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C4 Rates of Change question

Can only help me understand what I'm supposed to be doing in this question:

"At a given instant, the radii of two concentric circles are 8cm and 12cm. The radius of the outer circle is increasing at a rate of 1cm/s and the radius of the inner circle is increasing at a rate of 2cm/s.

Find the rate of change of the area enclosed by the two circles."

Thanks!

Reply 1

KD35

This is tough! is it from a solomon paper?

What I'd do is find an expression for the Area enclosed in terms of the radius. Find DA/DR
You have DR/DT so DR/DT*DA/DR

Reply 2

JerzyDudek

Where did you get this question? Sounds interesting. I would find the change in the area of each circle and then combine them but there might be a better method.

Reply 3

Chris-69

It's from my text book.

The answer ends up being "decrease 8π cm^2" but I'm not really of the method behind it. You basically have to find how the change in area between the big circle and small circle.

Reply 4

JerzyDudek

But you have

\dfrac{dr_1}{dt}

and

\dfrac{dr_2}{dt}

. I suppose you can just substract one from another, that would work.

Reply 5

JerzyDudek

Isn't it -8π? So anyway. You've got two radii and two areas.

A = \pi r^2

, so you can find

\dfrac{dA_1}{dr_1}

and

\dfrac{dA_2}{dr_2}

. You have got

\dfrac{dr_1}{dt}

(that's just how fast a radius is changing) and

\dfrac{dr_2}{dt}

. Then you can find how fast each area changes (that is

\dfrac{dA_1}{dt}

and

\dfrac{dA_2}{dt}

) using chain rule. Lastly, just substract one derivative from another.

(edited 10 years ago)

Reply 6

Chris-69

Original post by JerzyDudek

But you have

\dfrac{dr_1}{dt}

and

\dfrac{dr_2}{dt}

. I suppose you can just substract one from another, that would work.

Yeah that works, thanks!

I tried that originally but I multiplied the inner circle by 1 instead of 2 (basically used the rate of increase of the larger circle) and was getting the wrong answer. Did it again right and it worked haha!

Was a good question, though; haven't seen anything like it in a past paper.

Reply 7

FireGarden

I get the answer

A'(t) = -6\pi t + c

where c is some constant decided by the initial values of the two circles. I don't really know what to continue on with, if im honest.

I can post my working later if anyone would like it

(edited 10 years ago)

Reply 8

Chris-69

Original post by JerzyDudek

Isn't it -8π? So anyway. You've got two radii and two areas.

A = \pi r^2

, so you can find

\dfrac{dA_1}{dr_1}

and

\dfrac{dA_2}{dr_2}

. You have got

\dfrac{dr_1}{dt}

(that's just how fast a radius is changing) and

\dfrac{dr_2}{dt}

. Then you can find how fast each area changes (that is

\dfrac{dA_1}{dt}

and

\dfrac{dA_2}{dt}

) using chain rule. Lastly, just substract one derivative from another.

One issue; how would I know which to subtract from which? I ended up with

\dfrac{dA_1}{dt_1}

= 24π and

\dfrac{dA_2}{dt_2}

= 16π

Original post by FireGarden

I get the answer

A'(t) = -6\pi t + c

where c is some constant decided by the initial values of the two circles. I don't really know what to continue on with, if im honest.

I can post my working later if anyone would like it

That's seems way too complex for me to understand lol Is that a C4 method? And sure, post your working out and I can see if I understand :tongue:

(edited 10 years ago)

Reply 9

JerzyDudek

Original post by Chris-69

One issue; how would I know which to subtract from which? I ended up with

\dfrac{dA_1}{dt_1}

= 24π and

\dfrac{dA_2}{dt_2}

= 16π

The area of the region you need in the area of the big circle minus the are of the small one. So similarly you substract the rate of change of the small one from the big one. That's why I got a negative answer. It seems logical that if the inner circle "grows" at a higher rate, then the area between two circles will decrease.

Original post by FireGarden

I get the answer

A'(t) = -6\pi t + c

where c is some constant decided by the initial values of the two circles. I don't really know what to continue on with, if im honest.

I can post my working later if anyone would like it

Did you do diff. equations or something? Please post your working.

(edited 10 years ago)

Reply 10

Chris-69

Original post by JerzyDudek

The area of the region you need in the area of the big circle minus the are of the small one. So similarly you substract the rate of change of the small one from the big one. That's why I got a negative answer. It seems logical that if the inner circle "grows" at higher rate, than the area between two circles will decrease.

Ah yes, I see. Thanks.