Differentia Calculus

7/22/2019 Differentia Calculus

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Sample Calculus Problems

Last Revision: December 2, 2013

Part 1: Single Variable Functions 1

Part 2: Multi-Variable Functions 47

Part 3: Sequences and Series 67

Part 4: Vector Analysis 84


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Part 1: Single Variable Functions

1. Evaluate the limit lim4

3x 1+

. (Do not use LH opitalsRule.)

Solution:

x5x2 7x+ 10

lim4 3x+ 1

lim(4 3x+ 1)(4 + 3x+ 1)

x5x2 7x+ 10 =x5(x2 7x+ 10)(4 + 3x+ 1)

16(3x+ 1)= limx5(x2 7x+ 10)(4 +

15 3x

= limx5(x 5)(x2)(4 +

3 5 x

= limx5(x 5)(x2)(4 +

3

3x+ 1)

3x+ 1)

3x+ 1)

= lim

x5(x 2)(4 +

3=

3x+ 1)

(5 2)(4 + 35 + 1)

3=3 8

1=

8

Remark: = is the most frequently used verb in mathematics. It was introduced in

1557by Robert Recorde to avoid the tedious repetition of the words isequal to. It isimportant to use the equal sign correctly.

To introduce,the phantom equal sign, to avoid the tedious repetition of thesymbol =is not a good idea. The solution above should not go like:

lim4

3x 1

lim(4

3x 1 4+

3x 1

x5x2 7x+ 10 x5(x2 7x+ 10)(4 + 3x+ 1)lim

153xx5(x 5)(x2)(4 + 3x+ 1)lim

3(5x)


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x5(x 5)(x2)(4 + 3x+ 1)


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3

One must also not use other symbols, which have completely diff erent meanings, in place of=.The solution above should not go like:

lim4

3x 1

lim(4

3x 1 4 +

3x 1

x5x2 7x+ 10 x5(x2 7x+ 10)(4 + 3x+ 1)

153xlim

x5(x 5)(x2)(4 +

3x+ 1)

lQf lim 3(5 x)

x5(x 5)(x2)(4 + 3x+ 1)

The equal sign always standsbetween two things, although sometimes one of thesethings are at the end of theprevious line or at thebeginning of the next line. Thesolution above should not start like:

43x+ 1)(4 + 3x+ 1)= lim

(x5(x2 7x+ 10)(4 +

3x+ 1)

This begs the question: What is equal to lim (4

3x 1 4 +

3x 1+ ) ?x5(x2 7x+ 10)(4 + 3x+ 1)

The equal sign can be used between two functions when we deal with identities , like

x2 1x 1 =

or when we deal with equations , like

x + 1 for all x =/ 1

Find all x such that x2 = 4.

Therefore we can not just drop some of the limit signs in the solution above to make it looklike:

lim4

3x 1 (4

3x 1 4 +

3x 1

x5x2 7x+ 10 = (x2 7x+ 10)(4 + 3x+ 1)

3

(x 2)(4 + 3x+ 1)

3

(5 2)(4 + 3 5 + 1)

3= 3 8

1=

8

The equalities on the lines marked with are not correct. 3 is(x 2)(4 + 3x+ 1)

not equal to1

because, for instance, if we letx 1 then8

= =

3 1 1.

(x 2)(4 +

3x+ 1)

(1 2)(4 + 31 + 1) = 2 =/ 8


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4

2

=

2

= lim .

x c

x c

x c

x c

x22. Let m be the slope of the tangent line to the graph ofy =

x at the point (3,9).

Express m as a limit. (Do not compute m.)+

Solution: The slope m of the tangent line to the graph ofy =f(x) at the point(x0,f(x0)) is given by the limit

f(x) f(x0)

or equivalently by the limit

m limxx0 x x0

m limh0

f(x0 + h) f(x0 ).

h

Therefore twopossible answers are

x2(9) (3 + h) (9)

m = limx+2 3+h+2x3 x (3)

h0 h

3. Suppose that limf(x) =/ 0 and limg(x) = 0. Show that limf(x) does not exist.

xc xc xc g(x)

Solution: Assume that limf(x)

xc g(x)

product rule for limits we obtain

f(x)

exists, and letL

f(x)

limf(x)

. Then by the=xc g(x)

limf(x) = lim( g(x)) = lim limg(x) =L 0 = 0 .xc xc g(x)

xc g(x)xc

This contradicts the fact that limf(x) =/ 0. Therefore our assumption cannot

be true: limf(x)

xc g(x)

does not exist.

4. Suppose that limf(x) = 0 and there exists a constantK such that g(x)K for allx =/ c in some open interval containing c. Show that lim(f(x)g(x)) = 0.

Solution: We have f(x)g(x) = f(x) g(x)f(x)Kin some open intervalaround c. ThereforeKf(x)f(x)g(x) K f(x).

Nowapplying the Sandwich Theorem and using the fact that limf(x)= 0 we

obtain the result.


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x 0

x 0

+

x 1

2

r2

2 2 2

2

4

5. Determine the following limits if limf(x) =A and limf(x) =B.x0+ x0

a. limf(x2 x) b. lim (f(x2) f(x)) c. limf(x3 x)x0 x0 x0+

d. lim (f(x3) f(x)) e. limf(x2 x)x0 x1

Solution: a. If x < 0, then x2 > 0 and x > 0. Thereforex2 x > 0 forx < 0, andx2 x approaches 0 from the right asx approaches 0 from the left.limf(x2 x) =A.b. Sincex2 > 0 forx < 0,x2 approaches 0 from the right asx approaches 0

from the left. Hence lim (f(x2 ) f(x)) = limf(x2) limf(x) =A B.x0 x0 x0

c. For 0


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Then

lim a = limr3 /2

r0+

r0+ r

= lim (

r2 r4/4r3/2

(r +

r2 r4/4))

r0+r2 (r2 r4/4)

2 lim 1r0+

1r2/4)= 2 (1 +

102/4) = 4.ThereforeR approaches the point (4,0) as r 0+.


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7.* Use the formal definitionof the limit to show that lim1

2.

x1/2x=

Solution: Given > 0 we want to find > 0 such that

1 10 < x

2< =

x2< . ()

We will do this in two diff erent ways.

Solve the Inequality Method : First we solve1

2 forx.

x

1 .2 +

If 2= 0, that is if = 2, then2

12 x

1.

2 +

*Examples marked red are not part of the Fall 2013 syllabus.


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If 2< 0, that is if > 2, then2

12 x

2

1or

+

1

2>x .

Next we choose in such a way that everyx satisfying the condition 0


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x 3

3 2 3 2

in () holds. In all three cases choosing a delta such that 0 < 1/21/(2+)=

/(4 + 2) achieves this.

The Estimation Method : Suppose 0 < x1/2< for some > 0. Then1x2= 2

x

1/2

x2 . 0 we have to finda > 0 so that for allx wehave

0 < x(3)< = x4+ 7x1743< .

We have (x4 + 7x17)43 =x4 + 7x60 = (x + 3)(x33x2+ 9x20). Supposethat 0 < x(3) < and 1. Then 4 3 < x < 3 + 2.

In particular,

x

< 4. Therefore, using the Triangle Inequality, we obtain

x 3x + 9x20x + 3x + 9x+ 20 < 43

+ 3 42 + 9 4 + 20 = 168. Now ifwe choose to satisfy 0 < min{/168,1},then we have

x4+7x1743= x4+7x60= x+3x33x2+9x20< 168

whenever 0 < x(3) < . We are done.

168

168 =

9.* Suppose that for all 0 < < 1,

x1 0 such that

0 < x< = f(x) L< L/2.

Takex = /2. Then 0 < x< is true, but L= 0L= f(x) L< L/2is

not true. We have a contradiction.

On the other hand, ifL = 0, let = 1/2. Then there is a > 0 such that

0 1/, takex = 1/n. Then 0 < x< is

true, but 1 = 1= f(x)< 1/2 is not true. Again we have a contradiction.

Hence limf x cannot exist.x0

11. Show that the equation x2 10 =x sinx has a real solution.Solution: Consider the functionf(x) =x2 10x sinx. Thenf(0) = 10 < 0 andf

(10) = 10

2 10

10sin

(10

) = 90

10sin

(10

) 90

10 = 80 > 0. Note thatf is continuous on [ 0, 10]. Therefore we can apply the Intermediate ValueTheorem to the functionf on the interval [ 0, 10] for the value 0 to conclude thatthere is a point c in ( 0, 10) such thatf(c) = 0. This c is also a solutionof the given equation.

12. Show that at any moment there are two antipodalpoints on the equator of the Earth

with the same temperature.

Solution: First we will make a mathematical model of the problem. We willconsider the equator as a circle, and use the longitude as our coordinate . Wechoose the positive direction for to correspond to the East, measure in radians, and let it take any real value. So 45W corresponds to = /4,

= 7/4, and = 15/4, among other values. Note that +corresponds to the

antipode of the point corresponding to . We let T () denote the temperatureat . We have T (+2)= T () for all . We will assume that T is a continuous

function. We want to show that there is a c such that T (c + ) = T (c).

Consider the functionf() = T (+ ) T (). Note that since T is continuous,

f is continuous. Our quest to find a c such that T (c + ) = T (c) is equivalent


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to find a c such thatf(c) = 0. Iff(0) = 0, then we let c = 0 and we are done. Supposethatf(0) =/ 0. Observe thatf(0) = T () T (0) = T () T (2) =f(). In other words,f(0) andf() have opposite signs. Now we apply theIntermediate Value Theorem tof on the interval [0, ] for the value 0, and concludethat there is a point c in [0,] such thatf(c) = 0. We are done.Remark: It ispossible to show that at any moment there are two antipodal points onEarth with the same temperature and the same pressure.

13. Find all tangent lines to the graph ofy =x3 that pass through the point (2,4).

Solution: As dy/dx= d(x3)/dx = 3x2,the equation of the tangent line through

a point (x0, x3) on the graph isy x3 = 3x2(xx0). This linepasses through

0 0 0

2, 4) exactly when 4 x3 = 3x2(2x0), or in other words,x3 3x2+ 2 = 0.( 0 0 0 0We observe that x0 = 1 is a root of this polynomial. Therefore we have the

factorizationx3 3x2+ 2 = (x0 1)(x22x02). The roots of the quadratic0 0 0factor arex0 = 1 3.

Therefore the tangent lines toy =x3 at the points (1,1),(1 +

3, 10 + 6

3),

and (1 3,1063)pass through (2,4). The equations of these lines are

y = 3x2,y = (12 + 63)x(20 + 123),andy = (12 63)x(20 123),

respectively.


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x0

x 0

dx

2

2

3 2

dx

14. Evaluate the limit lim1 + sin2x2 cos3x

.

Solution:

x0

2 2

x3 tanx

3 2

2 2 3 2

lim 1 + sin x cosx

lim 1 + sin x 1 1cos x x

x

0 x3 tanx

=x0

x4

+

x4

tanx

= lim(((sinx2

21

) 1cos3x2 x+ ) )x0 x2

Now we observe that:

limsinx

1

1 + sin2x2 + 1 x4

tanx

x0

lim

x2=

1 1 1

x0

2 1 + sin2 02 + 1=

2

limx

1

and

x0 tanx=

lim1cosx

lim 1cosx2

1 cosx2

cos2x2

x0 x4=x0

(

x4

( + + ))

2sin2(x2/2)

limx0

1

x4

sin(x2/2)

(1 + cosx2 + cos2x2 ))2

2 2 2

=2

(lim1

2

x2/2

) lim(1 + cosx + cos x )

Therefore:

=2

1 3

3=

2.

2 2 3 2

lim 1 + sin x cosx

11 3

1 2.

x0 x3 tanx= (

2+

2) =

15. Find the equation of the tangent line to the graph ofy = sin2(x3/6) at the point

withx = 1.

dySolution: y = sin2(x3/6)

= 2sin(x3/6) cos(x3/6) 3x2/6.

Therefore,

dyx=1

= 2sin(

/6)

cos(

/6)

/2 =

3

.

4Since yx=1 = 1/4, using the point-slope formula we findthe equation of thetangent line as

y14 =

3


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4(x 1)


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h 0=

h 0

h 0

or, after some reorganization,

y =

3

x4

1

3.

4

16. Let

12x+x2 sin(

f(x) =

) if x =/ 0,

a. Findf(x)for allx.

0 if x = 0.

b. Show thatf is not continuous at 0.

Solution: a. For x =/

differentiation:0 we compute the derivative using the rules of

d 2(x) =dx

(2x+x

forx =/ 0.

sin(1/x))= 2 + 2xsin(1/x)+x2

cos(1/x)(1/x2)

Forx = 0 we must use the definitionof the derivative:

f(0)= lim

f(0 + h) f(0)

h1

limh0

2h + h2 sin(1/h)

h

= lim2 + limh sin( ) = 2 + 0 = 2h0 h0 h

Here we used the fact that limh sin(1/h) = 0 whose proof uses the Sandwich(or Squeeze) Theorem. Here is a recap of the proof: Since

sin(1/h)1 for all h =/0 we have hsin(1/h)= h sin(1/h)hfor all h =/ 0. Therefore1

hh sin(h) h for all h =/ 0.

As limh= 0 = lim(h), it follows by the Sandwich Theorem that

h0 h0

limh sin(1/h) = 0.

To summarize: 12 + 2xsin(

1) cos(

x) if x =/ 0,

f(x)=2 if x = 0.

b. Consider limf(x). We have lim2 = 2 and lim2xsin(1/x)= 0 as in partx0 x0 x0

(a). However limcos(1/x) does not exist. It follows that limf(x) does notx0

exist and hencef is not continuous at 0.x0


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2

dx2

3x

4

=

=

=

=

17. Findd y

(x,y)=(2,1)

ify is a diff erentiable function ofx satisfying the equationx3 +2y3 =

5xy.

Solution:

x3 + 2y3 = 5xy

d/dx

dy dy+ 6y

dx= 5y + 5x

dx ()

x = 2, y = 1

dy dy12 + 6

dx= 5 + 10

dx

dy

dx= 7

dy 7

dx=

4at (x, y) = (2,1)

Now we diff erentiate the equation marked () with respect to x to find the


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3x2 2

5x

=

=

=

second derivative.

dy dy+ 6y

dx= 5y + 5x

dxd/dx

2

dy 2 d2

y

dy dy

d2

y

6x+ 12y (dx

)

+ 6ydx2

= 5dx

+ 5dx

+ 5xdx2

dy 7x = 2, y = 1,

dx=4

2

7 d2y 7 d2y12 + 12(

4)

+ 6

dx2= 10

4+ 10

dx2

d2y

dx2=

125at x, y 2, 1

16( ) = ( )

5y 3x2Remark: An alternative approach is to solvey from (), viz. y =

6y2, and then

diff erentiate this with respect tox to findy.

18. A piston is connectedby a rod of length 14 cm to a crankshaft at a point 5 cm away from theaxis of rotation of the crankshaft. Determine how fast the crankshaft is rotating when the pistonis 11 cm away from the axis of rotation and is moving toward it at a speed of 1200 cm/sec.

Solution: LetP (x, y) and Q(a, 0)be the ends of the connecting rod as shown in the

picture. The axis of rotation of the crankshaft passes through the origin of the xy-plane and is perpendicular to it. The point P where the rod is connected tocrankshaft moves on a circle with radius 5 cm and center at the origin. The point Qwhere the rod is connected to the piston moves along the positivex-axis. is theanglebetween the ray OP and the positivex-axis.

y

P (x,y)

5 5

x2 +y2 = 25

Q(a, 0)

x


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2

2

2

2

The question is:

a = 11 cm and da

dt= 1200 cm/sec =

d

dt=?

We have

and

x2 +y2 = 52 (I)

(x a) +y = 142 . (II)At the moment in question a = 11 cm. Substituting this in (I) and (II) we

obtainx2 +y2 = 52 and (x 11 2 +y = 14 . Subtracting the second equationfrom the first gives 22x112 = 52 142, and solving forx we getx = 25/11 cm.Theny = 20

6/11cm.

Diff erentiating (I) and (II) with respect to time t we obtain

xdx dy

anddt

+y

dt

= 0 (III)

dx da dy(x a) (

dt

dt) +y

dt= 0 . (IV)

da

At the moment in question a = 11 cm,dt

= v = 1200 cm/sec, x = 25/11 cm

andy = 20

6/11cm. Substituting these in (III) and (IV) we get

and

dx5

dt+ 4

dy6 0 (V)

dt

dx146dt

+ 20 dy6 146v . (VI)

dt

Subtracting 5 times (V) from (VI) we find 121dx

146v, and hencedt

=

dx

dt=

146

v. Substituting this back in (V) givesdy 365

v.121 dt

=242

6

Now we are ready to computed

. Since tan dt =y/x,diff erentiation gives

xdy

ydx

sec2 ddt =

and using sec2 = 1 + tan2 = 1 + (y/x)

dt

dtx2

we obtain

2

dy dx

ddt

=

x ydt dt .x2 +y2

25 20

6 dx 146 dy 365Pluggingx =

11cm, y =

cm,

11 dt=

121v, and

dt=

242

6 v in this

formula gives

d 73v

1460 6radian/sec.

dt

=

110

6

=

11


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Remark:

1460 6radian/sec is

1460 6 60rpm or approximately 3105 rpm.

11 11

2

Remark: This problem has a shorter solution if we use the law of cosines. Start with

x2 + 52 2 5x cos= 142 and diff erentiate with respect to t to obtainddt

=5cos11 v .

5sin

Put x = 11 cm in the first equation to find cos= 5/11 and then sin= 4

6/11. Nowsubstituting these in the second equation gives the answer.

19. Determine how fast the length of an edge of a cube is changing at the moment when thelength of the edge is 5 cm and the volume of the cube is decreasing at a rate of 100 cm3/sec.

Solution: Let a denote the length of an edge of the cube, and V denote thevolume of the cube. Then we have V = a3. Differentiating with respect to

time t givesdV

3a2da

. SubstitutingdV

100 cm3/sec and a

5 cm for

dt=

dt dt= =

the moment in question, we obtain

da 4cm/sec. Therefore the length of

the edge is decreasing at a rate of43

dt=

3

cm/sec at that moment.

20. We measure the radius and the height of a cone with 1% and 2% errors, respectively.

We use these data to compute the volume of the cone. Estimate thepercentage error involume.

Solution: Let r, h, and V be the radius, the height and the volume of the cone,respectively.

2

2 2

dV dr dhV =

3rh = dV = rhdr r

3 3dh =

V= 2

r+

h.

dr dhSince the error in r is 1% we have

r1%. Similarly

h2%. Now using

the triangle inequality we obtain

dV dr dh dr dhV= 2

r+ h

2r+

h2 1% + 2% = 4%.

The error in volume is 4%.


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1

21. Find the absolute maximum value and the absolute minimum value off(x) =x4/3

x x1/3 on the interval [1, 6].

Remark: How to find the absolute maximum and the absolute minimum values of acontinuous functionf on a closed interval [ a, b] of finite length:

i. Computef

.

ii. Find the critical points off in (a, b).

iii. Add the endpoints a and b to this list.

iv. Compute the value off at each point in the list.

v. The largest value is the the absolute maximum and the smallest value is the absolute

minimum off on [a, b].

Solution: fx 4x1/3( ) =3

1 2 313x / . The derivative is not definedat x = 0,

thereforex = 0 is a critical point. Next we solvef(x)= 0.

In the equation4x1/3

3

1x2/33

= 0 we letz =x1/3 to obtain the equation

4z3 3z2 1 = 0. Sincez = 1 is a root, we have the factorization 4z3 3z2 1 =(z 1)(4z2 +z + 1).As the quadratic factor has no real roots,z = 1 is the onlysolution. Thereforex =z3 = 13 = 1, which belongs to the interval [1, 6],is theonly other critical point.

We havef(0) = 0,f(1) = 1,f(1) = 3, andf(6) = 5 61/36. Observe that

56

1/36 > 3

5

6

1/3

> 9

5

3 6 >

93 725 > 721.

We conclude that the absolute maximum and minimum values off(x) =x4/3

x x1/3 on the interval [1, 6] are 5 61/36 and 1, respectively.

22. Find the absolute maximum and the absolute minimum values off(x) =x2

x + 1

on the interval [ 0,).+x + 9

Remark: When looking for the absolute maximum and the absolute minimum values of acontinuous function on an interval that is not necessarily closed or of finite length, a modifiedversion of the algorithm above can be used.

In Step iv, if an endpoint does not belong to the interval, then we compute theappropriate one-sided limit of the function at that point instead of the value of thefunction.

In Step v , if the largest value (which can be ) occurs only as a limit, then weconclude that there is no absolute maximum. Similarly for the smallest value.


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2. The roots off(x) = 0 are

x

Solution: We computef(x) =

x2 + 2x8(x2 +x + 9)

x = 4 andx = 2. Only x = 2 is in the interval [ 0,). So our list is 0, 2, and

.

1 1 1 1f(0) =

9

,f(2) =

5

, and limf(x) = 0. Since

5

>

9

> 0, the absolute maximum

is1

5and there is no absolute minimum.

23. Suppose that a functionf satisfiesf(x)=f(x/2) for allx andf(0) = 1. Show that

iff(x0) = 0 for somex0 > 0, then there isx1 such that 0


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25. Sketch the graph ofy = 5x2/32x5/3.1

01 3

102 3

1

01 3

Solution: y = x / x /3 3

=3

(1 x). Therefore y > 0 on (0,1),andy < 0 on (, 0) and (1,).

1 0 4 3

2 0 1 3

2 0 4 3

y = 9x /

x /

9=

9x / (x + 1/2). Thereforey> 0 on (,1/2),

andy< 0 on (1/2, 0) and (0,).

These give us the following table of signs and shapes.

x 1/2 0 1y + 0 y + 0

inf.pt. loc.min. loc.max.

We compute they-coordinates of the important points to get 0, 0 for the local3

minimum, (1,3) for the local maximum, and (1/2, 3 2) for the inflection

point. Note that since3

2 > 1, we have 33

2 > 3. Also note that the function

is continuous at x = 0, but

10limy = lim ( x1/3(1 x)) =

andx0+ x0+ 3

10

limy = lim ( x1/3(1x)) = .x

0

Therefore (0,0) is a cusp.

x

0

3

Finally we find thex-intercepts: y = 0 5x2/32x5/3 = 0 2x2/3(52x)=

0x = 0 or x = 5/2.

(The graph is on the next page .)


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Now we use these data to draw the graph:

Remark: Here is the same graph drawn byMaple:


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26. Two corridors meet at a corner. One of the corridors is 2 m wide and the other one is 3 mwide. What is the length of the longest ladder that can be carried horizontally around thiscorner?

Solution: Length of the longest ladder will be equal to the absolute minimumvalue ofL in the picture.

y

2m L

3m

x

Using the relation 3/x=y/2 we obtain

1/2

2 1/2

2 1/2

2 1/2

L = (x2 + 9) + (4 +y ) = (x + 9) + (4 + (6/x) ) .

Therefore we have to minimize

L = (x2 + 9 1/2 (1+ 2

x) for 0


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=

2

27. Find the maximum possible total surface area of a cylinder inscribed in a hemisphere ofradius 1.

Solution: Let S be the total surface area of the cylinder, S = 2r2+ 2rh.1/2

We have h = (1 r2 )

. Hence we want to maximize

1/2

S = 2r2+ 2r(1r2) for 0 r 1.

1 h

r

First we find the critical points:1 dS

2r

1 r21/2 r2 1

r2 1/2 0

1/2 2 2dr

=

2 2

+ ( )

2 2

( )

4 2

= =

2r(1 r2 )

8 32= 2r 1 = 4r (1 r

) = (2r

1)

= 8r 8r

1+ 1 = 0 =

r2 =16

2 +

2

. Note that 2r(1 r2 1/2

2+

2

= 2r2 1 implies r2 , and therefore2

r24

and r =2

.

2 + 2

We get S = (1+

2)at the critical point r = , S = 0 at the endpoint

r = 0, and S = 2 at the endpoint r = 1. As

2> 1 we have 1 +

2> 2, and

the absolute maximum value occurs at the critical point.

The maximumpossible surface area of the cylinder is (1 + 2).


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25

28. A fold is formed on a 20 cm 30 cm rectangular sheet of paper running from the shortside to the long side by placing a corner over the long side. Find the minimum possiblelength of the fold.

Solution: Let ABC D be the sheet of paper and let P be the point on the edgeAB where the corner C is folded over. The fold runs from Q on the edgeBC to Ron the edge CD. Let S be the projection ofR on to the edgeAB. Let L be thelength of the fold QR and letx = CQ.

D R C

L x

Q

x

A S P B

Then P Q/P B = RP /RSby the similarity of the trianglesP BQ andRSP .2

HenceRP = 20x/ x2 (20 x) = 20x/ 40x400 andL2 =RQ2 =RP 2 +

P Q2 = 400x2/(40x400) +x2 =x3/(x 10). The largestpossible value ofxis 20 cm. The smallestpossible value ofx occurs whenRP = 30 cm; that is

when 20x/

40x400 = 30 orx = 4515 5 cm.Therefore we want to minimizeL whereL2 =x3/(x 10) for 4515

5x 20.

Differentiating 2LdL/dx= 3x2/(x10) x3 /(x 10 2 and setting dL/dx= 0we obtainx = 15 cm as the only critical point in the domain. For x = 15 cm

we haveL = 15

3cm.

At the endpointx = 20 cm we haveL = 20

2cm and at the endpointx =

45155cm we haveL = 15 1865cm.Since 15

3cm is the smallest of these values, we conclude that this is thesmallest possible value for the length of the fold.

Remark: If the question isposed for a w sheet of paper with w 22/3,then the length ofthe shortest possible fold is 3

3w/4. Therefore, for a A4 size paper with w = 210

mm and = 297 mm, the shortest fold is approximately 273 mm; and for a letter sizepaper with w = 8.5in and = 11 in, the shortest fold is approximately 11.04 in.


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26

n 2 2n 3 4n 1

3

.

8

129. Evaluate the limit lim (n(

2+

12

+ +1

2)) .

n (2n+ 1)

(2n+ 3)

1

(4n1)

Solution: Consider the functionf(x) =x2

on the interval [2, 4]. If we divide

this interval into n subintervals of equal length2

using thepointsx 2

n k= +

2k,

n0 k n, and choose our sample points to be the midpoints ck = 2 +

2k 1,

n1 k n, of these subintervals, the Riemann sum for these data becomes

n n 2k 1 2 n 2nf(ck )xk = f(2+

n)

n=

2

k=1 k=1 k=1 (2n+ 2k 1)and the definitionof the definiteintegral gives

n

lim 2n

2=

4 dx2

=1 4 1 1 1

] = + =

Therefore

nk=1 (2n

+2k

1) 2 x

1 1

x 2 4 2 4

1 1lim (n( (2n+ 1)

2+ +

( + )2)) = .

( )

30. Letf be a continuous function.

x2

a. Findf(4) if 0 f(t) dt =xsinx for allx.

f(x)2

b. Findf(4) if

0t

x2

dt =x sinx for allx.

d x2

dFTC1

Solution: a. 0

f(t) dt =x sinx = dx

0

f(t) dt =dx

(x sinx) =

f(x2 ) 2x= sinx +x cosx. Now lettingx = 2 we getf(4) = /2.f(x)

2 t3

f(x) f(x 3b.

0

t

dt =x sinx = ]

0=x sinx =

3

=x sinx. Hence

x = 4 givesf(4) = 0.

Remark: One might ask if such functions exist. In part (b), f(x) =unique function satisfying the given condition.

3

3xsinx is the

sinx

In part (a), f(x) = 2

x

+2

cos

x forx > 0, andf(0) = by continuity; but it

can be anything forx < 0 so long as it is continuous.


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27

0

31. Letf be a continuous function and let

1

g(x) =1

f(t) xtdt .Expressg(x) in terms off(x) for 1


30/103

28

2

dx

Remark: In part (a), if we let u = cos(x2), then we obtain the answer

2 2 1 2 2

x sin(x ) cos(x ) dx = 4

cos (x ) + C ;

1

and if we first observe that sin(x2) cos(x2) =

we obtain the answer

sin(2x2),and then let u = sin(2x2), then

2 21 2


cos(2x ) + C .

These are all correct answers. If we write

2 21 2 2

x sin(x ) cos(x ) dx =4

sin (x ) + C1

2 2 1 2 2


cos (x ) + C2

2 2 1 2


cos(2x ) + C3

1 1then C2 = C1 +

4and C3 = C1 +

8.

a f(x) a33. Show that

0 f(x) +f(ax) dx = 2 for anypositive continuous function on [0,a] .a f(x)

Solution: LetI = 0 f(x) +f(ax) dx .

Let u = a x. Then du =dx, x = 0

u = a, x = a

u = 0, and

a f(x) 0 f(a u) a f(a x)I =

0 f(x) +f(ax) dx = a f(au) +f(u) (du) = 0 f(x) +f(ax) dx .Therefore

a f(x)

a f(a x)2I=

0 f(x) +f(ax) dx + 0 dxf(x) +f(a x)= 0

af(x) +f(a x)

f(x) +f(a x)

a

andIa

.=2

= 0

dx = a

Remark: Here is an explanation of what is going on with no integral signs: Consider the

rectangle with a vertex at the origin, and sides along the positivex- and they-axes withlengths a and 1, respectively. The graph ofy =f(x)/(f(x) +f(a x)) is symmetric with respectto the center (a/2, 1/2) of this rectangle, and therefore divides it into two regionsof equal area. Since the area of the rectangle is a andI is the area of the lower half, we

ahaveI =

2.


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29

34. LetR be the region boundedby theparabolay =x x2 and thex-axis, and let V be thevolume of the solid generatedby revolvingR about thex-axis.

a. Express V as an integral using the disk method. (Do not compute! )

b. Express V as an integral using the cylindrical shell method. (Do not compute! )

Solution: The parabola intersects thex-axis at x = 0 andx = 1. Therefore wehave

1 12 2 2

for part (a).

V = 0

R(x)

dx = 0

(x x ) dx

For part (b), the cylindrical shell method gives ay-integral,

d

V = 2c

(radius of the shell)(height of the shell) dy

as the region is revolved about thex-axis. We have c = 0 and d = 1/4, the

y-coordinate of the highest point of the parabola. The radius of the shell isthe vertical distance from the red rectangle in the figure to thex-axis, which


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30

andx =

=

1

is y. The height of the shell is the horizontal length of the rectangle; that is, thediff erence between thex-coordinatesof the right and the left sides of the rectangle.By solving the equationy =x x2 forx, we find these values as

1 +

14y

2

1

1 4y, respectively. Hence

2

1/4

V = 20

y (

1 +

14y2

114y2

) dy .

135. Evaluate the limit lim (

n

2k/n).

n

n 1

n k=1

n

Solution: 2k/n is a Riemann sum f(ck )xk forf(x) = 2x on [0, 1]

k=1 nk

k=1

kfor the partitionxk =

n, 0 k n, and the sample points ck =

n, 1 k n.

Therefore,

1lim (

n

2k/n) =1 2x 1

2xdx = ] = .

n n k=10 ln2 0 ln2

Remark: Alternatively, we can use the formula for the sum of a finite geometric series

n

2k/n =2(n+1)/n21/n

1/n=

21/n

1/n,

k=1 2 1 2 1


33/103

31

.

and then

1lim (

n

2k/n)= lim

21/n

1/n

limt0+

t2t

tLH= lim

t0+

2t+ t ln2 2t 1ln2 2t =

nn k=1

nn(21)

2 1 ln2

36. Find the absolute maximum and the absolute minimum values off(x) = (x2 3)ex

on the interval [2, 2].

Solution: f(x) = 2xex + (x2 3)ex = (x2 + 2x3)ex. We want to solve

f(x)= 0.

(x2 + 2x3)ex = 0= x2 + 2x3 = 0= x = 1,3.

Since3 does notbelong to the interval [2, 2],the only critical point isx = 1.

Hence we are going to computef at the pointsx = 1,

2, 2.f(1) = 2e,f(2) = e2, andf(2) = e2. Since e > 1 we have e2 > e2 > 2e.Therefore the absolute maximum is e2 and the absolute minimum is 2e.

37. Find the absolute maximum and the absolute minimum values ofx1/x.

lnxSolution: Lety =x1/x.Then lny =

x we obtain

. Diff erentiating this with respect tox

1 dy d

d lnx

1x lnx 1

x

1lnx

d 1 x 1 x 1lnx

y dx=

dxlny =

dx x=

x2=

x2=

dxx /

=x /

x2.

Since 1lnx > 0 for 0 e, the absolute maximum

value ofx1/x occurs at x = e and is e1/e. x1/x has no absolute minimumvalue.

Remark: Although the reasoning above does not require it, let us also look at whathappens at the endpoints of the domain. Since

lim lny = limlnx

LH

lim 1/x

= 0,

x x x=x 1

we have limx1/x = limy = lim elny = e0 = 1. On the other hand, limx1/x = 0 as 0 = 0.x x x x0+


34/103

32

= lim =

4

1 1

sinx 1/x2

38. Evaluate the limit lim .x0 x

Solution: Since limsinx

1 and lim1

, this is limit has the indetermi-

nate form 1. x0 x=

x0x2=

sinx

sinx 1/x2

ln(x

)Let y = (

x) . Then lny =

0

x2. As x 0, this will have the

indeterminate form and we can use LHopitalsRule.0

ln(sinx) lnxlimlny = limx0 x0

LH

x0

x2cosx 1

sinxx

2x

limx0

x cosx sinx2x2sinx

LH

= limcosx x sinx cosx

= limsinx

x0 4xsinx + 2x2cosxsinx

x0 4sinx + 2xcosx

x= lim =

=

x 0 sinx

x+ 2cosx

4 + 2 6

Here applications of LHopitalsRule are indicated with LH. Then

lim(sinx 1/x

2

) = limy = limelny = e1/6

x

0 x x

0 x

0

using the continuity of the exponential function.


35/103

33

2 2

2

2 2

3

Remark: The first example of an application of LHopitals Rule in Guillaume FrancoisAntoine Marquis de LHopitalsbook Analyse des Infiniment Petits pour lIntelligence des

Lignes Courbes of 1696:

There is a typo. Can you find it?

cos 2x e2x2

39. Evaluate lim .x0

Solution:

sin4x

2x2

2x2 4

limcos(2x)e lim cos(2x)e x

x0 sin4x=x0

( x

4

sin4x)

2

= limcos(2x)e2x

x 4

(lim )x0

limx0

x4

cos(2x)e2x

x4

x0sinx

2

LH2sin(2x)+ 4xe2x

= lim

4x3

LH

x0

4cos(2x)+ 4e2x16x2e2x

= lim

12x2LH

x0

8sin(2x)48xe2x + 64x3e2xlimx0

sin(2x) 24x

2 8 2limx0

2

2e2x3x

4

+ x2e2x )

=32 + 0 =

3


36/103

34

lim

= lim

2

lim

Remark: It is easier to solve thisproblem using the Taylor series, which will be seen inCalculus II:

cos 2x e2x2

limx0 sin4x

2 4 6

2 2 2 3

2x) (2x) (2x)2

(2x ) (2x )

limx0

(1

(

2!+

4!

6!+

)

(1+

(

2x )+

x3 4(x

3!+ )

2!+ 3!

+

)

limx0

44

3x

x4

5

66

+

45x +

2 6

3x +

4

5

62

= lim

3+

45x +

x0

4=

3

2 2

1

3

x +

sinx ax3 x40. Find the value of the constant a for which the limit lim exists and

compute the limit for this value of a.

Solution: We have

x0 x5

sin(x+

ax3

)x

x0

LH

x0

x5

cos(x+ ax3)(1+ 3ax2)1

5x4LH

= lim

sin(x + ax3)(1+ 3ax2)

20x3

+ cos(x + ax3)(6ax)

LHx0 cos(x+ ax3)(1+ 3ax2)

sin(x + ax ) 3(1 + 3ax )(6ax)+ cos(x + ax )(6a)

3 3 2 3

= lim

60x2.

x0

The numerator of the fraction inside this last limit goes to 1 + 6a asx 0.

Therefore the limit does not exist unless a = 1/6. If a = 1/6 then

sin(x +x3/6)x

x0 x53 3 2 3

lim

x0

cos(x +x3/6)(1+x2 /2)sin(x +x /6) 3(1 +x /2)x+ cos(x+x /6)

60x2

is the sum of

3 2 3

limcos(x+x /6)(1 (1 +x /2) )

x0 60x2

limx0

cos(x+x3/6)(3/2+ 3x2/4+x4 /8)60

1= 40


37/103

35

sin(x + ax3 ) = (x + ax3)

dy

a2 a2

a2 y2

2 2

2 2

and3 2

limsin(x +x /6) 3(1 +x /2)x

x0

Hence

60x2

limx0

sin(x +x3/6)

x +x3 /6

3

limx0

(1 +x2/2)(1+x2 /6)

20

1=

20.

limsin(x +x /6) x 3

.

x0 x5=

40

Remark: Once again there are shorter ways of doing this. If we use the Taylor seriesthen

3 3 5

(x + ax3)

3!

(x + ax )+

5!

1 3 1 1 5=x + (a 6

)x+ (

120

2a) x +

and it is immediate that the limit exists exactly when a = 1/6 and then its value is 3/40.

41. Let b > a > 0 be constants. Find the area of the surface generated by revolving thecircle (x b 2 +y = a about they-axis.

Solution: We haver

2d dx

Surface Area = 2c

x

1 + ( ) dy

for a surface generated by revolving a curve about the y-axis. We havex =

b +

y andx = b y for the right and left halves of the circle,respectively. Then

dx yr

2

dx a

Hence

dy= = 1 + (

dy)

= .a2 y2

a aSurface Area = 2

a

(b +

a2 y2) dy

a2 y2a a+ 2

aa 1(b

a2 y2) dya2 y2

= 4aba

d

ya2 y2

a

y= 4ab arcsin(

a) ]

a

2


38/103

36

= 4ab (arcsin1arcsin(1)) = 4ab(2(

2)) = 4 ab


39/103

37

2

3

Remark: The surface generated by revolving a circle about a line (in the same plane)that does not intersect it is called a torus.

42. Find y(1) ifdy

2

dx=xy

andy(0)= 1.

Solution: We havedydx

=xy2

dy=

y2=

x dx

= dy

y2=

x dx =

1 1 2 2

y = 2x

+C. Theny(0)= 1

=

1 = C. Hencey = 2 .

This in turn givesy(1)= 2.

Remark: What would your answerbe if the question askedy(2)?

43. Nitrogen dioxide is a reddish-brown gas that contributes to air pollution, and alsogives the smog its color. Under sunlight it decomposes producing other pollutants,one of which is ozone. As nitrogen dioxide decomposes, its density decreases at a rateproportional to the square of the density. Suppose that the density of nitrogen dioxide

Q is 1/2 grams per liter at time t = 0 minutes and 6/25 grams per liter at time t = 3

minutes. Find Q when t = 15 minutes. (Assume that no new nitrogen dioxide is added tothe environment.)

Solution: We have dQ/dt = kQ2 where k is a positive constant. ThendQ/Q2 = k dt and integrating we obtain 1/Q = kt+C where C is a constant.Letting t = 0 minutes we find 1/(1/2) = 1/Q(0) = C, hence C = 2 liters pergram. Now letting t = 3 min gives 1/(6/25) = 1/Q(3) = k 32 and hence

k = 13/18 liters per gram per minute. Finally we obtain Q = 18/(13t+ 36).

Using this we compute the density of nitrogen dioxide as Q(15) = 6/77 grams

per liter after 15 minutes.

44. Evaluate the following integrals.

a. sin x sin2xdx

b. 1

e lnxx

dx


40/103

38

u

3 3

3

Solution: a. We use the identity sin2x= 2sincosx to obtain

sinx sin2xdx = sin x 2sinx cosx dx

4

= 2 sin

4

x cosx dx

= 2 u du25

=

5u + C

2 5=5

sinx + C

after the substitution u = sinx, du = cosx dx.

b. The substitution u =x, du = dx/(2

x)gives:

e lnx

e

1 x dx = 41 lnu due

= 4[ulnu u]1

= 4( e ln e e + 1)

= 42e

Remark: There are other ways of doing these. Here are some:

Solution: a. Use the double-angle formula sin2x = (1 cos2x)/2and then thesubstitution u = 1cos2x,du = 2sin2xdx :

3

3 1cos2x sinx sin2xdx = (

1

2)

3/2

sin2xdx

=4

2 u du

1

=10

2u

1

5/2+ C

5/2

= 102(1 cos2x) + COr you can run this by simply saying u = sin2x, du = 2sinx cosx dx = sin2xdx

and:

sinx sin2xdx = u3/2du

2 5/2=52 5=5

sin

+ C

x + C


41/103

39

= [ ]

=

3

3

Or use the identity sin3x = (3sinx sin3x)/4and the trigonometricproduct

to sum formulas:

sin x sin2xdx =

11

4(3sinx sin3x)sin2xdx

=

4

(3sinx sin2xsin3xsin2x)dx1

=8

(3(cosx cos3x)(cosx cos5x))dx

1=

8 (cos5x3cos3x+ 2cosx) dx

1 1 1=

40sin5x

8sin3x+

4sinx + C

Or, if you are willing to go complex, use the identity sin= (eiei)/(2i) :

3

sin x sin2xdx = (

eix eix2i )

e2ix e2ix

dx2i

1=

16 (e

5ix+ e5ix3e3ix3e3ix+ 2eix + 2eix) dx1

=16(

1

e5ix e5ix5i

1

e3ix e3ixi

+ 2

1

eix eixi

) + C

=40

sin5x8

sin3x+4

sinx + C

b. Do integration by parts:

e lnx e 1 x dx = 1 lnx d(2e

2 x lnx1

x)

1

e dx2 x

xe dx

2 e 21 x e

= 2

e 4[ x]1= 2

e

4

e + 4

= 42e

Or do the other integration by parts first,

e lnx e1

x

dx = 1

1x

d(x lnx x)

ex lnx x

= [ ]1

1+

2 1

ex lnx xdx

x3/2

1 e lnx e= 1 +

21

x

dx [x]

1

= 2

e+

1

2 1

e lnx


42/103

40

x

dx


43/103

41

u

0

0

2 2

x x

x t

and then from

solve for

1

21

e lnx x

dx = 2 e

e lnx

1

x

dx = 42

e .

Or first do the substitutionx = eu, dx = eu du, and then do an integration by parts:

e lnx1

x

dx = 0

1 u ue e du

1

= 0

u e

1

u/2du

u/2

= 0

u d(2e )

1 1

= [2ueu/2]

20

1

eu/2du

= 2e1/24[eu/2]

= 2e1/24e1/2+ 4= 42e

45. Evaluate the following integrals:

x

a. e dx

b.

c.

1x2 dxdx

(x + 1)

Solution: a. We first do a change of variablex = t2, dx = 2tdt,

e dx = 2 te dt

and then do an integration by parts, u = t, dv = et dt = du = dt, v = et :

= 2tet2 et dt= 2tet2et+ C= 2

xe

2e + C


44/103

42

1 1 cos2

+ 1+ 1

+ 1

b. We use the trigonometric substitutionx = sin,

2

2. Then dx =

cosd and

x2 = sin2 = = cos = cos as cos0 for

2

2.

1 x2 dx = coscosd

2

= cos

1

d

=2

(1 + cos2) d1 1

=2

+4

sin2+ C

1 1=

2+

2sincos+ C

1 1

2

=2

arcsinx +2x

1x + C

1x

1x2

c. We use the trigonometric substitutionx = tan ,

2< 1, this integral converges.

b. The surface area formula gives

dx = 1 x2 .

1 1 2

Surface Area = 2

1y

1 + (y ) dx = 21 x 1 + x4 dx .We have

11

1 10 forx

1, and

dx. The surface area is

x+

x4

x

1 x=

infiniteby the Direct Comparison Test.


48/103

46

n

t

n

Remark: We want to get the surface S painted for a reasonable, finite price. We offer the jobtoPainter1 andPainter2.

Painter1 says: It cannot be done. S has infinitearea, it cannot be painted withfinite amount ofpaint.

Painter2 says: It can be done.D has finite volume. We can fill the inside ofS with volume(D) cubic units of paint and let the excesspaint run out.

Whom should we believe?

49. Let n be a nonnegative integer. Show that 0

t

Solution: We use induction on n.

et dt = n! .

Let n = 0. Then 0

e

dt = 1 = 0!.

n1 t

Let n > 0 and assume that 0

t

gives

e dt = (n 1)!. Integration by parts

0

t etdt = lim c

tnet dt0

c

cc

= lim([tnet] + n tn1et dt)c 0

0

c

= lim [tnet] + n tn1et dtc

0 0

n1 t

= n 0

t e dt

= n(n 1)!

= n!

where lim cnec 0 can be seen after n applications of LHopitalsc

Rule.

Remark: The Gamma function is definedby

x1 t

(x)= 0

t

e dt

forx > 0. It can be shown that the improper integral on the right converges if and only ifx > 0.Note that 0 is also a badpointbesides for 0 0. This relation (x) = (x+ 1)/x can be used repeatedly to definethe Gamma

function for all real numbers which are not nonpositive integers.

Since (n+ 1) = n! for all nonnegative integersn, we can use the Gamma function to definethe factorials of all real numbers which are not negative integers byx! = (x+ 1).In particular,


49/103

47

=

c

c

1 1/2 t

u2

(2)!= (1/2)=

0t

e dt = 20

e

du = 2 2

= .

The volume of an n-dimensionalball with radius r is n/2rn/(n/2)!. Check this formula for n =1 (the interval [r, r]), n = 2 (the disk with radius r), and n = 3 (the sphere withradius r). The case n = 4 will be seen in Calculus II.

50. Show that 0

lnx

x2 + 1dx = 0.

Solution: First of all, by definition,

lnx0 x2 + 1

1

dx = 0

lnx

x2 + 1

dx + 1

lnxdxx2 + 1

and the improper integral on the left converges if and only if both basicimproper integrals on the right converge.

Consider

1

hand,

lnxx2 + 1

dx . Note that 0 lnxx2 + 1

lnxx2

forx 1. On the other

lnx t

1 x2

dx = 0

tec

limc

0

c

limc

0

dt

tet dt

td(et)

c c

= lim ([ tet] + et dt)c 0

0c

= lim (cec [et] )c 0

= lim (cec ec + 1)

= 1

where we used the substitutionx = et, dx = et dt, followed by the definitionof theimproper integral, then an integration by parts and finally the limit

lim cec = lim

LH

lim 1

= 0.c cec

=cec

lnxThereforeby the Direct Comparison Test the improper integral 1

converges.

dxx2 + 1

1

Now the convergence of 0

lnx

x2 + 1dx follows as

1 lnx 1dx =

ln(1/u)2

du

= lnu

du

0 x2 + 1 (1/u) + 1 u

2

1 1 + u2


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45

.

where we used the change of variablex = 1/u,dx = du/u2,and this also gives:

lnx0 x2 + 1

1

dx = 0

lnxx2 + 1

dx + 1

lnxdxx2 + 1

= 1

lnx

x2 + 1

dx + 1

lnx

x2 + 1dx = 0

Remark: Instead of computing 1

lnx

x2dx, one can show its convergence using the

lnx 1 dx

comparison 0 x2

x3/2

forx 1 and the fact that 1

is convergent asp 3 2 1.x3/2

Here the fact thatlnx 1

forx

1 can be seen as follows: Considerf x

x lnx

x2x3/2

( ) =

on [1,). Thenf(x)= 1/(2x)1/x= ( x 2)/(2x)andx = 4 is the only critical point.

Sincef(x)< 0 forx < 4 andf(x)> 0 forx > 4,f(4) =

4ln4 = 22ln2 = 2(1ln2) > 0must be the absolute minimum value off on [1,), and we are done.

51.* Determine whether the improper integral 0

Solution: By definition,

dx

ex ex converges or diverges.

dx 1

0 ex ex = 0

1

ex ex + 1dx

ex exand the given integral converges if and only if both of the integrals on the righthand side converge.

1

Let us consider 0

dx

ex ex first. Since we have the linearizationex ex (1 +x) (1 x) = 2x,

centered at x = 0, we expect 1/(exex) tobehave like 1/(2x)near the bad

point0, and therefore this integral to diverge.

In fact, we have

L = lim1/(exex)

= lim

LH

lim1 1

x0+ 1/x

1 dx

x0+ ex ex = x0+ ex + ex 2Since 0


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46

c

x 0

e

Remark: The other improper integral on the right hand side converges. We have

L = lim1/(exex)

= lim1

= 1.

Since 0


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47

x

dx

also converges

by the Limit

Comparison

Test.

1e1/x

x

dx

converges.


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47

Part 2: Multi-variable Functions

1. Consider thepointP (3,5, 1) and the lineL x = 2t1,y = t+2,z = 2t; < t


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48

# have a normal nperpendicular toP Q = i j + 3k . Therefore we can take nto be

# Ii j kI

I

n = n P Q = II

3 4 1I = 11i10j7k,I

I1 1 3I

I I

or rather n = 11i+ 10j + 7k . This gives the equation of the plane as

11 (x 2) + 10 (y 3) + 7 (z (1)) = 0,

or 11x+ 10y + 7z = 45.

3. Let c be a constant. Show the anglebetween the position and the velocity vectorsalong the curve r = ect costi + ect sintj , < t


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49

r= ((ect cost) + (e sint) ) = e ,v= ((c ect cost ect sint) + (c e sint + e cost) )

= + 1e ,ct ct

2 ct ct

ct ct

2 1/2 c2 ct

ct ct 2t

r v = e cost (c e

cost e sint) + e

sint (c e

sint + e cost) = ce .


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50

+y

,

Therefore, if is the anglebetween r and v, we have

cos=r v

=ce2t c = rv ect c2 + 1ect c2 + 1

and we conclude that is constant.

4. a. Show that lim xy2

0 .(x,y)(0,0)x6 +y2

=

b. Show that limxy

does not exist.(x,y)(0,0)x6 +y2

Solution: a. We have 0 y2 x6 +y2 for all (x, y). Hence

xy20

x62= x

y2

x6 +y2x1 = x

for all x, y 0, 0 . Since lim(x,y)(0,0)

x

= 0, the Sandwich Theorem gives

2

limxy

0 .

(x,y)(0,0)x6 +y2=

b. The limit along thex-axis is

lim xy

lim x 0

lim0 0,

(x,y)(0,0)along the x-axis

x6 +y2=x0x6 + 02

=x0

=

whereas the limit along they =x line is

lim xy

lim x x

lim1

1 .

(x,y)(0,0)along the line y =x

x6 +y2=x0x6 +x2

=x0x4 + 1

=

Since these two limits are different, the two-variable limit lim xydoes not exist by the Two-Path Test.

(x,y)(0,0)x6 +y2

Remark: Let a, b,c, d be positive constants. It can be shown that the limit

a b

lim x

y

c d(x,y)(0,0) x + yexists if

a b1 , and does not exist otherwise.

c+

d>

5. Determine all values of the constant > 0 for which the limit

limx2y3

3

exists.(x,y)(0,0) x

+ y


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51

3

/3

2

2 3

2

2

Solution: We observe that

2 3

3 2 3

limx y3

=lim(y ) y

3 3

y

= lim1

lim .(x,y)(0,0)

along the curve x =y3

x+ y

y0 y

+ y

y0 y

y0 1 + y9

lim

1

9

is 1 if

> 9 and 1/2 if

9. On the other hand lim

yy0 does

y0 1 + y ynot exist. Therefore the limit of

x yalong the curvex

y3 does not

x+ y

=

x2y3exist, and consequently the two-variable limit lim

3 does not exist

either for 9.(x,y)(0,0) x

+ y

Now we will show that the limit is 0 if < 9. We have

x2y3

x/y 2 3 3 3 30 3

= (

3 3 y / y /

x

+

y

(x

/y

/ ) + 1

for all (x,y)2=/ (0,0). Here we used the fact that if t 1, then t2 t3

and hencet

t3 + 1

1, and if 0 < t < 1, then t2 1 gives

t 0 and

a local maximum for a < 0. On the other hand, (0,0) = 9a2 implies that(0,0) is a saddlepoint for a =/ 0.

Now we look at the sole critical point (0,0) in the case a = 0. As (0,0) = 0,

the second derivative test fails in this case. If we restrictf(x, y) =x3 +y3 tothex-axis we getf(x, 0) =x3 . Since this single variable function does not have

a local maximum or minimum atx = 0,f(x, y) cannot have a local maximumor minimum at (0,0) either. We conclude that (0,0) is a saddle point whena = 0.

12. Find the absolute maximum and minimum values of the functionf(x, y) = 2x3+

2xy2x y2 on the unit diskD = {(x, y) x2 +y2 1}.


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56

1

2

Solution: We first find the critical points off(x, y) in the interior ofD. At

a critical point we havefx = 6x2 + 2y2 1 = 0 andfy = 4xy2y = 0. Thesecond equation implies thaty = 0 orx = 1/2. Substituting these into the firstequation we obtainy = 1/ 6 in the first case, and no solution in the second case. Therefore the criticalpoints are (x, y) = (1/

that both of thesepoints lie in the interior ofD.

6, 0) and (1/

6, 0).Note

Now we look at the boundary ofD, that is, the unit circlex2 +y2 = 1. We

can solvey asy = 1x2 , 1 x 1. These solutions correspond to theupper and lower semicircles. Thenf(x, 1x2) =x2 +x 1 for 1 x 1,and

df x,

x2

2x 1 0 x 1

. This gives the critical points

dx( 1 ) = + =

=

2

(x, y) = (1/2,

3/2)and (1/2, 3/2) of the restriction off to the boundaryofD. We must also include the endpoints x = 1 andx = 1, in other words,the points (x, y) = (1,0) and (1,0) in our list.

Hence the absolute maximum and the absolute minimum off onD occur atsome of the points

(1/ 6, 0), (1/ 6, 0), (1/2, 3/2), (1/2, 3/2), (1,0), (1,0) .

The values off at thesepoints are

1

2 5 5

,3 3 3

, , , 1, 1,3 4 4

respectively. Therefore the absolute maximum value is 1 and the absolute

minimum value is 5/4.


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57

2+y2

Remark: Here are two more ways of dealing with the critical points of the restriction of

f to theboundary ofD.

In the first one we parametrize theboundary,which is the unit circle,byx = cost,y = sint,

< t


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58

1,1/2,1/2,13/27,13/27 , respectively. Therefore the absolute maximum is1 and the absolute minimum is 1.


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59

2

2

14. Evaluate the following integrals:

1 13

a. 0

y

sin(x ) dx dy

b. R

y ex2

dA whereR = {(x, y) 0 y x}2

c. 0

0

2yy xyx2 +y2

dx dy

dy dxd.

00 (x2+y

2 2 + 1

Solution: a. We will first express the iterated integral as a double integral

and then reverse the order of integration. Thex-integral goes from x =y to

x = 1 as shown by the red line segments in the figure in thexy-plane. Then

they-integral goes fromy = 0 toy = 1.

Therefore the intervals of thex-integral trace out the regionR bounded by

the parabolay =x2, the linex = 1, and thex-axis. Note that thex-integralsare always from left to right in the interval 0 y 1. Hence we have

1 13 3

0

y

sin(x ) dx dy = R

sin(x ) dA .

Now weexpress this double integral as an iterated integral with they-integralfirst. The green line segment in the figure shows the interval of integration for

they-integral which goes fromy = 0 toy =x2. Then:


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60

3

= lim ([te ]0 +

c

2

2

2

2 2

t t

1 13 3

0

y

sin(x ) dx dy = R

sin(x ) dA

1 x23

= 0

0

sin(x ) dy dx

13 y=x

2

= 0

sin(x )y]y=0 dx

1

= 0

sin(x )x dx

1 3 1

= 3

cos(x )]0

2

=3

b. We integrate with respect toy first to obtain

x

R

yex

2 dA = 0

0

y

ex2 dy dx

1

3 x2

=3

0

x e dx

1 t

=6 0

te dt

1=

6

where we used the integration by parts

0 te

dt = 0 td(e )c

t c

c0

c

et

c

dt)

limcec

lim[et]0

LH

= lim 1 lim(ec 1)

in the last step.

cec= 1

c

c. This time we will use the polar coordinates. To do so we first determinethe region of integrationR.

In the iterated integral the x-integral goes fromx = 0 to x =

y y2 asshown by the red line segment in the figure. Sincex =

y y2 = x2 =

2y y = x + (y 1) = 1 , x = y y gives the right semicircle of the2 2 2 2 2 2x2 2

circle + (y 1) = 1. On the other hand, they-integral goes fromy = 0 to

2

y = 2. ThereforeR is right half of the diskx2 + (y 1)2

1. Note that the

polar equation of the circlex2 + (y 1) = 1 is r = 2sin, and to obtain theright semicircle we vary from = 0 to = /2.


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61

2

0 2

2

y 2 2

Hence:

2 2yy xydx dy = xy dA

00 x2 +y2 Rx2 +y2

/2

= 0

0

/2

2 sin r cos r sin

r2r dr d

2 sin

= 0

0

sincosr dr d

/2 r2r=2 sin

= sincos[ ]r=0

dr d

/23=

02sin

cosd

sin4/2

= ]0

1=

2

d. Again we use the polar coordinates. This time the integration region isR = {(x, y) x 0 andy 0},that is, the first quadrant.

0

0

dy dx2 2 2( + ) + 1

= R (x2

1dA

+y ) + 1/2

= 0

0

r dr dr4 + 1


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62

2

4

4

2

= 0

/2 1

lim arctan r22 c

r c)]

r=0d

= 0

2

/2d

4

=8

15. Consider the triple integralD

(x21

+y +z2 2dV whereD is the regionbounded

)

by the cylinderx2 +y2 = 1 on the sides and by the hemispherez =

x2 y2 at thebottom. Express this integral in terms of iterated integrals in

a. Cartesian coordinates,

b. cylindrical coordinates,

c. spherical coordinates, and

d. evaluate the integral in the coordinate system of your choice.

Solution: We observe that the projection of D to the xy-plane is the unit diskx2 + y2 1. A vertical line passing through a point of the unit disk enters theregion at a point on the hemisphere and remains in the region from

there on. The equations of the hemisphere z =

x2 y2 and the cylinderx2 +y2 = 1 are z =

4r2 and r = 1, respectively,in cylindrical coordinates.

The answers to parts (a),

1

2dV =

1 1x 1 2 dz dy dx,D (x2 +y2 +z2)

and (b),

1 1x2 4x2y2 (x2 +y2 +z2)

1 2

2dV =

1

1 2 r dz dr d,D (x2 +y2 +z2) 0 0 4r2 (r2 +z2)immediately follow from these observations.

To do part (c) we further observe that

the equations of the hemisphere and the cylinder are= 2 andsin= 1,respectively, a ray starting at the origin enters the region at a point on the hemisphere and

leaves the region at a point on the cylinder, and such a ray intersects the region exactly when 0 /6. (The ray passes througha point on the intersection circle when = /6.)


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63

1

2

24

Therefore,

1

2dV =

2 /6

csc

2 sinddd.

D (x2 +y2 +z2) 0 0 2

4

We use this last iterated integral for the computation (d):

12

dV = 2 /6

csc 1

sinddd

D (x2 +y2 +z2)

0 0

2 /6

2 2

1 =csc = 0

0

[

] =

sindd

2 /6 1= 0

0

(sin +2

) sindd

2 /6 1cos2 sin= 0

0

( 2

+2

) dd

2 sin2 cos=/6

= 0 [

2+

] d=02

3 3 1=

0(

12+

8

4+

2)d

3= (16

4)


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64

2

16. The region enclosed by the lemniscate r2 = 2cos(2) in the plane is the base of a solidright cylinder whose top isboundedby the spherez =

2r2. Find the cylinders

volume.

Solution: Let R be the the region enclosed by the lemniscate in the firstquadrant. By symmetry the volume is

4

R

2r2 /4dA = 40

0

2 cos 2

2r2 r dr d

/4 1 r=2 3/22 cos 2 4 /4 3/2 3/2= 4

0[

3(2 r )

]r=0

d=3 0

(2(2 2cos2)

) d

8

2 /4 2 3/2 2

2 32 /4 3

=3

0

(1(2sin ) ) d= 3 3 0 sin d2

2 32 /4 2 2

2 32 1

=3

3 0

(1 cos ) sind=3

3 1

(1 u ) du2

2 32

u31 2

2

/ 2

32 2 5

= 3

3 [u

3 ] = 3

3 (3

)

2

2

1/ 2 6 2

64 40

2=

3

9+

9.


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65

17. Consider the iterated integral

/2 1 2r2

in cylindrical coordinates.

0

0

r

dz dr d

a. Change the order of integration into dr dz d.

b. Express the integral in spherical coordinates with order of integration ddd.

Solution: a. z = r is the conez2 =x2 +y2, andz = 2r2 is the paraboloidz =2 x2 y2. These surfaces intersect along a circle that is also the curve ofintersection of the cylinder r = 1 and the horizontal planez = 1. Since

0 r 1 and 0 /2, the integration region D is the region in the first

octantboundedby theparaboloid on the top and the cone at the bottom.


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66

2

When we integrate with respect to r first, we will be moving along a curve on whichzand are constant. These two conditions describe a horizontal plane and a planecontaining the z-axis, respectively. Therefore the r-integration is along a ray

perpendicular to thez-axis like the blue ones in the figure. Such a ray starts on thez-axis inD, and leavesD through the cone if 0 z 1, and through theparaboloid if1 z 2. Therefore,

/2 1 2r2 /2 1 z

0

0

r

dz dr d= 0

0

0

dr dz d

/2 2 2z

+ 0

1

0

dr dz d

where we used the fact that z = 2r2 and r 0= r =2z .

b. We have/2 1 2r2 1

0

0

r

dz dr d=D r

dV

as dV = r dz dr d in cylindrical coordinates where D is the region described inpart (a).

Since we want to integrate with respect to first, we will be moving along thecurves on which and are constant. These two conditions describe a spherewith center at the origin and a half-plane whose spine is the z- axis,respectively. Therefore the -integration takes place along a vertical semicirclesubtended by a diameter along the z-axis and with center at the origin like thegreen ones in the figure. Such a semicircle starts on the positivez-axis in D,and leave the region of integration intersecting the cone

for 0

2and the paraboloid for

2 2. The upper half of the

conez = r has the equation = /4 in spherical coordinates. On the other

hand,z = 2r2 = cos= 2(sin) = cos= (1 + 42 7)/(2) fora point on theparaboloid as /4. Therefore,

1 1

D r dV =Dsin dV/2

2 /4

= 0

0

0

ddd

/2 2 arccos((1+ 42 7)/(2))

+ 0

20

ddd

where we substituted dV = 2 sin ddd for the volume element inspherical coordinates.


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0 2y+3 x +y x 2y18. Evaluate the iterated integral1

y

2e

(x 2y) dx dy .Solution:

(x,y) (u, v)19. Let (u, v) and (x, y) be two coordinate systems. Show that

(u, v)

(x,y)= 1.

Solution: By definition,

(x,y) xu xv x y x y

and

(u, v)=

yu yv=

u v v u

(u, v) ux uy

u v

u v .

(x, y) = vx vy

=

x y

y x


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68

,

Therefore,

(x,y) (u, v)x y x y u v u v

(u, v)

(x, y)= ( u v

v u )( x y

y x)

=xu uxyv vy +xv vxyu uy xu uyyv vx xv uxyu vy=xu uxyv vy +xv vxyu uy xu uyyv vx xv uxyu vy

+xv vxyv vy +xu uxyu uy xv vxyv vy xu uxyu uy= (xu ux +xv vx) (yu uy +yv vy )

(xu uy +xv vy ) (yu ux +yv vx)

=xxyy xy yx= 1 10 0= 1

20. Evaluate the double integral R e x2/ydA whereR = {(x, y) x2

y

x .

Solution: We change the coordinates to u =y2/x,v =x2/y. In this coordinatesystem the region of integration becomes G = {(u, v) 0 u 1 and 0 v 1}.

We have

(u, v) ux uy y2/x2 2y/x

3= (x, y) 1 1

= =(x, y)

= vx vy

=

2x/y x2/y2=

(u, v)

(u, v) 3(x,y)

and the change of variables formula gives

x2/y

v (x,y) 1 1

v 1 e 1

Re

dx dy =G

e

(u, v)

du dv = 0

0

e

3

dv du =3

.


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66

Iz

Remark: We can multiply differentials. The rules are du du = 0 = dv dv and dv du =du dv in any uv-coordinate system. For instance, when changing from Cartesian to

polar coordinates, we have

dx dy = d(r cos) d(r sin)

= (cosdr r sind)(sindr + r cosd)=r sin2ddr + r cos2 dr d

= r(sin2 + cos2 ) dr d

= r dr d

and in Example 19 we have

du dv = d(y2/x)d(x2/y)

= (2y/xdy y2/x2dx)(2x/ydx x2 /y2 dy)= 4dy dx + dx dy

= 4dx dy + dx dy

= 3dx dy,

hence dx dy1

du dv . This can be used to keep track of how the area element changes= 3

under a coordinate change,but note that the sign of the factor in front must be correctedby hand so that it is positive on the region of integration.

21. Compute the Jacobian(x, y ,z)

(,,)

(x, y, z) are the Cartesian coordinates.

where(,,)

are the spherical coordinates and

Solution: We havex =sincos,y =sinsin,z =cos. Therefore

Ix x xI(x, y, z) I IIy y yI

(,,)

=II

z zII

Isincos coscos sinsinII I= Isinsin cossin sincosII

cos sin 0IIsinsin cossin

= (sin sin)

cos sin

sincos coscos

(sincos)

cos sin

sincos coscos

+ 0 sinsin cossin

= (sin sin)(sin)(sin2 + cos2 )

(sincos)(cos)(sin2 + cos2 )

+ 0

=2 sin(sin2 + cos2 )

=2 sin

where we used the cofactor expansion with respect to the third column.


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67

2

3 1

Part 3: Sequences and Series

1. Let a1 = 1, a2 = a3 = 2, a4 = a5 = a6 = 3, a7 = a8 = a9 = a10 = 4, and so on. That is,

an 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, . . . . What is a2013?Solution: an = k if 1 + 2 + + (k 1) < n 1 + 2 + + k. In other words, an =k if k(k 1)/2 < n k(k + 1)/2. Since for k = 63, k(k 1)/2 = 1953 and k(k + 1)/2= 2016, we have a2010 = 63.

8n+ 1 1

Remark: A more explicit formula an = 2

can be obtained.

2. Curve1 is an equilateral triangle with unit sides. For n 2, Curven is obtained from

Curven1 by replacing the middle third of every edge with the other two sides of theoutward pointing equilateral triangle sitting on it. LetLn be the length of Curven and

An be the area of the region enclosedby Curven. FindLn,An, limLn and limAn.n n

Solution: Let en and dn denote the number of edges and the length of each

edge of Curven, respectively. Since en = 4en1 and dn = dn1/3for n 2, and

e1 = 3 and d1 = 1, we find that en = 34n1 and dn = 1/3n1 for n 1. It followsthatLn = endn = 3 (4/3 n1 for n 1, and limLn = lim 3 (4/3 n1 =.

n n

On the other hand, the nth region is obtained by adjoining enst

1 equilateraltriangles of side length dn1/3to the (n 1) region. Therefore,

An =An1+ en1

3(dn1/3)

n1

2

n2

and then

4=A

+ 3 4 4 ( 3n1)

1 4 n2 4 n3An =

4

3((

9)

3

+ (9)

+ + 1)+A1

for n 2. SinceA1 = , this gives4

3

3 4 n1

3

for n 1. We obtain

An =20

(1(9)

) +4

3

3

3

2

3limAn

n 20+

4=

5.


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68

Curve1 : Curve2 :

Curve3 : Curve4 :

Remark: A magicianputs 1ball into an empty box at t = 0 sec, takes out 1ball at t = 1/2 sec, puts2balls at t = 3/4 sec, takes out 1ball at t = 7/8 sec, puts 3balls at t = 15/16 sec, takes out 1ball at t= 31/32 sec, and so on. Then she challenges you to guess how manyballs there are in the box at t = 1 sec.

You ask for the advice of your friends.

Friend1 says: The net result of 2k + 1st and 2k + 2nd steps for k 1 is to put atleast one more ball into the box. There are infinitelymany balls in the box at t = 1sec.

Friend2 says: Imaginethat the balls are labeled as ball1, ball2, ball3, and so onwith invisible ink which only the magician can see. Then sheputs ball1at t = 0 sec, takes out ball1 at t = 1/2 sec, puts ball2 and ball3 at t = 3/4 sec, takesout ball2 at t = 7/8 sec, puts ball4, ball5, and ball6 at t = 15/16 sec, takes outball3 at t = 31/32 sec, and so on. For any given k, ballk is not in the box at time t= 1 sec,because it was taken out at time t = 11/22k1 sec.There are no balls in the box at time t = 1 sec.

What do you think?


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69

n=1

n=1

n=1

n

1

3. Let the sequence {an} be definedby a1 = 1 and an =1sequence converges and find its limit.

1

+ an1

for n 1. Show that the

Solution: Observe that if 0 < an < an+1, then 0 < 1 + an < 1 + an+1 andan+1 = 1/(1 + an) > 1/(1 + an+1) = an+2 . Similarly, if an > an+1 > 0, then

1+

an > 1+

an+1 > 0 and an+1 = 1/(1+

an) < 1/(1+

an+1) = an+2 . Sincea1 = 1 > 1/2 = a2, it follows that

1 a a a a a a1

.= 1 > 3 > 5 >> 6 > 4 > 2 =2

Therefore, the sequence {a2n} is increasing and bounded from above by

1, and the sequence {a2n1} is decreasing and bounded from below by1/2. By the Monotonic Sequence Theorem, then both of these sequences are

convergent.

Let lim a2n1=L and lim a2n =M . Taking the limit of a2n+1 = 1/(1 + a2n) asn n

n we obtainL = 1/(1 +M ), and taking the limit of a2n = 1/(1 + a2n1) as

n we obtainM = 1/(1 +L). FromM +M L = 1 andLM +L = 1 it followsthat M =L. Therefore the sequence {an}

converges toL =M .Finally, L2 +L 1 = 0 givesL = ( 51)/2 orL = ( 5 + 1)/2. Since 0 < anfor all n 1, we must have 0 L . HenceL = (

5 1lim an = .

51)/2 is the limit and

n 2

Remark: This result is sometimes expressed symbolically in the form:

51 1

2=

11 +

11 +

11 +

11 +

11+

+

Remark: Note that xn =Fn/Fn+1 for n 1 where Fn 1, 1, 2, 3, 5, 8, . . . is the Fibonaccisequence. So we showed that limFn/Fn+1 = (

51)/2.

Remark: It can be showed with a little bit more work that any sequence satisfying the given recursion relation converges to ( 51)/2 if a1 =/ (1+ 5)/2 and a1 =/ Fn+1/Fn forany n 1. In the first of these cases the sequence is constant and in the second an = 1and an+1 is undefined.


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n1

1

2

2

0

4. Let the sequence {xn} be definedbyx0 = 2 andxn =

xn1

2+x

1

n1

for n 1.

a. Find the limit of this sequence assuming it exists.

b. Show that the limit exists.

Solution: a. We assume that the limit L = limxn exists. Then

2

xn =xn1

2+x

1

n1

for n 1= xnxn1=x

2+ 1 for n 1

2n1

2 L

22

= lim(xnxn1) = lim 2+ 1= L

=

2+ 1= L = 2

as limxn1= limxn =L . ThereforeL = 2 orL = 2.

x0 = 2 > 0 and ifxn > 0 thenxn+1 =xn/2+ 1/xn > 0+ 0 = 0. Henceby induction

xn > 0 for all n 0. It follows that L = limxn 0. We conclude that the limit

is

2.

b. In part (a) we proved thatxn > 0 for all n 0. Therefore {xn} is boundedfrom below.

Now we will showby induction on n that

2


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n! n 7

n 3

lim

x 3

= =

n!

=

n

n

3

n6

n

n

1

5. Determine whether each of the following series converges or diverges.

n lnn

n+1a. (21/n21/(n+1)) b.n=1

nn=2

c. (1)n=1

cos(n

)

nn 1 5n2nd. e.

n=1

n=1 +

n sinn

f. n

n=1

1 sinng.

n lnn ln= (lnn)h.

2n=1

n n n

Solution: a. We havesn = (21/k21/(k+1))= 21/k21/(k+1) = 221/(n+1).

k=1 k=1 k=1

Hence limsn = 1. Therefore the series converges and (21/n21/(n+1))= 1.

nn=1

b. We have an = n lnn/3n and

n+1

liman+1 (n + 1) ln(n + 1)/3

=

1= lim n + 1 lim ln(n + 1)

n an nLH

n lnn/3n

3 n n n lnn

1=

31 lim

ln(x + 1)

lnx=

1lim

x

1/(x+ 1)

1/x=

1 1

31 =

3.

Since= 1/3 < 1, the series

n=2

n lnn

3nconvergesby the Ratio Test.

c. Since an = (1n+1

cos(/n), we have an= cos(/n) for n 2, andlim an= lim cos(/n) = cos0 = 1. Therefore lim an =/ 0, and the seriesn n n

divergesby the nth Term Test.

d. We have an = nn/n!and

an+1n + 1 n+1

n/(n + 1)! 1

limn an

lim

(n nn/n!

= lim (1+ )= e .

nnSince= e > 1, the series divergesby the Ratio Test.

n=1

e. We observe that

L limn

1/(n +

n sinn)

1/n

limn

+

1sinn

n

1=

1 + 0

= 1 .

1Since 0


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1

+

n sinn

diverges

by the

Limit

Compariso

n Test.


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5

7

n 3

n

n n

n

2

n! 1

n

f. The geometric series (5/7)n=1

converges as r = 5/7 = r = 5/7 < 1.

nn

We also haveL = lim(

2n)/(7n6n) = lim 1(2/5)n

=

10 = 1. Sincen

5n/7n

5n2n n1(6/7) 100


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Remark: The following was abonus problem onMoodle in Spring 2010 Math 102 course.

Problem K: In class we showed that 0.12= 0.121212.. . = 12/99 as an application

of the geometric series.

We learn the following trick in elementary school: Ifx = 0.121212. . . , then 100x=

12.121212. . . ; their diff erence gives 99x= 12 and thereforex = 12/99. Of course

when we see this in elementary school, no one talks about convergence.

That is what we are going to do in this problem. We will forget about convergence as weknow it, and take a trip to Tersonia.

In Tersonia after they teach the students about integers and rational numbers, theycome to the decimal representations of numbers. As you know our decimal expansionshave the form

dndn1. . . d2d1d0.d1d2. . .where each di is in {0,1, . . . , 9}. We can have infinitelymany nonzero digits after the

decimal point, but we must have only finitely many nonzero digits before the decimalpoint. In Tersonia they do just the opposite. Their decimal expansions have the form. . . t3t2t1t0.t1t2. . . tn

where each ti is in {0,1, . . . , 9}.Note that there is no minus sign. They can haveinfinitelymany nonzero digitsbefore the decimal point, but they can only havefinitely many digits after the decimalpoint.

Take a few minutes to convince yourself that Tersonians can add and multiply theirdecimal expansionsjust like we do.

Why no minus sign? Well, because Tersonians dontneed it. Negative numbers arealready there. For instance, consider the numbery = 12.0 = . . . 121212.0.

Then 100y= . . . 121200.0 and 99y = 12. Therefore y = 12/99 . So in fact 12.0

is a negative number.

Here are some problems from Tersonian Elementary School Mathematics Book :

a.1

2=

? b.

1

3=

? c.

1

7=

? d. 1 =?

e. Find two nonzero numbersA andB such that A B = 0.

Part (e) was later turned into a programming challenge. A Java applet that computes

the last n digits of A and B when their last digits are given can be found at

http://www.fen.bilkent.edu.tr/~otekman/calc2/ters.html .
http://www.fen.bilkent.edu.tr/~otekman/calc2/ters.htmlhttp://www.fen.bilkent.edu.tr/~otekman/calc2/ters.html


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n

n

.

We have cn =

2

n

2

.

n x

n

6. Find the radius of convergence of thepower series (1)n=0

(2 + 1)(n2.

+ 1)

Solution: We will give two diff erent solutions.

Solution 1: Here we will use the Ratio Test.

n xn

We have an = (1)

(2n+ 1)(n2+ 1)

an+1

and

limn an

n+1xn+1

= lim(1)

(2n+1

+ 1)((n

n

+ 1) + 1)

n x

1 (2n+ 1)(n2+ 1)xn+1

limn

(2n+1

+ 1)((n+ 1 2x

+ 1)

(2n+ 1)(n2+ 1)

= lim (1 + 2n

n2 + 1

2 )x

n

x=

2

2 + 2n (n + 1) + 1

If x< 2, then= x/2< 1 and thepower series converges absolutely by theRatio Test. On the other hand, if x > 2, then= x/2> 1 and the power

series diverges by the Ratio Test.

It followsby the definitionof the radius of convergence that R = 2.

Solution 3: Here we will use the radius of convergence formulas.

(1 n

(2n+ 1)(n2+ 1)

, and the radius of convergence formula gives

1

R= lim cn+1 = lim 1/((2n+1+ 1)((n+ 1) + 1))

n cn n 1/((2n+ 1)(n2+ 1))= lim (

1 + 2n

n2 + 1 12

) =

ThereforeR = 2.

n 2 + 2n (n + 1) + 1 2


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n n

n

2

2

2

n

.

n n

n

n

n n

n

n x

2n+1

7. Consider thepower series (1)n=0

(2 + 1)(n2.

+ 1)

a. Find the radius of convergence of thepower series.

b. Determine whether thepower series converges or diverges at the right endpoint of itsinterval of convergence.

c. Determine whether thepower series converges or diverges at the left endpoint of itsinterval of convergence.

x2n+1Solution: a. We have a 1

(2n+ 1)(n2+ 1)and

= lim an1/n

x 2n+1 1/n

= lim (

)n (2n+ 1)(n2+ 1)

2

1/n

= lim

x

xn2 (1 + 2n 1/n

2

n1/n (1 + n2 1 n

x=

2

If x 2, then= x /2 > 1 and thepower series divergesRoot Test; and if 2by the Root Test. ThereforeR =

2.

b. At x =

2we have

2n+1

n x

2n+1 n ( 2)

(1)

n=0(2 + 1)(n2 + 1)

= (1)n=0

(2 + 1)(n + 1)

n 2n

= 2 (1)n=0

(2 + 1)(n2.

+ 1)

Consider the corresponding absolute value series

n=0 (2

2n

+ 1)(n2+ 1). Since

2n 10 < 1(2n+ 1)(n2+ 1) n2

2nn 1 n

2

converges,n

n=0 (2 + 1)(n2 + 1)convergesby the Direct Comparison Test; and

then the power series at x =

2 converges absolutely by the Absolute

Convergence Test.

c. At x = 2 we have

2n+1

n x2n+1

n ( 2)

(1)n=0

(2 + 1)(n2 + 1)= (1)

n=0(2 + 1)(n2 + 1)

n n+1 2= 2 (1)

n=0 (2 + 1

)(n

.+ 1

)


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n=0

n=0

n

n=0

This is just 1 times the series we considered in part (b), and therefore itconverges absolutely.

8. Determine the radius of convergence and the interval of convergence of thepower series

xnn+1

.3n + (1)

Also determine the type of convergence at each point of the interval of convergence.

Solution: As cn =3n

1

1 n+1, the formulas for the radius of convergence

gives+ ()

1

R = lim

cn+1= lim

1/(3(n+ 1) + (1 n+2)

n+1= lim

3 + (1n+1

/n

n+2

= 1.

cn 1/(3n+ (1) ) 3 + 3/n+ (1) /nThereforeR = 1. Hence we have absolute convergence for x < 1 anddivergence for x> 1.At x = 1 we have

xn

n+1=

1

n+1.

Since

3n + (1)

n+1

n=03n + (1)

L = lim1/(3n+ (1) ) = lim

1 1n+1

=n 1/n

n3 + (1) /n 3

1is a positive real number and the harmonic series diverges, the power

n=1

series diverges by the Limit Comparison Test atx = 1.

At x = 1 we have

xn

n+1=

(1)n+1

.

Let un =3n

3n + (1)

1

1 n+1.

n=03n + (1)

+ ()

i . u

1n =

3n 1n+1 > 0 for all n 1.

+ (

)

n+1n+2

ii. 0 < 3n + (1) 3n + 1 < 3n + 2 3(n+ 1) + (1) for all n 1. Henceun > un+1 for n 1.

1iii. lim un = lim n+1 = 0.

n n3n + (1)


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1

3n

( )=1

R

n

n

)

It follows that the series 3n

( 1 n 1

convergesby the Alternating Series

n=0+ () +

1

Test. Its absolute value series is the same as the series 1+

n=1 + (1 n 1and we showed that this series diverges. Hence the power series convergesconditionally at x = 1.

To summarize, the radius of convergence isR = 1, the interval of convergenceis [1, 1),thepower series converges absolutely at every point of (1,1),and

it converges conditionally atx = 1.

1 3 5 (2n1) n9. Consider thepower seriesf(x) = 1 + n=1

2 4 6 x .(2n)

a. Show that the radius of convergence of thepower series isR = 1.

b. Determine the behavior of this power series at the endpoints of its interval ofconvergence.

c. Show that 2(1 x)f(x)=f(x) for x< 1.d. Solve this diff erential equation to show thatf x

1x for x< 1.

e. Show that 1 3 5 (2n1) x2n+1

for x< 1.arcsinx =x +

n=1 24

6

(2n)

2n

+

1

1 3 5 (2n1)Solution: a. We have cn =

of convergence formula2 4 6 (2n)

for n 1. We use the radius

1 cn+11 3 5 (2n1)(2n + 1)

2 4 6 (2n)(2n+ 2) 2n + 1limn

cn

lim

n1 3 5

(2n1) = lim = 1n2n + 2to obtainR = 1.

2 4 6 (2n)

1 3 5 (2n1)b. At x = 1 we have the series 1 +

2 4 6 . Since

2n

1 3 5 (2n1) n=1

3 5

( )

2n1 1 12 4 6 (2n) = 2 4 2n 2 2n > 2n

1for n > 1 and the harmonic series

diverges, we conclude that the series

n=1atx = 1 diverges by the Direct Comparison Test.


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2

n

2 2

2

2

2 2

2 2

2

2 2

2

4

n 1 3 5 (2n1)At x = 1 we obtain the alternating series 1 + (1)

2 4 6 2n

1 3 5 (2n1) n=1 ( )with un =

2 4 6 (2n) . We have (i) un > 0 for all n 0, and we alsohave (i i) un > un+1 for all n 0 as un/un+1 = (2n+ 2)/(2n+ 1) > 1 for n 0.On the other hand,

u21 3 5 (2n1)

2=

1 3 3 5

(2n3)(2n 1)

2

2n 1 1 1 1. Therefore 0 < un < 1/ 2n for n > 1, and the Sandwich Theoremgives (i i i) lim un = 0. We conclude that the series atx = 1 converges by the

Alternating Series Test. The convergence is conditional as we have alreadyseen that the absolute value series diverges.

c. For

x

< 1 we have

1 3 5 (2n1) nf(x) = 1 + n=1

2 4 x .6 (2n)Diff erentiating this we get

1 1 3 5 (2n1) n1f(x)= + n=2

4 6 (2n x2) 2for x< 1. Therefore

1 3 5 (2n1) n

and

2xf(x)=x + n=2

6 (2n

x2)

3 1 3 5 (2n+ 1) n2f(x)= 1 + x + n=2

2 4 x6 (2n)for x< 1. Taking the diff erence of these two, we obtain

1 1 3 5 (2n1) 2n + 1 n2(1 x)f(x)= 1 + x + n=2

4

6

(2n

2)

(2n

1)x

1 1 3 5 (2n1) 1 n= 1 + x + n=2 4 6 (2n2) 2nx

1 1 3 5 (2n1) n= 1 + x + n=2 2 4 x

6 (2n)

1 3 5 (2n1) nfor x< 1.

= 1 + n=1

=f(x)

2 4 x6 (2n)


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x

x

d. For x< 1,

f(x) 1

2(1x)f(x)=f(x) =f(x)

=2(1 x)

1

= lnf(x)= 2

ln1x+ C for some constant CA

= f(x)=1 for some constant A .

1Now substitutingx = 0 gives 1 =f(0) =A. Therefore,f(x)=

1 for

x< 1.e. We have

1 =

1x

1 + n=1

1 3 5 (2n1)xn

2 4 6 (2n)for x< 1. Substitutingx

2 forx gives

1 =

1x2

1 + n=1

1 3 5 (2n1)x2n

2 4 6 (2n)

for x2< 1. Integrating this we obtain 1 3 5 (2n1) x2n+1

arcsinx =x + 2 4 6

2n

+ Cn=1 ( ) 2n + 1

for

x

< 1. Substitutingx = 0 gives 0 = arcsin0 = C, and hence C = 0. Thus

1 3 5 (2n1) x2n+1for x< 1.

arcsinx =x + n=1 2 4 6 (2n)

2n + 1

Remark: It can be shown that

1 =

1xfor 1 x < 1, and

1 + n=1

1 3 5 (2n1)xn

24 6

(2n)

1 3 5 (2n1) x2n+1for x1. arcsinx =x + n=1 2 4 6 (2n)

2n + 1


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=

3

3n

4 4

=

2

10. Consider thepower series n=1

n(2x1 3n+1

5n.

a. Find the radius of convergence of thepower series.

b. Find the sum of thepower series explicitly.

Solution: a. an =n(2x1 3n+1

5ngives

an+1

lim

n

an

3(n+1)+1

n+1

n + 1)(2x1) /5 lim

(

n

n 1= lim

n(2x1)2x1

+1/5n

n n 53

2x1 .

5

If 2x1 3/5 < 1, then< 1 and the series converges by the Ratio Test; and

if 2x1 3/5 > 1, then> 1 and the series diverges by the Ratio Test. Since

3

31 52x1 /5 < 1 x

2 0 for n 4, we have n! 2n(n3)/2

0 for n 2= (n + 1)(2n+ 1) > 22(2n + 1) for n 2. Hence

un > un+1 for n 2.


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4

5

2 3 4 5

2 3 4 5

n=0

n

4niii. lim un = lim (

2) = 0.

n n n! 2n + 1

The series satisfies the conditions of Alternating Series Test for n 2. Since

223/(23 11!) = 32768/3586275 0.009 < 0.01, it follows by the Alternating

Series Estimate that the sum

2x2

23 25 219 221 122237581820

e

dx 23

+2!5

+ 9!19

+10!21

=13749310575

0.89

2

approximates 0

e

x2dx with error less than 0.01.

13. Find the exact value of

n

n=0

1.

(2n+ 1)

Solution: On one h

Differentia Calculus

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