Nope.
A = πr^2
dA = 2πr dr
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?
6 answers
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
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.
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
i did
.01(2pi)
.02(pi) in ^2
= .06 in ^2
4 answers