Q. I forgot the formula. If anyone has the correct formulas for area and perimeter please answer.
A. The area of a circle is:
(pi)(r)^2
Circumference (the "perimeter" of a circle) is:
2(pi)(r)
http://en.wikipedia.org/wiki/Circle#Length_of_circumference
(pi)(r)^2
Circumference (the "perimeter" of a circle) is:
2(pi)(r)
http://en.wikipedia.org/wiki/Circle#Length_of_circumference
How do i find the arc length of the following function?
Q. Polar Function
r=a*sin(theta) where a>0
Formula for finding arc length of polar function
L=int( sqrt( f(theta)^2 + f '(theta)^2), theta, alpha, beta) where alpha<=theta<=beta
Answer is a*pi
How do i solve this problem? I have no clue how to find a*pi as the answer.
r=a*sin(theta) where a>0
Formula for finding arc length of polar function
L=int( sqrt( f(theta)^2 + f '(theta)^2), theta, alpha, beta) where alpha<=theta<=beta
Answer is a*pi
How do i solve this problem? I have no clue how to find a*pi as the answer.
A. The graph of r = asinÎ is a circle of radius a/2 centered at (0, a/2). The curve is traversed exactly once by allowing 0 ⤠Π< Ï. (Note that the arclength is the circumference 2Ï(a/2) = aÏ.)
f ² (Î) = a²sin²Πand [f '(Î)]² = a² cos²Î.
So â(r² + (dr/dÎ)²) = â[a²(sin²Π+ cos²Î)] = â(a²) = |a|. Since a > 0, the integrand is just a.
So your arclength is
Ï
â« a dÎ = aÏ.
0
This is about as nice an arclength problem as you can ever expect to see.
f ² (Î) = a²sin²Πand [f '(Î)]² = a² cos²Î.
So â(r² + (dr/dÎ)²) = â[a²(sin²Π+ cos²Î)] = â(a²) = |a|. Since a > 0, the integrand is just a.
So your arclength is
Ï
â« a dÎ = aÏ.
0
This is about as nice an arclength problem as you can ever expect to see.
How do you find the area of a circle with the middle as square shape?
Q. Ok, so you have to find the area of the shaded shape which is the circle.The circle has a radius of 4cm and the squares' (which is in the middle of the circle) length is 4x4. Please help me, It's due for Monday.
A. Use the formula Ïr^2 to find the circle.
Ï x (4)^2 = 50.27
The area of the square can be found by 4 x 4 = 16
subtract both of them to find the shaded area.
50.27 - 16 = 34.27
Ï x (4)^2 = 50.27
The area of the square can be found by 4 x 4 = 16
subtract both of them to find the shaded area.
50.27 - 16 = 34.27
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