The graph of

Answers for Calculus A Final Examination

1.

Consider a segment of the curve described by the equation x²⁼³ +y²⁼³ = 1 in the rst quadrant (^i.e., when 0^ x^1 and 0^y^1).

(a) Find the length of the curve.

(b) Find the area of the surface generated by revolving the curve about the x-axis.

(c) Find the centroid of the curve. From that, what can you say about the area of the surface generated by revolving the curve about the x =^;1 line.

Sol : (a) View y as a function of x: y = (1^;x²³)³², for 0 ^x^1.

dy = 32(1^;x²³)¹²
(^;2 3x^;

1

3)dx =^;(x^{;23 ;}1)¹²dx ds =^pdx² +dy² =

q

1 +x^{;23 ;}1dx = x^;13dx So the length of the curve = ^{Z 1}

0

x^;13dx = 32x²³







 1

0 = 32

(b) The area of the surface generated by revolving the cureve about the x-axis is

Z

1

0

2y

r

1 + (dydx)²dx = ^{Z 1}

0

2(1^;x²³)³²x^;13dx (let t = x²⁼³)

=

Z

1

0

2(1^;t)³²
3

2dt (dt = 2

3x^;1=3dx)

= 3^{Z 1}

0

(1^;t)³²dt

= 3 [^;2

5(1^;t)⁵²]







 1

0

= 65

(c) Suppose that the coordinate of the centroid is (m^x;m^y), then m^x =

R

x=1

x=0 xds 3=2 = 2

3

Z

1

0

x²⁼³dx = 23
3 5 = 2

5

By Pappus's Theorem, the area of the surface generated by revolving the curve is S = 2rL = 2[25^;(^;1)]32 = 21

This value is the same as 5

Z

1

0

2[(1^;y²⁼³)³⁼²+ 1]y^;1=3dy (^Z 2(x + 1)ds)

2.

Considerf(x) = x^x,x > 0, evaluate f⁰(x), f⁰⁰(x), and lim^x!0+f(x). Give a full discussion of the monotonicity, concavity, local extrema of1 f(x), and sketch the graph.

Sol : f(x) = x^x =e^xlnx.

f⁰(x) = e^xlnx
d

dx(xlnx) = x^x(lnx + 1) f⁰⁰(x) = (lnx + 1)
d

dxx^x+x^x
d

dx(lnx + 1) = x^x(lnx + 1)²+x^x;1 lim

x!0

+f(x) = lim

x!0

+e^xlnx = exp( lim

x!0

+xlnx) = exp( lim

x!0 +

lnx x^;1)

= exp( lim

x!0 +

1=x

;x^;2) (by L'H^opital's rule)

= exp( lim

x!0

+(^;x)) = 1

Since x > 0, x^x > 0, we know that f⁰⁰(x) > 0 for all x > 0, ^i.e.,f is always concave up.

f⁰(x) > (<)0 ^,(lnx + 1) > (<)0^,lnx > (<)^;1^,x > (<)e^;1. Hencef is increasing if x > e^;1, and is decreasing if 0< x < e^;1.

f⁰(x) = 0 ^,x = e^;1. Since f⁰⁰(x) > 0 for all x > 0, we know that f attains minimum at x = e^;1.

Note that since the limit of f(x)=x as x tends to innity does not exists, there is not aysmptote.

The graph of

f(x) = x^x

:

−0.5 0 0.5 1 1.5 2

−0.5 0 0.5 1 1.5 2

x y

O

y = f(x) = x^x

(e⁻¹, f(e⁻¹))

3.

Find the limits

(a) lim

x!1 +

 1

x^;1 ^; 1 lnx



; (b) lim

x!0 +

exp(^;1=x²)

x :

Sol : (a) lim

x!1 +

 1

x^;1 ^; 1 lnx



= lim

x!1 +

lnx^;(x^;1) (x^;1)lnx

= lim

x!1 +

1=x^;1

lnx + (x^;1)=x (by L'H^opital's rule)

= lim

x!1 +

1^;x xlnx + x^;1 2

= lim

x!1 +

;1

lnx + x
1=x + 1 (by L'H^opital's rule)

= ^;1

(b) Let y = 1=x, then as x tends to 0⁺2, y tends to innity. Then lim

x!0 +

exp(^;1=x²)

x = lim

y !1

exp(^;y²)

= lim 1=y

y !1

exp(yy ²)

= lim

y !1

2y exp(y1 ²) (by L'H^opital's rule)

= 0

4.

You are sitting in a classroom next to the wall looking at the blackboard at the front of the room. The blackboard is 12 ft long and starts 3 ft from the wall you are sitting next to. How far from the front of the room should you sit so that your viewing angle of the blackboard is maximal (see the included gure below)?

you blackboard

12 ft

3 ft

x

α

Figure 1: Figure for Problem #4

Sol : As the gure shows, let x be the distance between I and the wall. Then  = f(x) = tan^;115

x ^;tan^;1 3 The problem is equivalent to maximize. x

f⁰(x) = 1

1 + (15=x)² d dx(15

x )^; 1

1 + (3=x)² d dx(3

x) = ^;12x²+ 540 (x² + 9)(x²+ 225) f⁰(x) = 0 ^,;12x²+ 540 = 0^,x²= 45 ^,x =^3^p5. Here we choose x = 3^p5.

For x < 3^p5,f⁰(x) > 0, and for x > 3^p5,f⁰(x) < 0.

So f attains its maximum at x = 3^p5, ^i.e., the viewing angle is maximal when I sit 3^p5-ft from the front of the room.

5.

In an oilrenery a storage tank contains 2000 gallons of gasoline that initially has 100lb of an additive dissolved in it. In preparation for winter weather, gasoline containing 2lb of additive per gallon is pumped into the tank at the rate of 40 gal/min. The well-mixed solution is pumped out at a rate of 45 gal/min. Find the amount of additive in the tank 20 min after the process starts. 3

Sol : Suppose at timet (min) there are x(t) lb additive in the tank. Then x(0) = 100.

At timet, the rate of additive pumped into the tank is 40
2 = 80 lb/min, and since there are 2000 +(40^;45)t = 2000^;5t gallons of gasoline in the tank, and the rate of additive pumped out of the tank is 45
^2000;5tx(t) . Hence we have the dierential equation:

dxdt = 80^;45
x

2000^;5t = 80^; 9x

400^;t with initial condition x(0) = 100

Then dx

dt + 9

400^;tx = 80⁾ d

dt x

(400^;t)⁹ = 80

(400^;t)^{9 )} (400x^;t)⁹ =

Z 80

(400^;t)⁹dt = 10(400^;t)^;8+C , or x = 10(400^;t) + C(400^;t)⁹ x(0) = 100 implies C = ^100;40004009 =^;39004009. Hence

x(20) = ^;3900

400⁹
(400^;20)⁹+ 10
(400^;20)

= ^;3900
(380400)⁹ + 10
380

= ^;3900
0:95⁹+ 3800 (^)

 1342 (lb)

Rmk:

The teacher said that if you have got the correct formula, then you will obtain the points if you have (*).

6.

Evaluate the following integrals:

(a)

Z

x³⁼²lnx dx; (b)

Z

3

0

x²

p9 +x² dx;

(c)

Z

1

1

 x

2(x²+ 1) ^; 1 2x



dx; (d) ^Z x + 1

(x^;1)²(x²+ 2) dx:

Sol : (a) Set u = lnx, dv = x³⁼²dx, then v = ²⁵x⁵⁼², hence

Z

x³⁼²lnxdx = lnx
2 5x^5=2;

Z 2 5x⁵⁼²

 d dx lnx



dx

= 25x⁵⁼²lnx^; 2 5

Z

x⁵⁼²
1 xdx

= 25x⁵⁼²lnx^; 2 5

Z

x³⁼²dx

= 25x⁵⁼²lnx^; 2 5
2

5x⁵⁼²+C

= 25x⁵⁼²lnx^; 4

25x⁵⁼²+C

(b) Setx = 3sinhy, then dx = 3cosh ydy, and x = 0^!y = 0, x = 3^!y = ln(1+^p2), hence

Z

3

0

x²

p9 +x²dx = ^{Z ln(1+}

p

2)

0

9sinh²y

q9(1 + sinh²y)
3coshydy 4

= 9

Z

ln(1+

p

2)

0

sinh²y

3cosh
3coshydy

= 9

Z

ln(1+

p

2)

0

sinh²ydy

= 9

Z

ln(1+

p

2)

0

cosh2y^;1

2 dy

= 92

1

2 sinh2y





 ln(1+

p

2)

0

;y^ln(1+

p

2)

0



= 92

"

e^2ln(1+p2);e^;2ln(1+p2)

4 ^;ln(1 +^p2)

#

= 92

"

(1 +^p2)^2;(1 +^p2)^;2

4 ^;ln(1 +^p2)

#

= 92^hp2^;ln(1 +^p2)ⁱ

(c) ^{Z 1}

1

 x

2(x²+ 1) ^; 1 2x



dx = lim

t!1 Z

t

1

 x

2(x²+ 1) ^; 1 2x



dx

= lim

t!1

"

ln(x²+ 1) 4







 t

1

;

lnx 2







 t

1

#

= lim

t!1

14

ln(t²+ 1)^;ln2^;lnt^2

= 14 lim^t!1ln t²+ 1 2t²

= 14 ln



lim

t!1

t² + 1 2t²



= 14 ln1

2 =^;ln2 (d) Assume that 4

x + 1

(x^;1)²(x²+ 2) = a

x^;1 + b

(x^;1)² + cx + dx²+ 2 Then [ax+(b^;a)](x²+2)+(x^;1)²(cx+d) = x+1

Then

8

>

>

<

>

>

:

a + c = 0

;a + b^;2c + d = 0 2a + c^;2d = 1 2(b^;a) + d = 1

) 8

>

>

<

>

>

:

a =^;19 b = ²³ c = ¹⁹

d =^;59 Hence

Z x + 1

(x^;1)²(x²+ 2)dx =

Z



;

9(x1^;1) + 2

3(x^;1)² + x^;5 9(x² + 2)



dx

= ^;1

9 ln(x^;1)^; 2 3
1

x^;1 + 1 18

Z 2xdx x²+ 2 ^; 5

9

Z dx x²+ 2

= ^;1

9 ln(x^;1)^; 2 3
1

x^;1 + 1

18 ln(x²+ 2)^; 5

9^p2 tan^;1 x

p2 + C

= ^;1

9 ln(x^;1)^; 2 3
1

x 1 + 1

18 ln(x²+ 2)^; 5^p2

18 tan^;1 x

p2 + C 5

7.

(a) Find the Taylor formula for the logarithmic function ln(1^;t) about t = 0,

(b) LetF(x) =^R0x[ln(1^;t)=t] dt, nd a polynomialthat willapproximate F(x)through- out the interval [0;1=5] with an error of magnitude less than 10^;3.

Sol : (a) Letf(t) = ln(1^;t), then f^{(k )}(t) =^;(k^;1)!(1^;t)^;k, then by Taylor formula, there exists a  between 0 and t such that

f(t) =^Xn

k =0

f^{(k )}(0)

k! t^k + f⁽ⁿ⁺¹⁾()

(n + 1)! tⁿ⁺¹=^Xn

k =1

;t^k

k ^; tⁿ⁺¹

(n + 1)(1^;)^;(n+1) (b) From (a), for t²(0; 0:2), there exists an  ²(0; t) such that

ln(1^;t)

t =^Xn

k =1

;t^{k ;1}

k ^; tⁿ

(n + 1)(1^;)^;(n+1)

To approximateF(x) in (0; 0:2) with error less than 10^;3, choose n large enough so that the integral of remainder term is less than 10^;3. Since









;

tⁿ

(n + 1)(1^;)⁽ⁿ⁺¹⁾











5ⁿ⁺¹tⁿ (n + 1)4ⁿ⁺¹









 Z

x

0

"

ln(1^;t)

t ^;

n

X

k =1

;t^{k ;1} k

!#

dt











 Z

0:2

0 









ln(1^;t)

t ^;

n

X

k =1

;t^{k ;1} k

!









dt

 Z

0:2

0

tⁿ

(n + 1)(1^;)ⁿ⁺¹dt^{Z 0:2}

0

5ⁿ⁺¹tⁿ

(n + 1)4ⁿ⁺¹dt^ 5ⁿ⁺¹0:2ⁿ⁺¹ 4ⁿ⁺¹(n + 1)^{2 } 1

4ⁿ Choose n^ 3

log¹⁰4 ^4:9829 , or n = 5 We have

F(x) =^{Z x}

0

ln(1 +t) t dt^

Z

x

0 5

X

k =1

;t^{k ;1} k dt =

5

X

k =1

;x^k

k² =^;x^; x² 4 ^; x³

9 ^; x⁴ 16 ^; x⁵

25 with error less than 14⁶6^{2 }6:782^10^;6:

Rmk :

From the analysis, when n ^ 3, the error is less than 1:563^10^;4. So for all n ^ 3 will get the credit.

6