Example find the area between one loop of the cycloid

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Calculus: Early Transcendental Functions
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Chapter 13 / Exercise 53
Calculus: Early Transcendental Functions
Edwards/Larson
Expert Verified
Example Find the area between one loop of the cycloid x = θ - sin( θ ) , y = 1 - cos( θ ) and the x -axis. One loop means from cusp to cusp, so we will take 0 θ 2 π . The area can be written 2 π x =0 ydx where we are thinking of y as a function of x . So after substituting for x we get 2 π x =0 ydx = 2 π θ =0 y dx = 2 π θ =0 1 - cos( θ ) 1 - cos( θ ) = 2 π θ =0 1 - 2 cos( θ ) + cos( θ ) 2 = 2 π θ =0 1 - 2 cos( θ ) + 1 + cos(2 θ ) 2 = 2 π θ =0 3 2 - 2 cos( θ ) + cos(2 θ ) 2 = 3 2 θ - 2 sin( θ ) + sin(2 θ ) 4 2 π 0 = 3 π. For arclength problems, we use ds = 1 + dy dx 2 dx = 1 + dy dt dx dt 2 dx dt dt = dx dt 2 + dy dt 2 dt Example Find the arclength of one loop of the cycloid x = θ - sin( θ ) , y = 1 - cos( θ ) 72
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Calculus: Early Transcendental Functions
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Chapter 13 / Exercise 53
Calculus: Early Transcendental Functions
Edwards/Larson
Expert Verified
This is 2 π θ =0 ds = 2 π θ =0 1 - cos( θ ) 2 + sin( θ ) 2 = 2 π θ =0 1 - 2 cos( θ ) + cos( θ ) 2 + sin( θ ) 2 = 2 π θ =0 2 - 2 cos( θ ) = 2 π θ =0 4 sin θ 2 2 = 2 2 π θ =0 sin θ 2 = 2 2 π θ =0 sin θ 2 = - 4 cos θ 2 2 π θ =0 = 8 . Example Find the area inside the astroid x = a cos( θ ) 3 , y = a sin( θ ) 3 . Figure 35. The astroid curve 73
The curve is traversed once anticlockwise for 0 θ 2 π . We will work out the area of the piece in the first quadrant (shown in blue) and multiply by 4 . So the area is 4 a x =0 ydx = 4 0 θ = π 2 a sin( θ ) 3 dx = 4 a 0 θ = π 2 sin( θ ) 3 3 a cos( θ ) 2 ( - sin( θ ) = 12 a 2 π 2 θ =0 sin( θ ) 4 cos( θ ) 2 = 12 a 2 π 2 θ =0 1 - cos(2 θ ) 2 2 1 + cos(2 θ ) 2 = 3 2 a 2 π 2 θ =0 1 - cos(2 θ ) - cos(2 θ ) 2 + cos(2 θ ) 3 = 3 2 a 2 π 2 θ =0 1 - cos(2 θ ) 2 since cos(2( π 2 - θ )) = - cos(2 θ ) the odd power term integrate to zero, = 3 2 a 2 π 2 θ =0 1 - 1 + cos(4 θ ) 2 = 3 a 2 π 8 . 1.21 Still more on parametric curves - Feb 27 Example Find the area enclosed by the parametric curve x = t 2 - 2 t , y = t and the y -axis. 74
Figure 36. The parametric curve x = t 2 - 2 t , y = t , 0 t 2 . Not as shown in class! The y -axis is x = 0 , so the curve meets the y axis when t 2 - 2 t = 0 , i.e. for t = 0 and t = 2 . Persumably the relevant section of the curve is given by 0 t 2 . A quick check shows that dx dt t =0 = - 2 so the curve starts off in the second quadrant for small values of t . The desired area is then 2 t =0 ( - x ) dy = 2 t =0 ( - x ) dy dt dt = 2 t =0 (2 t - t 2 ) 1 2 t - 1 2 dt = 1 2 2 t =0 2 t 1 2 - t 3 2 dt = 2 3 t 3 2 - 1 5 t 5 2 2 t =0 = 8 2 15 Example Find the area enclosed by the parametric curve x = 1 + e t , y = t - t 2 and the x -axis. 75
Figure 37. The parametric curve x = 1 + e t , y = t - t 2 , 0 t 1 The x -axis is y = 0 and the curve meets this for t = 0 and t = 1 , so we expect the relevant range to be 0 t 1 . When t = 0 we have ( x, y ) = (2 , 0) and when t = 1 we have ( x, y ) = (1 + e, 0) . Also y 0 for 0 t 1 , so the desired area is 1 t =0 ydx = 1 t =0 y dx dt dt = 1 t =0 ( t - t 2 ) e t dt We give two ways of finding the integral. First, using integration by parts. 1 t =0 u ( t - t 2 ) dv e t dt = ( t - t 2 ) e t 1 t =0 - 1 t =0 v e t du (1 - 2 t ) dt and integrating by parts again = 1 t =0 u (2 t - 1) dv e t dt = (2 t - 1) e t 1 t =0 - 1 t =0 v e t du 2 dt 76
= e + 1 - 2 e t 1 t =0 = e + 1 - 2 e + 2 = 3 - e The other method is the method of undetermined coefficients and depends on knowing at the outset that ( t - t 2 ) e t dt = ( At 2 + Bt + C ) e t + constant of integration for some constants A , B and C . Differentiating gives

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