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UIUC MATH 241 - Worksheet_011513_sol

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Tuesday, January 15 ∗ Solutions ∗ A review of some important calculus topics1. Chain Rule:(a) Let h(t ) =sin¡cos(tan t)¢. Find the derivative with respect to t .Solution.dd t(h(t )) =dd t(sin(cos(tan t )))=cos(cos(tan t )) ·dd t(cos(tan t ))=cos(cos(tan t )) ·(−sin(tan t)) ·dd t(tan t )=cos(cos(tan t )) ·(−sin(tan t)) ·sec2t(b) Let s(x) =4px where x(t) = ln¡f (t)¢and f (t) is a differentiable function. Findd sd t.Solution. Fromd sd t=d sd x·d xd t, we getd sd t=14x3/4·f0(t )f (t).But we need to make sure thatd sd tis a single variable function of f , sod sd t=14£ln(f (t))¤3/4·f0(t )f (t).2. Parameterized curves:(a) Describe and sketch the curve given parametrically byx = 5sin(3t)y =3cos(3t )for 0 ≤ t <2π3.What happens if we instead allow t to vary between 0 and 2π?Solution. Note that³x5´2+³y3´2=sin2(3t ) +cos2(3t ) =1.So this parameterizes (at least part of) the ellipse¡x5¢2+¡y3¢2=1.By examining differing values of t in 0 ≤ t ≤2π3, we see that this parametrization travelsthe ellipse in a clockwise fashion exactly once.t = 0 : (x(0), y(0)) = (0,3)t = π/6 : (x(π/6), y(π/6)) = (5,0)t = π/3 : (x(π/3), y(π/3)) = (0,−3)t = π/2 : (x(π/2), y(π/2)) = (−5,0)!4 !2 2 4!3!2!1123Figure 1: Ellipse.If we let t vary between 0 and 2π, we will traverse the ellipse 3 times.(b) Set up, but do not evaluate an integral that calculates the arc length of the curve describedin part (a).Solution. Arc lengths =Zbasµd xd t¶2+µd yd t¶2d t=Z2π30p(15cos(3t))2+(−9sin(3t))2d t .(c) Consider the equation x2+ y2= 16. Graph the set of solutions of this equation in R2andfind a parametrization that traverses the curve once counterclockwise.Solution. If we let x = 4 cos t and y = 4 sin t , then x2+y2=(4cos t)2+(4sin 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 = π/2 : (x(π/2), y(π/2)) = (0,4)t = π : (x(π), y(π)) = (−4,0)t = 3π/2 : (x(3π/2), y(3π/2)) = (0,−4)t = 2π : (x(2π), y(2π)) = (4,0)!4 !2 2 4!4!224Figure 2: Circle.To ensure that we travel the curve only once, we restrict t to the interval [0,2π). So theparametrization isx = 4cos ty =4 sin twhen 0 ≤ t ≤ 2π.3. 1st and 2nd Derivative Tests:(a) Use the 2nd Derivative Test to classify the critical numbers of the function f (x) =x4−8x2+10.Solution. First, we find the critical points of f (x).f0(x) = 4x3−16x.f0(x) = 0 when 4x3−16x =4x(x2−4) = 4x(x−2)(x+2) =0. Hence f0(x) = 0 when x =0, x =2or x =−2.Now apply the 2nd Derivative Test to the three critical points. From f "(x) =12x2−16, weget: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) = s4+4s3−1.Solution. First, find the critical points of h(s).h0(s) =4s3+12s2.Then h0(s) =0 when 4s3+12s2=4s2(s +3) = 0. So h0(s) =0 when s = 0 and s =−3.For the 1st Derivative Test, we need to determine if h is increasing or decreasing on theintervals (−∞,−3), (−3,0) and (0,∞).On (−∞,−3) choose any test point (for example, choose s = −1000). The sign of h0(s) =4s3+12s2<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 h0(s) = 4s3+12s2>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 h0(s) = 4s3+12s2>0 on this interval. Hence h(s) is increasing on (0,∞).Since at s = −3 the function changes from decreasing to increasing, the function musthave 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) =s4+4s3−1.Solution. When s = −3, h"(−3) = 36 > 0. A local min occurs when s = −3 by the 2ndDerivative Test.When s = 0, h"(0) = 0. The 2nd Derivative Test is inconclusive. The graph of y = h(s) hasno concavity at (0, h(0)). Without more information (the 1st Derivative Test), we are un-able to identify (0,h(0)) as a local max, min or a point of inflection.4. Consider the function f (x) = x2e−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 synonymouswith the equation of the tangent line or the 1st order Taylor polynomial. Hence, f0(x) =2xe−x+x2(−e−x).Since f0(0) = 0, the tangent line has no slope at (0, f (0)) = (0,0). The equation of the tan-gent 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 isT2(x) = f (0) +f0(0)1!(x −0) +f "(0)2!(x −0)2.Since f "(x) =2e−x−4xe−x+x2e−x, we compute that f "(0) =2. HenceT2(x) = 0 +01!(x −0) +22!(x −0)2= x2.(c) Explain how the second-order Taylor polynomial at x = 0 demonstrates that f must havea local minimum at x =0.Solution. The second-order Taylor polynomial is the best quadratic approximation to thecurve y = f (x) at the point (0, f (0)). Since T2(x) = x2clearly has a local minimum at (0,0),and (0,0) is the location of a critical point of f , then f must also have a local minimum at(0,0).5. Consider the integralZp3π02x cos(x2)d x.(a) Sketch the area in the x y-plane that is implicitly defined by this integral.Solution. The shadow area in the following picture is the area defined by the integral.0.5 1.0 1.5 2.0 2.5 3.0!6!4!224Figure 3: 5(a).(b) To evaluate, you will need to perform a substitution. Choose a proper u = f (x) and rewritethe integral in terms of u. Sketch the area in the uv-plane that is implicitly defined by thisintegral.Solution. Let u = x2. Then du = 2xdx, so the integral becomesZp3π02x cos(x2)d x =Z3x0cosudu.2 4 6 8!1.0!0.50.51.0Figure 4: 5(b).(c) Evaluate the integralZp3π02x cos(x2)d x.Solution.Zp3π02x cos(x2)dx =Z3x0cosudu =hsinuiu=3πu=0=sin(3π) −sin0 =


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UIUC MATH 241 - Worksheet_011513_sol

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