# Rate of Change

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.