Question

If f is the focal length of a convex lens and an object is placed at a distance p from the lens, then its image will be at a distance q from the lens, where f, p, and q are related by the lens equation
1/f=1/p+1/q
What is the rate of change of p with respect to q if q=2 and f=4? (Make sure you have the correct sign for the rate.)

Answer 1

You set f=4.
The express p in terms of q, i.e. p(q)=function of q.

To get you started:

1/4=1/p+1/q, so 1/p=1/4-1/q.

Better looking: \(\frac{1}{p}=\frac{1}{4}-\frac{1}{q}\).

Now try to write the right hand side as one fraction.
After that is done, you can write p in terms of q, by "turning the fraction upside down" (take the reciprocal).

Then calculate the derivative of p with respect to q and set q=2.

Go and try!

Answer 2

the right hand side would be -1/4

Answer 3

i got -4 as my final answer

Answer 4

Me too:

\(\dfrac{1}{f}=\dfrac{1}{4}-\dfrac{1}{q}=\dfrac{q-4}{4q}\)

\(f=\dfrac{4q}{q-4}\)

\(\dfrac{df}{dq}=\dfrac{4(q-4)-4q}{(4-q)^2}=-\dfrac{16}{(q-4)^2}\)

\(\dfrac{dp}{dq}(2)=-\dfrac{16}{(2-4)^2}=-4\)