To see the graph of the parametric equations x 5 t 6

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To see the graph of the parametric equations x = (5 t 6 17 t 3 + 13 y ) 1 / 5 , y = t, we executed the Mathematica command ParametricPlot[ { h[t], t } , { t, 1.3, 1.6 } ]; with the result shown next. -1 -0.5 0.5 1 1.5 2 x -1 -0.5 0.5 1 1.5 y Next we found that f ( t ) = 30 t 5 51 t 2 + 13 5(5 t 6 17 t 3 + 13 t ) 4 / 5 and f ( t ) = 2(75 t 10 1530 t 7 + 3315 t 5 867 t 4 663 t 2 338) 25(5 t 6 17 t 3 + 13 t ) 9 / 5 . It follows with the aid of Newton’s method that f ( t ) = 0 when t ≈ − 0 . 488, when t 0 . 528, and when t 1 . 104. Hence the graph of the equation x 5 = 5 y 6 17 y 3 + 13 y has vertical tangents at the three points ( 1 . 338784051 , 0 . 488418117) , (0 . 80783532 , 1 . 103631606) , and (1 . 349152308 , 0 . 528310640) . The horizontal tangents will occur when the denominator in f ( t ) = 0; that is, at (0 , 0 . 812678591) and at (0 , 0) 994
(numbers with decimal points are approximations). Finally, Newton’s method yields the zeros of the numer- ator of f ( t ) and—also checking the zeros of its denominator—we find that the graph has inflection points at (0 , 0 . 813073457) , (2 . 513563956 , 1 . 516333702) , (0 , 0) , (0 . 992940442 , 1 . 004409592) , (1 . 185188240 , 1 . 258427457) , (3 . 735641965 , 2 . 388360391) . The second and last of these don’t appear on the preceding graph, but the graph appears to be a straight line in their vicinity, so showing more of the graph isn’t much use. Section 9.5 9.5.1: The area is A = 1 1 (2 t 2 + 1)(3 t 2 ) dt = 1 1 (6 t 4 + 3 t 2 ) dt = 6 5 t 5 + t 3 1 1 = 2 · 6 5 + 1 = 22 5 . 9.5.2: The area is A = ln 2 0 ( e t )(3 e 3 t ) dt = ln 2 0 3 e 2 t dt = 3 2 e 2 t ln 2 0 = 6 3 2 = 9 2 . 9.5.3: The area is A = π 0 sin 3 t dt = π 0 ( sin t cos 2 t sin t ) dt = 1 3 cos 3 t cos t π 0 = 4 3 . 9.5.4: The area is A = 1 0 3 e 2 t dt = 3 2 e 2 t 1 0 = 3 2 ( e 2 1) 9 . 5835841484. 9.5.5: The area is A = π 0 e t sin t dt = 1 2 e t (sin t cos t ) π 0 = 1 2 ( e π + 1) 12 . 0703463164 . See Example 5 of Section 7.3 for the evaluation of the antiderivative using integration by parts. 9.5.6: The area is A = 1 0 (2 t + 1) e t dt = (2 t 1) e t 1 0 = e ( 1) = e + 1 3 . 718281828459 . 995
See Example 3 of Section 7.3 for the evaluation of the antiderivative using integration by parts. 9.5.7: The volume is V = 1 1 π (2 t 2 + 1) 2 · 3 t 2 dt = π 1 1 (12 t 6 + 12 t 4 + 3 t 2 ) dt = π 12 7 t 7 + 12 5 t 5 + t 3 1 1 = 179 35 + 179 35 π = 358 35 π 32 . 1340048567 . 9.5.8: The volume is V = ln 2 0 π ( e 2 t ) · 3 e 3 t dt = π ln 2 0 3 e t dt = 3 π e t ln 2 0 = 3 π 9 . 424777960769 . 9.5.9: The volume is V = π 0 π (sin t ) 5 dt = π π 0 (1 2 cos 2 t + cos 4 t ) sin t dt = π 1 5 cos 5 t + 2 3 cos 3 t cos t π 0 = π 8 15 + 8 15 = 16 15 π 3 . 351032163829 . 9.5.10: The volume is V = π π 0 e 2 t sin t dt = π 1 5 e 2 t (2 sin t cos t ) π 0 = π 5 e 2 π + π 5 = π 5 ( e 2 π + 1 ) 337 . 087648741765 . See Example 5 of Section 7.3 for the technique of finding the antiderivative using integration by parts. 9.5.11: The arc-length element is ds = ( t + 4) 1 / 2 dt . Hence the length of the curve is L = 12 5 ( t + 4) 1 / 2 dt = 2 3 ( t + 4) 3 / 2 12 5 = 128 3 18 = 74 3 24 . 6666666667 . 9.5.12: The arc-length element is ds = ( t 2 + t 4 ) 1 / 2 dt = t ( t 2 + 1) 1 / 2 dt . Thus the length of the curve is L = 1 0 t ( t 2 + 1) 1 / 2 dt = 1 3 ( t 2 + 1) 3 / 2 1 0 = 2 2 1 3 0 . 6094757082 . 9.5.13: The arc-length element is ds = (cos t + sin t ) 2 + (cos t sin t ) 2 dt = 2 dt . Therefore the length of the curve is L = π/ 2 π/ 4 2 dt = t 2 π/ 2 π/ 4 = π 2 4 1 . 1107207345 .

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