# Integraldisplay t 1 t 0 parenleftbig 5 4 t 5 t 5 t 2

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integraldisplay t 1 t 0 parenleftBig 5 - 4 t - 5 t + 5 t 2 - 1 t 3 parenrightBig dt = bracketleftBig 5 t - 2 t 2 - 5 ln t - 5 t + 1 2 t 2 bracketrightBig t 1 t 0 . But the graph of x ( t ) = t - 1 t , y ( t ) = 4 t + 1 t intersects y = 5 when 4 t + 1 t = 5 , i.e. , when 4 t 2 - 5 t + 1 = (4 t - 1)( t - 1) = 0 .
qazi (kaq87) – HW01 – berg – (55290) 10 Thus t 0 = 1 / 4 while t 1 = 1, so area( A ) = parenleftBig - 3 2 - 5 ln 1 parenrightBig - parenleftBig - 87 8 - 5 ln parenleftBig 1 4 parenrightBigparenrightBig . Consequently, area( A ) = 75 8 - 5 ln 4 . keywords: parametric curve, area, 016 10.0points The curve traced out by a point P on the circumference of a circle as the circle rolls along a straight line shown in P x y is called a Cycloid and the shaded region is the region, A , below an arch . If the circle has radius R , the cycloid is given parametrically by x ( t ) = R ( t - sin t ) , y ( t ) = R (1 - cos t ) . Find the area of A when R = 5. 017 10.0points Which one of the following integrals gives the length of the parametric curve x ( t ) = t 2 , y ( t ) = 2 t , 0 t 12 . 1. I = integraldisplay 12 5(1 - cos t ) dx , where x = x ( t ) = 5( t - sin t ) , dx = 5(1 - cos t ) dt . Thus after a change of variable from x to t we see that area( A ) = 25 integraldisplay 2 π 0 (1 - cos t ) 2 dt = 25 integraldisplay 2 π 0 (1 - 2 cos t + cos 2 t ) dt = 25 integraldisplay 2 π 0 parenleftBig 3 2 - 2 cos t + 1 2 cos 2 t parenrightBig dt . Consequently, area( A ) = 75 π . keywords: parametric curve, area, cycloid
qazi (kaq87) – HW01 – berg – (55290) 11 The arc length of the parametric curve ( x ( t ) , y ( t )) , a t b is given by the integral I = integraldisplay b a radicalBig ( x ( t )) 2 + ( y ( t )) 2 dt . But when x ( t ) = t 2 , y = 2 t , we see that x ( t ) = 2 t , y ( t ) = 2 . Consequently, the curve has arc length = 2 integraldisplay 12 0 radicalbig t 2 + 1 dt . 018 10.0points Which integral represents the arc length of the astroid shown in x y and given parametrically by x ( t ) = 3 cos 3 t , y ( t ) = 3 sin 3 t . 1. I = 3 integraldisplay 2 π 0 cos t sin t dt 2. I = 3 integraldisplay π 0 | cos t sin t | dt 3. I = 9 integraldisplay 2 π 0 | cos t sin t | dt correct 4. I = 3 integraldisplay 2 π 0 | cos t sin t | dt 5. I = 9 integraldisplay 2 π 0 cos t sin t dt 6. I = 9 integraldisplay π 0 | cos t sin t | dt Explanation: The arc length of the parametric curve ( x ( t ) , y ( t )) , a t b is given by the integral I = integraldisplay b a radicalBig ( x ( t )) 2 + ( y ( t )) 2 dt . But when x ( t ) = 3 cos 3 t , y ( t ) = 3 sin 3 t we see that x ( t ) = - 9 sin t cos 2 t , y ( t ) = 9 cos t sin 2 t , in which case radicalBig ( x ( t )) 2 + ( y ( t )) 2 = radicalBig (9 cos t sin t ) 2 (cos 2 t + sin 2 t ) = | 9 cos t sin t | . Consequently, I = 9 integraldisplay 2 π 0 | cos t sin t | dt . keywords: arc length, parametric curve, as- troid trig functions, definite integral