If the radius of a circle changes from 1.00in. to 1.01in. estimate the change in the area.
Would the answer be .03 in^2?
Nope.
A = πr^2
dA = 2πr dr
I'm not understanding
You know that dA/dr = 2πr, right?
dA/dr is in fact a fraction (remember the difference quotient?)
That gives the formula I cited.
Think of it this way. The circumference of the circle is just a line segment of length 2πr.
So if r increases by a small amount (dr), the area of the circle increases by dA = the area of the rectangle of length 2πr and width dr
lol I dont I just can't seem to wrap my head around it. Can you start the problem off for me maybe?
start it off? I did it all for you.
Consider the circle. It is a disc with a circumference C.
If you wrap a thin ribbon around it, of thickness dr, then the area of that ribbon, if unwound is just length*width = C*dr
So the area of the circle has just increased by C*dr = 2πr dr
As the circle gets larger, its area grows faster and faster, as longer and longer ribbons have to be used.
ok so I got an asnwer of .06 in ^2
i did
.01(2pi)
.02(pi) in ^2
= .06 in ^2
To estimate the change in area when the radius of a circle changes, we can use the formula for the area of a circle:
A = π * r^2
Given that the radius changes from 1.00in to 1.01in, we can calculate the areas of two circles with these radii.
For the initial circle with radius 1.00in:
A1 = π * (1.00in)^2
For the final circle with radius 1.01in:
A2 = π * (1.01in)^2
To estimate the change in area, we subtract the initial area from the final area:
ΔA = A2 - A1 = π * (1.01in)^2 - π * (1.00in)^2
Calculating this expression, we find:
ΔA ≈ 3.14 * (1.01in)^2 - 3.14 * (1.00in)^2
Simplifying further:
ΔA ≈ 3.14 * (1.0201in^2) - 3.14 * (1.0000in^2)
ΔA ≈ 3.196314in^2 - 3.140000in^2
ΔA ≈ 0.056314in^2
Therefore, the estimated change in area is approximately 0.056in^2, not 0.03in^2.