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Unformatted text preview: on the homework. Problem 10.2.14: Find the points on the parametric curve x ( t ) = t + ln t, y ( t ) = tln t where the curve is concave up ( y ±± > 0). Solution: dy dx = 11 t 1 + 1 t = t1 t + 1 . So d dx ± dy dx ¶ = d dt ± t1 t + 1 ¶ d dt ( t + ln t ) , which simpliﬁes to 2 t ( t + 1) 3 . Now we check signs of the numerator (changes from minus to plus at t = 0) and the denominator (changes from minus to plus at t =1) to see that the curve is concave up at the points where t <1 and where t > ....
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 Summer '05
 Dodson
 Calculus, Arc Length, Derivative, Polar Coordinates, Mathematical analysis, Parametric equation, Vectorvalued function

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