1 Tuesday, January 2 Solutions A review of some important calculus topics 1. Chain Rule: (a) Let h(t) = sin ( cos(tan t) ). Find the derivative with r...

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= 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 ) 2π for 0 ≤ t < . y = 3 cos (3t ) 3 What happens if we instead allow t to vary between 0 and 2π? Solution. Note that

³ x ´2 5

+

³ 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 ≤ 2π , we see that this parametrization travels 3 the 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)

3 2 1

!4

!2

2

4

!1 !2 !3

Figure 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 described in part (a). Solution. Arc length b

Z s=

µ

a

Z =

s

0

2π 3

p

dx dt

¶2

dy + dt µ

¶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 = 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

!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, 2π). So the parametrization is x = 4 cos t when 0 ≤ t ≤ 2π. 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 the location of a critical point of f , then f must also have a local minimum at (0, 0).

Z 5. Consider the integral

p 3π

2x cos (x 2 ) d x.

0

(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.

4 2

0.5

1.0

1.5

2.0

2.5

3.0

!2 !4 !6

Figure 3: 5(a).

(b) To evaluate, you will need to perform a substitution. Choose a proper u = f (x) and rewrite the integral in terms of u. Sketch the area in the uv-plane that is implicitly defined by this integral. Solution. Let u = x 2 . Then d u = 2xd x, so the integral becomes Z

p 3π

Z

2

2x cos(x )d x =

0

3x

cos ud u. 0

1.0

0.5

2

4

6

8

!0.5

!1.0

Figure 4: 5(b). p 3π

Z (c) Evaluate the integral

2x cos (x 2 ) d x.

0

Solution. Z 0

p 3π

2

2x cos(x )d x =

3x

Z 0

h

cos ud u = sin u

iu=3π u=0

= sin(3π) − sin 0 = 0.

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