# 1. (12 Points) Compute each of the following limits. Justify your answers.

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Math11 Fall 2009 Exam #2 Friday October 30, 2009 Professor D. Benedetto

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(a) lim x→0 3x − x3 sin(7x) (c) xlim→∞ 8x4 − 17 3x4+ 2009x (b) lim x→0 2 tan(3x) x (d) xlim→∞ x7 − 4x + 7 x2 + 9 (a) lim x→0 3x − x3 sin(7x) = limx→0 x(3 − x2 ) sin(7x) = limx→0 x sin(7x) · limx→0(3 − x 2 ) = 1 7xlim→0 7x sin(7x) · 3 = 3 7· 1 · 3 = 3 7 (b) lim x→0 2 tan(3x) x = 2 limx→0 sin(3x) x · limx→0 1 cos(3x) = 2 · 3 limx→0 sin(3x) 3x · 1 = 6 · 1 · 1 = 6 (c) lim x→∞ (8x4 − 17)(x14) (3x4 + 2009x)(x14) = lim x→∞ 8 −x174 3 +2009x3 = 8 3 (d) lim x→∞ (x7 − 4x + 7)(1 x2) (x2 + 9)(x12) = lim x→∞ x54x+ 7 x2 1 +x92 = ∞

### 2.

(20 Points) Differentiate each of the following functions. You do not need to simplify your

(a) f (x) = (9 − x2)8(x3− 6x)9 (c) f (x) = √ sin x (x − cos x)2 (b) f (t) = sin3  t √ t  (d) f (x) = 1  tan(7x) + 1 x 57 (a) f′(x) = (9 − x2 )8 9(x3 − 6x)8 (3x2 − 6) + (x3 − 6x)9 8(9 − x2 )7 (−2x) (b) Note (don’t have to though): f (t) = sin3 t

√ t  = sin3 (√t) =⇒ f′(t) = 3 sin2 (√t) cos(√t) 1 2√t (c) f′(x) = (x − cos x) 2 1 2√sin xcos x − √

sin x(2)(x − cos x)(1 + sin x) (x − cos x)4 (d) f′(x) = −5 7  tan(7x) + 1 x −127 [sec2 (7x)(7) −x12]

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### 3.

(10 Points) Find the absolute maximum and absolute minimum value(s) of the function G(x) = 5x x2+ 1 on the interval [0, 2]. First G′(x) = (x 2 + 1)5 − 5x(2x) (x2+ 1)2 = 5x2 + 5 − 10x2 (x2+ 1)2 = 5 − 5x2 (x2+ 1)2 So critical numbers are when 5 − 5x2

= 0 or when x = ±1. Notice that x = −1 is not in our interval of interest, so we evaluate f at x = 1 here and at the endpoints x = 0 and x = 2 and use the Closed Interval Method:

f(1) = 5

2 ABSOLUTE MAX VALUE f(0) = 0 ABSOLUTE MIN VALUE f(2) = 2

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(10 Points) Let f (x) = x3

− 3x + 3. For this function, discuss domain, vertical and horizontal asymptote(s), interval(s) of increase or decrease, local extreme value(s), concavity, and inflection point(s). Then use this information to present a detailed and labelled sketch of the curve.

• f(x) has domain (−∞, ∞)

• It is a polynomial, continuous everywhere, and so has no vertical asymptotes. • There are no horizontal asymptotes for this f since limx

→∞f(x) = ∞ and limx→−∞f(x) = −∞. • First Derivative Information

We compute f′(x) = 3x2

− 3 and set it equal to 0 and solve for x to find critical numbers. The critical points occur where f′ is undefined (never here) or zero. The latter happens when x = ±1. As a result, x = ±1 are the critical numbers. Using sign testing/analysis for f′,

 t t -−1 1 f′ f ⊕ ⊖ ⊕ ր local ց ր

max localmin or our f′ chart is

x (−∞, −1) (−1, 1) (1, ∞)

f′(x)

f(x) ր ց ր

So f is increasing on (−∞, −1) and on (1, ∞); and f is decreasing on (−1, 1). Moreover, f has a local max at x = −1 with f(−1) = 5, and a local min at x = 1 with f(1) = 1.

• Second Derivative Information

Meanwhile, f′′ is always defined and continuous, and f′′ = 6x = 0 only at our possible inflection point x = 0. Using sign testing/analysis for f′′,

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 t -0 f′′ f ⊖ ⊕ T S infl. point or our f′′ chart is x (−∞, 0) (0, ∞) f′′(x) f(x) T S

So f is concave down on (−∞, 0) and concave up on (0, ∞), with an inflection point at x = 0. • Piece the first and second derivative information together

 t t t -−1 0 1 f ր ց ց ր f T T S S f     local

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### 5.

(18 Points) Let f (x) = x

2 − 9 x2

− 4

. For this function, discuss domain, vertical and horizontal asymptote(s), interval(s) of increase or decrease, local extreme value(s), concavity, and inflection point(s). Then use this information to present a detailed and labelled sketch of the curve.

Take my word for it that (you do not have to compute these) f′(x) = 10x (x2 − 4)2 and f′′(x) = −10(x 2 + 4) (x2 − 4)3 . • f(x) has domain {x|x 6= ±2} • Vertical asymptotes at x = ±2.

• Horizontal asymptote at y = 1 for this f since limx

→±∞f(x) = 1. • First Derivative Information

We compute f′(x) = 10x (x2

− 4)2 and set it equal to 0 and solve for x to find critical numbers. The critical points occur where f′ is undefined or zero. The latter happens when x = 0. The derivative is undefined when x = ±2, but those values are not in the domain of the original function. As a result, x = 0 is the critical number. Using sign testing/analysis for f′,

 t t t -−2 0 2 f′ f ⊖ ⊖ ⊕ ⊕ ց ց localր ր min or our f′ chart is

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x (−∞, 0) (0, ∞)

f′(x)

f(x) ց ր

So f is decreasing on (−∞, 0); and f is increasing on (0, ∞). Moreover, f has a local min at x = 0. • Second Derivative Information

Meanwhile, f′′= −10(x 2

+ 4) (x2

− 4)3 is never zero. Using sign testing/analysis for f′′ around the vertical asymptotes,  t t -−2 2 f′′ f ⊖ ⊕ ⊖ T S T or our f′′ chart is x (−∞, −2) (−2, 2) (2, ∞) f′′(x) f(x) T S T

So f is concave up on (−2, 2) and concave down on (−∞, −2) and (2, ∞). • Piece the first and second derivative information together

 t t t -−2 0 2 ց ց ր ր f f T S S T f  local   min

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(7 Points) Find the equation of the tangent line to the curve y3 − xy 2 + cos(xy) = 2 at the point (0, 1). Differentiating gives: 3y2dy dx−  2xydy dx+ y 2  − sin(xy)  xdy dx + y 

= 0, and at (0, 1), this means 3dy dx −  0dy dx+ 1  − sin(0)  0dy dx+ 1  = 0 which implies dy dx = 1

3. So the tangent line is y − 1 = 1

3(x − 0), that is, y= 1

3x+ 1 .

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(15 Points) A conical tank, 14 feet across the entire top and 12 feet deep, is leaking water. The

radius of the water level is decreasing at the rate of 2 feet per minute. How fast is the water leaking out of the tank when the radius of the water level is 2 feet?

**Recall the volume of the cone is given by V = 1 3πr

2 h The cross section (with water level drawn in) looks like:

• Diagram 7 B B B B B BB        w a t e r -r 6 ? 12 6 ? h

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-• Variables

Let r = radius of the water level at time t Let h = height of the water level at time t

Let V = volume of the water in the tank at time t Find dV dt =? when r = 2 feet and dr dt = −2 ft min • Equation relating the variables: Volume= V = 1

3πr 2

h

• Extra solvable information: Note that h is not mentioned in the problem’s info. But there is a relationship, via similar triangles, between r and h. We must have

r 7 = h 12 =⇒ h = 12r 7 After substituting into our previous equation, we get: V = 1 3πr 2 12r 7  = 4 7πr 3

• Differentiate both sides w.r.t. time t. d dt(V ) = d dt  4 7πr 3  =⇒ dVdt = 4 7π· 3r 2 ·drdt =⇒ dVdt = 12 7 πr 2dr dt • Substitute Key Moment Information (now and not before now!!!):

dV dt = 12 7 π(2) 2 (−2)

• Solve for the desired quantity: dV dt = − 96π 7 ft3 min

• Answer the question that was asked: The water is leaking out of the tank at a rate of 96π7 cubic feet every minute.

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(8 Points) Let f be a differentiable function with f

6  = 1 and f′π 6  = −2 Let h(x) = sin x f(x). Compute h ′π 6  . Compute h′(x) = f(x) cos x − sin xf′(x)

[f (x)]2 h′(π 6) = f(π 6) cos π 6 − sin π 6 f′ π6  f π 6 2 = 1 · √3 2 − 1 2(−2) 1 = √ 3 2 + 1 = 2 +√3 2

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************************************************************************************** BONUS PROBLEMS: THESE ARE OPTIONAL!

Feel free to attempt either of the following two bonus problems, but ONLY if you are completely done with the original part of the exam, problems 1-8.

************************************************************************************** Bonus 1: A kite 50 feet high is being blown horizontally at 10 feet per second. When the kite has blown horizontally 12 seconds, how fast is the angle between the string and the ground changing?

The picture at arbitrary time t is: child 50 x kite     y θ x 50 • Variables

Let x = distance kite has travelled horizontally at time t Let y = distance between kite and child at time t

Find dθ

dt =? when x = 120 feet (since 12 sec.passed at a rate of 10 feet per second) and dx

dt = 10 ft sec • Equation relating the variables:

The trigonometry of the triangle yields tan θ = 50 x. • Differentiate both sides w.r.t. time t.

d dt(tan θ) = d dt  50 x  =⇒ sec2θdθ dt = − 50 x2 dx dt.

• Substitute Key Moment Information (now and not before now!!!):

At the key instant when y = 120, using the original equation, we have y = p(50)2+ (120)2 =

16900 = 130. Or notice we have a multiple of a 5 − 12 − 13 triangle. Therefore, sec θ = hyp

adj = 130 120  130 120 2 dθ dt = − 50 (120)2 · (10). • Solve for the desired quantity:

dθ dt = −500(120)2 (120)2(130)2 = − 500 16, 900 = − 5 169 rad sec

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• Answer the question that was asked: The angle is decreasing at a rate of 1695 radians every second. Bonus 2: Differentiate g(x) = (x + 7x−6) √ 2x + 1 cos2 (6x) sin(3x) . f′(x) =   sin(3x)  (x + 7x−6) 

2x + 1(2) cos(6x)(− sin(6x))(6) +cos 2 (6x)(2) 2√2x + 1  +√2x + 1 cos2(6x)(1 − 42x−7)  −(x + 7x−6)2x + 1 cos2 (6x) [cos(3x)(3)]   sin2 (3x).

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