Rate of Change for the area of the circle

[Johny 5]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[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.

 

[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?

 

[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.

 

[Johny 5]

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

 

[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. Homework 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)

 

[Rake-MC]

No that is not correct.

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

 

[Johny 5]

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

 

[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?

i think we made a mistake because we took out 2pi that was on the top...
so the answer should be
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 I am assuming i have to leave 2pi on top for this instance too...

 

[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.