The radius of a circle is r units. By how many units should the radius be increased so that the area increases by b square units?

I think the area of a circle formula is needed.

A = (pi)r^2

How do I get started?

The radius of a circle is r units. By how many units should the radius be increased so that the area increases by b square units?

I think the area of a circle formula is needed.

A = (pi)r^2

How do I get started?
You're getting kind of comfortable in asking for help before you start the problem. Let's try this: You have the area formula and you want to increase it by b units. So the new area is A + b, right? Can you go from there?

In the future, please let us know what work you've been able to do, or at least give some indication of what you have tried.

-Dan

Originally Posted by topsquark
You're getting kind of comfortable in asking for help before you start the problem. Let's try this: You have the area formula and you want to increase it by b units. So the new area is A + b, right? Can you go from there?

In the future, please let us know what work you've been able to do, or at least give some indication of what you have tried.

-Dan
Do I also increase the radius by b units?

(A + b) = pi(r + b)^2

Do I now solve for r?

No, the area is to increase by b square units. That does NOT mean the radius will increase by b units. If we take the original radius t be r then we have $\displaystyle A= \pi r^2$. If the new radius, such that the area is increased by b square units, is r' then we have $\displaystyle A+ b= \pi r'^2$. You want to find r'- r.

Originally Posted by HallsofIvy
No, the area is to increase by b square units. That does NOT mean the radius will increase by b units. If we take the original radius t be r then we have $\displaystyle A= \pi r^2$. If the new radius, such that the area is increased by b square units, is r' then we have $\displaystyle A+ b= \pi r'^2$. You want to find r'- r.
Are you saying to take the derivative of r?

Are you saying to take the derivative of r?
NO!.
You know, you should have made up a problem yourself:
as example, take a circle radius 5 and a circle radius 12;
then the radius difference is 7.
Calculate the area of both circles, then work backward
with the area difference and the radius increase:
means you'll be able to "see" what's going on, instead of
making wild guesses backed by no work!

Using the above example:
p = pi
d = area difference (~373.85)

p(r + x)^2 - p(r^2) = d
Solve resulting quadratic for x : you'll get x = 7.

Originally Posted by DenisB
NO!.
You know, you should have made up a problem yourself:
as example, take a circle radius 5 and a circle radius 12;
then the radius difference is 7.
Calculate the area of both circles, then work backward
with the area difference and the radius increase:
means you'll be able to "see" what's going on, instead of
making wild guesses backed by no work!

Using the above example:
p = pi
d = area difference (~373.85)

p(r + x)^2 - p(r^2) = d
Solve resulting quadratic for x : you'll get x = 7.
What's with the NO!? Please, skip my questions. Are you that upset with my questions?

What's with the NO!? Please, skip my questions. Are you that upset with my questions?
So you're allowed to spout off at us when you don't like our answers but no one is allowed to spout off at you when they feel you aren't paying attention.

I see.

Originally Posted by romsek
So you're allowed to spout off at us when you don't like our answers but no one is allowed to spout off at you when they feel you aren't paying attention.

I see.
I am doing the best I can to understand everyone who helps me here.

The radius of a circle is r units. By how many units should the radius be increased so that the area increases by b square units?

Let $\Delta r =$ radius increase

$A = \pi r^2$

$A+b = \pi(r+\Delta r)^2$

$\dfrac{A+b}{\pi} = (r+\Delta r)^2$

$\dfrac{A+b}{\pi} = r^2 + 2r \cdot \Delta r + (\Delta r)^2$

$0 = (\Delta r)^2 + 2r \cdot \Delta r + r^2 - \dfrac{A+b}{\pi}$

$0 = (\Delta r)^2 + 2r \cdot \Delta r + r^2 - \dfrac{A+b}{\pi}$

$\Delta r = \dfrac{-2r + \sqrt{(2r)^2 - 4\left(r^2 - \dfrac{A+b}{\pi}\right)}}{2}$

$\Delta r = \dfrac{-2r + \sqrt{4r^2 - 4r^2 + \dfrac{4(A+b)}{\pi}}}{2}$

$\Delta r = \sqrt{\dfrac{(A+b)}{\pi}}-r$

$A=\pi r^2$

$A+b=\pi R^2 \implies R= \sqrt{\dfrac{A+b}{\pi}}$

$\Delta r = R-r = \sqrt{\dfrac{A+b}{\pi}} - r$

edit ... took the LONG road to Dallas the first time.

Originally Posted by skeeter
$A=\pi r^2$

$A+b=\pi R^2 \implies R= \sqrt{\dfrac{A+b}{\pi}}$

$\Delta r = R-r = \sqrt{\dfrac{A+b}{\pi}} - r$

edit ... took the LONG road to Dallas the first time.
Very clear calculation.