Homework Help: Rate of Change for the area of the circle

1. Oct 2, 2008

Johny 5

1. The problem statement, all variables and given/known data
The radius of a circle is decreasing at a constant rate of 0.1 centimeter per second. In terms of the circumference C, what is the rate of change of the area of the circle, in square centimeters per second?
A. -(0.2)piC
B. -(0.1)C
C. - {(0.1)C}/(2pi)
D (0.1)^2C
E. (0.1)^2piC

EDIT: i changed answer b from -(0.2)C to -(0.1)C i wrote it wrong.
EDIT: i changed the decreasing at a constant rate of 0.2 to 0.1.
2. Relevant equations

3. The attempt at a solution

Last edited: Oct 2, 2008
2. Oct 2, 2008

Rake-MC

I'm not going to answer your question right off the bat, because then you wouldn't learn anything. This is almost identical to the previous question you asked. Here is a starter:

$$\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt}$$

dr/dt = -0.2

You can do the rest yourself.

3. Oct 2, 2008

Johny 5

im guessing dr/dt is not -0.2 instead its -0.1
so:
dA/dt = dA/dr * dr/dt
dA/dt = dA/dr * -(0.1)

C = 2(pi)r
r = C/2pi

dA/dt = C/2pi * -(0.1)
dA/dt = - {(0.1)C/2pi} ?
so its C?

4. Oct 2, 2008

Rake-MC

Looks good to me, you can clean it up by writing it as $$-\frac{C}{20 \pi}$$
See? You did it entirely yourself.

5. Oct 2, 2008

Johny 5

thanks you thnk i could ask another question within' the same topic because i don't think i should start another topic..

6. Oct 2, 2008

Johny 5

1. The problem statement, all variables and given/known
if d/dx{f(x)} = g(x) and if h(x) = x^2, then d/dx{f(hx))}

2. Relevant equations

3. The attempt at a solution
f'(x) = g(x)
d/dx [f(h(x))] = g(h(x))
h(x) = x^2
so is this correct?
d/dx [f(h(x))] = [f'(h(x))] (h'(x)) = g(x^2)(2x) = 2xg(x^2)

Last edited: Oct 2, 2008
7. Oct 2, 2008

Rake-MC

No that is not correct.

Use the chain rule again, $$\frac{d}{dx}[f(h(x))] = f'[h(x)] h'(x)$$

8. Oct 2, 2008

Johny 5

yea i edited my post after reading that in my book :p thanks :)

9. Oct 2, 2008

Johny 5

i think we made a mistake because we took out 2pi that was on the top...
dA/dt = 2piC/2pi * -(0.1)
dA/dt = -(0.1)C
which is option B
can anyone comfirm this?...
i had a simular problem that r = 10 and i left 2pi on the top so im assuming i have to leave 2pi on top for this instance too...

10. Oct 2, 2008

Rake-MC

Yes. I have no idea how I missed that. You're right.

The reasoning behind it is that dr/dt = -0.1, dA/dr = 2pi*r
somehow I didn't pick up that you let dA/dr = C/2pi
thus making your dA/dt = C/2pi * -0.1
Obviously it should be 2pi*r*-0.1 which of course is -0.1C
How clumsy of me.