gauss.vaniercollege.qc.ca/web...da050b4dd1.png, and http://gauss.vaniercollege.qc.ca/web...12a8ea5921.png are related by the lens equation
http://gauss.vaniercollege.qc.ca/web...be8d879c61.png
What is the rate of change of http://gauss.vaniercollege.qc.ca/web...da050b4dd1.png with respect to http://gauss.vaniercollege.qc.ca/web...12a8ea5921.png if http://gauss.vaniercollege.qc.ca/web...b139a78e91.png and http://gauss.vaniercollege.qc.ca/web...8c532d6091.png?

(My work is attached)

Any help would be greatly appreciated!
Thanks in advance!
October 16th 2009, 04:17 PM
rn443

Quote:

Originally Posted by s3a

Question:

If http://gauss.vaniercollege.qc.ca/web...faee426371.png is the focal length of a convex lens and an object is placed at a distance http://gauss.vaniercollege.qc.ca/web...da050b4dd1.png from the lens, then its image will be at a distance http://gauss.vaniercollege.qc.ca/web...12a8ea5921.png from the lens, where http://gauss.vaniercollege.qc.ca/web...faee426371.png, http://gauss.vaniercollege.qc.ca/web...da050b4dd1.png, and http://gauss.vaniercollege.qc.ca/web...12a8ea5921.png are related by the lens equation
http://gauss.vaniercollege.qc.ca/web...be8d879c61.png
What is the rate of change of http://gauss.vaniercollege.qc.ca/web...da050b4dd1.png with respect to http://gauss.vaniercollege.qc.ca/web...12a8ea5921.png if http://gauss.vaniercollege.qc.ca/web...b139a78e91.png and http://gauss.vaniercollege.qc.ca/web...8c532d6091.png?

(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.
October 18th 2009, 10:31 AM
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 :(.
October 18th 2009, 07:09 PM
rn443

Quote:

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.