Tuesday January 15 Solutions A review of some important calculus topics 1 Chain Rule a Let h t sin cos tan t Find the derivative with respect to t Solution d d h t sin cos tan t dt dt d cos cos tan t cos tan t dt d tan t dt cos cos tan t sin tan t sec2 t cos cos tan t sin tan t b Let s x p 4 ds x where x t ln f t and f t is a differentiable function Find dt Solution From ds dt dd xs dd xt we get 1 ds f 0 t 3 4 d t 4x f t But we need to make sure that ds dt is a single variable function of f so ds 1 f 0 t d t 4 ln f t 3 4 f t 2 Parameterized curves a Describe and sketch the curve given parametrically by x 5 sin 3t 2 for 0 t y 3 cos 3t 3 What happens if we instead allow t to vary between 0 and 2 Solution Note that x 2 5 y 2 3 sin2 3t cos2 3t 1 So this parameterizes at least part of the ellipse x 2 5 y 2 3 1 By examining differing values of t in 0 t 2 we see that this parametrization travels 3 the ellipse in a clockwise fashion exactly once t 0 x 0 y 0 0 3 t 6 x 6 y 6 5 0 t 3 x 3 y 3 0 3 t 2 x 2 y 2 5 0 3 2 1 4 2 2 4 1 2 3 Figure 1 Ellipse If we let t vary between 0 and 2 we will traverse the ellipse 3 times b Set up but do not evaluate an integral that calculates the arc length of the curve described in part a Solution Arc length b Z s a Z s 0 2 3 p dx dt 2 dy dt 2 dt 15 cos 3t 2 9 sin 3t 2 d t c Consider the equation x 2 y 2 16 Graph the set of solutions of this equation in R2 and find a parametrization that traverses the curve once counterclockwise Solution If we let x 4 cos t and y 4 sin t then x 2 y 2 4 cos t 2 4 sin t 2 16 More over as t increases this parametrization traverses the circle in a counterclockwise fashion t 0 x 0 y 0 4 0 t 2 x 2 y 2 0 4 t x y 4 0 t 3 2 x 3 2 y 3 2 0 4 t 2 x 2 y 2 4 0 4 2 4 2 2 4 2 4 Figure 2 Circle To ensure that we travel the curve only once we restrict t to the interval 0 2 So the parametrization is x 4 cos t when 0 t 2 y 4 sin t 3 1st and 2nd Derivative Tests a Use the 2nd Derivative Test to classify the critical numbers of the function f x x 4 8x 2 10 Solution First we find the critical points of f x f 0 x 4x 3 16x f 0 x 0 when 4x 3 16x 4x x 2 4 4x x 2 x 2 0 Hence f 0 x 0 when x 0 x 2 or x 2 Now apply the 2nd Derivative Test to the three critical points From f x 12x 2 16 we get f 0 16 0 so y f x is concave down at the point 0 f 0 So a local max occurs at 0 10 f 2 32 0 so y f x is concave up at the point 2 f 2 So a local min occurs at 2 6 f 2 32 0 so y f x is concave up at the point 2 f 2 So a local min occurs at 2 6 b Use the 1st Derivative Test and find the extrema of h s s 4 4s 3 1 Solution First find the critical points of h s h 0 s 4s 3 12s 2 Then h 0 s 0 when 4s 3 12s 2 4s 2 s 3 0 So h 0 s 0 when s 0 and s 3 For the 1st Derivative Test we need to determine if h is increasing or decreasing on the intervals 3 3 0 and 0 On 3 choose any test point for example choose s 1000 The sign of h 0 s 4s 3 12s 2 0 on this interval Hence h s is decreasing on 3 On 3 0 choose any test point for example choose s 1 The sign of h 0 s 4s 3 12s 2 0 on this interval Hence h s is increasing on 3 0 On 0 choose any test point for example choose s 1000 The sign of h 0 s 4s 3 12s 2 0 on this interval Hence h s is increasing on 0 Since at s 3 the function changes from decreasing to increasing the function must have obtained a local min at s 3 At s 0 neither a max or a min occurs in the value of h c Explain why the 2nd Derivative test is unable to classify all the critical numbers of h s s 4 4s 3 1 Solution When s 3 h 3 36 0 A local min occurs when s 3 by the 2nd Derivative Test When s 0 h 0 0 The 2nd Derivative Test is inconclusive The graph of y h s has no concavity at 0 h 0 Without more information the 1st Derivative Test we are unable to identify 0 h 0 as a local max min or a point of inflection 4 Consider the function f x x 2 e x a Find the best linear approximation to f at x 0 Solution Recall that in Calc I and II the best linear approximation is synonymous with the equation of the tangent line or the 1st order Taylor polynomial Hence f 0 x 2xe x x 2 e x Since f 0 0 0 the tangent line has no slope at 0 f 0 0 0 The equation of the tangent line is y 0 b Compute the second order Taylor polynomial at x 0 Solution By definition the second order Taylor polynomial at x 0 is T2 x f 0 f 0 0 f 0 x 0 x 0 2 1 2 Since f x 2e x 4xe x x 2 e x we compute that f 0 2 Hence T2 x 0 2 0 x 0 x 0 2 x 2 1 2 c Explain how the second order Taylor polynomial at x 0 demonstrates that f must have a local minimum at x 0 Solution The second order Taylor polynomial is the best quadratic approximation to the curve y f x at the point 0 f 0 Since T2 x x 2 clearly has a local minimum at 0 0 and 0 0 is …
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