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Area under a curve Suppose the function y = h(x) is nonnegative and continuous on [α, β], which implies that the area bounded by the graph of h and x-axis on [α, β] equals
103. Find the area under one arch of the cycloid x = 3(t − sin t), y = 3(1 − cos t) (see Example 5).
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Calculus: Early Transcendentals (3rd Edition)
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- Write parametric equations for a cycloid traced by a point P on a circle of radius a as the circle rolls along the x -axis given that P is at a maximum when x=0.arrow_forwardDIFFERENTIATION OF PARAMETRIC EQUATIONS d²y 1.) of y = sinInß and x = In(Inß) dx2 и 2.) dx2 of y = costu – sin*u and x = 2cos? 2 3.) of s = and t = k-1 k+1 d?x 4.) of x = (02 – 1)2 and y = 403 d²y 5.) dx2 of x = 1- Int and y t-Int %D 2arrow_forwardA curve Cis given by the parametric equations z= 2t –t2 and y = (t- 2)* wheret > 0. 1. The curve has a vertical tangent line at the point O(0,-2) O(1,0) O(0, -1) O(0, 4) 2. Is the point (-8, 8) lies on the curve C? O(0,0) O(1,-1) OYes ONOarrow_forward
- _1. The Cartesian equation of the curve given in parametric equations x = e' – e-t , y = e' + e¬t is (a) x² – y² = 4. (c) x² – y² = 1. (d) y² – x² = 1. (b) y² – x² = 4. _2. The equation of the tangent line to the curve x = -2t² + 3, y = t³ – 8t at the point where t = -2 is (a) x – 2y +7 = 0. (b) x + 2y – 7 = 0. (c) 2x – y + 7 = 0. (d) None of these. _3. The slope m of the tangent line at any point on the curve r = 2 cos 0 is (b) – cot 20. (a) – tan 20. (c) tan 20. (d) None of these. _4. The graph of the equation r² = 8 cos20 is a (a) limacon (c) circle (b) cardioid. (d) lemniscate. _5. The distance between the points A(-2, –3,1) and B(6,9,-3) is (b) 2/14 units . (a) 4v14 units . (c) 8/14 units . (d) 4/7 units.arrow_forwardSecond derivative Assume a curve is given by the parametric equations x = f(t) and y = g(t), where f and g are twice differ- entiable. Use the Chain Rule to show that f'(t)g"(t) – g'(t)f"(t) y"(x) = (f'(1))³arrow_forwardFind the values of h, k, and a that make the circle (x - h)2 + (y - k)2 = a2 tangent to the parabola y = x2 + 1 at the point (1, 2) and that also make the second derivatives d2y/dx2 have the same value on both curves there. Circles like this one that are tangent to a curve and have the same second derivative as the curve at the point of tangency are called osculating circles.arrow_forward
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