MTH 162 Midterm: MTH 162 University of Rochester Fall 09Exam2sol

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31 Jan 2019
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Z sec3(x) dx = sec(x) tan(x) + ln| sec(x) + tan(x)| 0 where u = e x du = e xdx where u = tan so du = sec2 d and 1 + u2 = sec . Solution: (a) we have dx dt dy dt dy dx. = 3 cos t sin2 t cos t sin2 t sin t cos2 t. The tangent line is horizontal when this derivative is 0, namely when t = 0 and t = . The tangent line is vertical when the derivative is unde ned, namely at t = /2 and t = 3 /2. Solution: (b) the slope of the tangent line is 1 when t = /4, 3 /4, 5 /4 and 7 /4. Solution: (c) at t = /4 we have x = y = 2/4 and dy/dx = 1, so the equation for the tangent line is y 2/4 x 2/4 y .

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