# Math Help - Weird calculus problem involving

1. ## Weird calculus problem involving

Okay, so I have this question that has been killing me. This is it:

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=4 and f=1 ?

So I've tried about a million approaches and I never get the right answer. I know you're supposed to show how you've attempted this problem, but I've tried so many ways I don't know which one to type on here. Basically, I'm hoping somebody will be able to give me a boot in the right direction because I actually haven't a clue as to how to even begin attacking this.
Any hints?
Thanks!

2. We are given $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$, where

f=1 and q=4. We must find dp/dt.

$\frac{1}{p}+\frac{1}{q}=\underbrace{1}_{\text{beca use f=1}}$

Differentiate wrt time:

$\frac{-1}{p^{2}}\cdot \frac{dp}{dt}-\frac{1}{q^{2}}\cdot \frac{dq}{dt}=0$

Solve for dp/dt:

$\frac{dp}{dt}=\frac{-1}{q^{2}}\cdot \frac{dq}{dt}\cdot p^{2}$

We know q=4 and we can find p by $1=\frac{1}{p}+\frac{1}{4}$

$p=\frac{4}{3}$

$\frac{dp}{dt}=\frac{-1}{16}\cdot \frac{dq}{dt}\cdot \frac{16}{9}=\frac{-1}{9}\cdot \frac{dq}{dt}$

We do not know dq/dt. Were you given that?. Assuming I read the problem correctly.

3. No, we don't have that information. I can't seem to figure out how to figure it out without anything such as dq/dt or df/dt. I can't find a way around this lack of info...
Thanks for the help though

4. So I just figured it out. In case anybody is curious, here's how it's done:
Use the steps suggested by galactus:

"f=1 and q=4. We must find dp/dt.

"

Then differentiate with respect to q.
1/p=-(1/q)+1
(-1/(p^2))(dp/dq)=1/(q^2)
dp/dq=(1/q^2))/(-1/(p^2))
Simplifies to dp/dq=-(p^2)/(q^2)

if we solve

for p, we end up with p=[1/(1-(1/q))]

plug that into
dp/dq=-(p^2)/(q^2)
for p and plug in q=4 and we end up with (-1/9)

Yay!
Thanks for your help. You basis was vital in helping me figure this out. Sooo happy!!

5. df/dt = 0 because it is not changing.

As we can see, dq/dt is 9 times dp/dt.

EDIT: yes, good approach. I just differentiated wrt time. The problem said q. My bad. Anyway, it leads to the same result.