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ishpiro | College Teacher | (Level 1) Educator

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This is the graph of `r = 1 -sin(theta)` . The start and end point is (0, 1). The curve is traversed in counterclockwise direction.

The length of this curve is

`L = int_0 ^ (2pi) sqrt(r^2 + ((dr)/(d(theta)))^2) d(theta)`

`(dr)/(d(theta)) = -cos(theta)`

The expression under the radical will be, if the Pythagorean identity is used to replace `(sin(theta))^ 2+(cos(theta))^2` with 1:

`(1 - sin(theta))^2 + (-cos(theta))^2 = 1 - 2sin(theta) + 1 `

`=2-2sin(theta) =2(1 - sin(theta))`

The next step to simplify this would be to use the half-angle identity, with sine replaced by cosine:

`2(1 - sin(theta)) = 2(1 - cos(pi/2 - theta)) = 2*2sin^2(pi/4 - theta/2)`

Then the square root of this expression is `2 sin(pi/4 - theta/2)` . Using it to replace the radical in original integral, we get

`int_0 ^ (2pi) 2sin(pi/4 - theta/2) d(theta)`

Use the substitution `u = pi/4 - theta/2` to evaluate the integral.

`du = -1/2 d(theta)`

New limits are `pi/4 ` and `pi/4 - (2pi)/2 = -(3pi)/4`

`int_(pi/4) ^ (-(3pi)/4) 2 sinu (-2 du)=4int_(-(3pi)/4) ^ (pi/4) sinu du = `

`=-4cosu |_(-(3pi)/4) ^( pi/4) = `

`-4(cos(-(3pi)/4) - cos(pi/4))= -4(-(sqrt 2)/2- (sqrt2)/2) = 4sqrt2` 

The length of the given cardioid is `4sqrt(2)` .

 

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