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
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 . If the new radius, such that the area is increased by b square units, is r' then we have . You want to find r'- 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
r = initial radius (5)
d = area difference (~373.85)
x = radius increase (?)
p(r + x)^2 - p(r^2) = d
Solve resulting quadratic for x : you'll get x = 7.
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
$A = \pi r^2$
$A+b = \pi(r+\Delta r)^2$
$\dfrac{A+b}{\pi} = (r+\Delta r)^2$
$\dfrac{A+b}{\pi} = r^2 + 2r \cdot \Delta r + (\Delta r)^2$
$0 = (\Delta r)^2 + 2r \cdot \Delta r + r^2 - \dfrac{A+b}{\pi}$
$0 = (\Delta r)^2 + 2r \cdot \Delta r + r^2 - \dfrac{A+b}{\pi}$
$\Delta r = \dfrac{-2r + \sqrt{(2r)^2 - 4\left(r^2 - \dfrac{A+b}{\pi}\right)}}{2}$
$\Delta r = \dfrac{-2r + \sqrt{4r^2 - 4r^2 + \dfrac{4(A+b)}{\pi}}}{2}$
$\Delta r = \sqrt{\dfrac{(A+b)}{\pi}}-r$