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L’Hopital’s Rule (62) Note that the quotien is still indeterminate at x = π/2. We removed this indeterminacy by cancelling the factor − cos x. 3 4 2 8 U seL'H opital'sRuleto evaluate lim . 2 20 x x x x 2 3 2 3 12 lim 6 1 3 7 4 2 4 x x x 2 2 4 Evaluate lim . sin x x x 2 2 4 lim cos 4 x x x 2 /2 cos Evaluate lim . 1 sin x x x /2 /2 2cos sin lim lim 2sin co 2 s x x x x x x 2cos sin cos x x x A lw ayscheck for indeterm inateform . 0 0
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### Transcript of L’Hopital’s Rule (62) Note that the quotient is still indeterminate at x = π/2. We removed this...

L’Hopital’s Rule (62)3

42

8Use L'Hopital's Rule to evaluate lim .

2 20x

x

x x

2

32

3 12lim

6

13 74 2 4x

x

x

2

2

4Evaluate lim .

sinx

x

x

2

2 4lim

cos

4x

x

x

2

/ 2

cosEvaluate lim .

1 sinx

x

x

/ 2 / 2

2cos sinlim lim 2sin

co2

sx x

x xx

x

Note that the quotient

2cos sin

cos

x x

x

is still indeterminate at x = π/2. We removed this indeterminacy by cancelling the factor − cos x.

Always check for

indeterminate form.

0

0

32

Show that converges and compute its value.dx

x

Step 1. Integrate over a finite interval [2, R].

22 2 2

22

23

1 1 1 1

2 2 2 4

1 1

8 2

RR

R

dxx x R

x R

Step 2. Compute the limit as R → ∞.

3 22

1

8

1 1lim lim

8 2

R

R R

dx

x R

Improper Integrals (63)

1

Determine whether converges.dx

x

11

n n0 l ll nR

R

dxx

xRR

1

lim lim lnR R

R

dxR

x

Improper Integrals (63)

9

0

?dx

x 1

f xx

The integral is improper because the integrand has an infinite discontinuity at x = 0.

lim b b

R aa R

f x dx f x dx

9

91/ 2 1/ 2

0 0 06

The integral converge

lim lim 2 lim 2

s

3

.

RR R RR

x dx x R

Improper Integrals (63)

Solve 0.3 4 with initial condition 0 1.y y y y

We'll rewrite 0.3 4 as 1.2 14

dy yy y y y y y

dt

0

0

11 /kt

ydy y Aky y t C

dt A e C y A

1.2

thus, 1.2 & 4

4so,

1 /t

k A

ye C

0

0

1

3

yC

y AC

1.2

4

1 3 ty

e

Logistics Growth (64)

0.40.4

1000 500 1

500 or1 9 500 1000 9

tt

ee

0.4 ln 99

0.5.493 s

4 yearte t

0.4

0.4 & 1000

0.4 11000

1000

1 /t

k A

dP Py

dt

ye C

0

00.4

1000

1 9

1

9 ty

yC

y A eC

Deer Population A deer population grows logistically with growth constant k = 0.4 year−1 in a forest with a carrying capacity of 1000 deer.(a) Find the deer population P(t) if the initial population is P0 = 100.(b) How long does it take for the deer population to reach 500?

0

0

1 & 1 /kt

ydy y Aky y C

dt A e C y A

ktyCe

y A

Solve for t

`

Same formula for any

point in the plane.

(64)

2

1

2

1

7 1 2 50 5

A B

x x xx xx

Finding the Constants

2?

7 10

dx

x x

2 52 5

2 5 2 5

1 5 2

x x A Bx x

x x x x

A x B x

1

2 1 33

15 1 3

3

x A A

x B B

2

1 1

7 10 3 2 3 5

dx dx dx

x x x x

1 1ln 2 ln 5

3 3x x C

Multiply by 2 5x x

Let a # which drops a parameterx

Partial Fractions

Partial Fractions (65)

Not enough for PRGM

Finding the Constants

2?

7 10

dx

x x

Partial Fractions (65)

Calculator:

APPS; A+ Calc; Calc Pack II; Partial Fractions

2

1 / 3 1 / 3 1 1

7 10 2 5 3 2 3 5

dx dx dxdx

x x x x x x

Awesome Program!

2 2 2 8 2 1 2 1 2 8x A x x B x x C x x

2 2 1 1

1 2 8 2 6 1 2 8 6 2

x dx dx dxdx

x x x x x x

2

1 2

2

81 2 8 22

A B C

x x

x

xx x x

2 2

?1 2 8 2

xdx

x x x

11 3 6 3

64 18 3 6 1

12 6 3 12

6

x A A

x B B

x C C

1ln 4

2

1 1ln 1 ln 2

6 6x xx C

Multiply by the

denominator of

our integrand.Let a # which

drops a parameter.

x

Partial Fractions (65)

2 2

?1 2 8 2

xdx

x x x

2

1 2

2

81 2 8 22

A B C

x x

x

xx x x

2 2 2 8 2 1 2 1 2 8x A x x B x x C x x

11 3 6 3

64 18 3 6 1

12 6 3 12

6

x A A

x B B

x C C

2 2 1 1

1 2 8 2 6 1 2 8 6 2

x dx dx dxdx

x x x x x x

1 1 1

ln 1 l1 1

ln 1 ln 26 6

1n 2 l

22 6n 2 8

6x

dux x C x x

uC

2 8 2u x du dx

Partial Fractions (65)

Finding the Constants Partial Fractions (65)

Calculator:

APPS; A+ Calc; Calc Pack II; Partial Fractions

Awesome Program!

2 2

?1 2 8 2

xdx

x x x

2 2 1 / 6 1 1 / 6

1 2 8 2 1 2 8 2

1 1

6 1 2 8 6 2

xdx

x x x x x x

dx dx dx

x x x

THEOREM 2 Vector-Valued Derivatives Are Computed Componentwise A vector-valued function r(t) = x (t), y (t) is differentiable iff each component is differentiable. In this case,

Calculate r”(3), where r(t) = ln t, t .

2

1 1' ,1 , ''

1'' 3 ,

0

09

,r t r tt t

r

' ' , 'd

r t r t x t y tdt

Velocity & Acceleration Vectors (66)

The Derivative as a Tangent Vector Plotting Tangent Vectors Plot r (t) = cos t, sin t together with its tangent vectors at and . Find a parametrization of the tangent line at .

/ 4t 3 / 2

/ 4t

' sin ,cosr t t t

2 2 3' , , ' 1,0

4 2 2 2r r

tangent line is parametrized by

2 2At , ,

4 4 2 2t r

and thus the

2 2 2 2, ,

4 2 24 2 2'r t rL t t

is the Unit Circlet r t

0 0 0Tangent at : 'r t L t r t t r t

x

y

4t

3

2t

Terminal Point

Velocity Vector

(66)

Parametric Acceleration (67)

,

' , ' and " , "

r t x t y t

v t x t y t a t x t y t

cos ,sin

sin ,cos and cos , sin

r t t t

v t t t a t t t

A particle travels along the path c (t) = (2t, 1 + t3/2). Find:(a) The particle’s speed at t = 1 (assume units of meters and minutes).

2 2Speed ' '

dsx t y t

dt

1/23' 2, '

2x t y t t

94

42.5 m/min

ds

dt

Parametric Speed (68)

Parametric Equations: Given a Velocity Vector, Find the Position Vector (69)

v t dt s t c 1 2, is constant vectorc c c

2 21 2

1 11, , ,

2 2t dt t t c t c t c

210 2,1 1, 2, 1

2s t dt t t

21, and 0 2,1

2, 12

1 s t t tv t t s

The particle stops when 0

' 0 an ' 0d

v t

x t y t

When Does a Particle on a Parametric Curve Stop? (70)

,

' , ' and " , "

r t x t y t

v t x t y t a t x t y t

1

1

The slopes of a tangent vector at

will be .

t

a t

Find the Slope of the Tangent to the Vector at t1 (71)

,

' , ' and " , "

r t x t y t

v t x t y t a t x t y t

/2 /22 2

0 0

/2/2

00 0

0

14sin 8 sin

2

4 1 cos 2

2

4 2 cos

2 2 sin 2 2 0 0

d d

d udu

u

The equation r = 4 sin θ defines a circle of radius 2 tangent to the x-axis at the origin. The right semicircle is “swept out” as θ varies from 0 to

Use Theorem 1 to compute the area of the right semicircle with equation r = 4 sin θ.

/ 2.By THM 1, the area of the right semicircle is

21.

2K f d

2 1

sin 1 cos 22

2 2u du d

Yesterday we graphed 2 cos .r a

,0a

Trig Identity...

Area Inside the Polar Curve (72)

Sketch r = sin 3θ and compute the area of one “petal.” Rectangular r f

r varies from 0 to 1 and back to 0 as θ varies from 0 to .

3

Polar Curve

r varies from 0 to -1 and back to 0 as θ varies from 2

to .3 3

r varies from 0 to 1 and back to 0 as θ varies from 2

to .3

21

2K r d

/3 /3

0 0

/3 /3

0 0

22

00

21 1 1 cos 6

2 2 2

1 1cos

sin 3

1

64 4

1 1cos sin

12 24 12 24 2

d d

d d

udu u

2 1 cos6sin 3

2

6 6u du d

Area Inside the Polar Curve (72)

sin cos/

/ cos sin

drrdy dy d d

drdx dx d rd

Find an equation of the line tangent to the polar curve r = sin 2θ when

4.

3

3sin 1 the point will be in the 4th quadrant.

2r

2 22 0

2 22cos2 sin cos

2cos2 cos sin 2 22 0

1

2 2

rr

rr

2 21

2 2y x

2cos

2

2sin

2

x r

y r

slope

1 1y y m x x

Slope of the Tangent to a Polar Curve (73)1m

Euler’s Method (74)n nt

0

1

2

3

4

5

1 1 1,n n n ny hF t y y

Let y (t) be the solution of sin cos , 0 0.y t y y

Use Euler’s Method with time step h = 0.1 to approximate y (0.5).

APPS; Calc Pack II; Euler's Method; ' sin cos Y ** Use ALPHA Y

Table; 0; 0; 0.1; 6

Y x

Hit enter to get the output for each step.

Use concavity to determine under or overestimate (graph).

0.5 0.98 is an underestimatey

Wow!

Let y (t) be the solution of sin cos , 0 0.y t y y

Use Euler’s Method with time step h = 0.1 to approximate y (0.5).

When h = 0.1, yk is an approximation to y (0 + k (0.1)) = y (0.1k), so y5 is an approximation to y (0.5). It is convenient to organize calculations in the following table. Note that the value yk+1 computed in the last column of each line is used in the next line to continue the process.

Thus, Euler’s Method yields the approximation y (0.5) ≈ y5 ≈ 0.1.

0

1

2

3

4

0

0.1

0.2

0.3

0.4

t

t

t

t

t

0

1

2

3

4

0

0

0.01

0.03

0.06

y

y

y

y

y

kt ky , sin cosk k k kF t y t y 1 ,k k k ky y hF t y

sin 0 cos0 0

sin 0.1 cos0 0.1

sin 0.2 cos 0.01 0.2

sin 0.3 cos 0.03 0.3

sin 0.4 cos 0.06 0.4

1

2

3

4

5

0 0.1 0 0

0 0.1 0.1 0.01

0.01 0.1 0.2 0.03

0.03 0.1 0.3 0.06

0.06 0.1 0.4 0.10

y

y

y

y

y

c

c

c c c c c c c

cc c c c c c

c c c c c c

cc c c c c

c c c c c c

c

c

c

Is this an overestimate or underestimate?Scatter Plot is concave up!

2

0

Let , and let be the solution of

, satisfying 2 3. Let 0.1 be the time

step in Euler's Method and set 2 3.

F t y t y y t

y F t y y h

y y

1 0

2 1 1

3 2 2 4 5

(a) Calculate 2,3 .

(b) Calculate 2.1, .

(c) Calculate 2.2, and continue computing , and .

(d) Find approximations of 2.2 and 2.5 .

y y hF

y y hF y

y y hF y y y

y y

APPS; Calc Pack II; Euler's Method

** Let and use ALPHA Y for t x y

Yes!

24 4 4, 3.58171 0.1 2.4 3.58171 3.799539y hF t y

21 1 1, 3.1 0.1 2.1 3.1 3.231y hF t y

22 2 2, 3.231 0.1 2.2 3.231 3.3919y hF t y

23 3 3, 3.3919 0.1 2.3 3.3919 3.58171y hF t y

2

0

Let , and let be the solution of

, satisfying 2 3. Let 0.1 be the time

step in Euler's Method and set 2 3.

F t y t y y t

y F t y y h

y y

1 0

2 1 1

3 2 2 4 5

(a) Calculate 2,3 .

(b) Calculate 2.1, .

(c) Calculate 2.2, and continue computing , and .

(d) Find approximations of 2.2 and 2.5 .

y y hF

y y hF y

y y hF y y y

y y

n nt

0

1

2

3

4

5

1 1 1,n n n ny hF t y y

2

2.1

2.2

2.3

2.4

2.5

3

20 0 0, 3 0.1 2 3 3.1y hF t y

cos ?x xdx Try writing as .cos 'x uvx

, ' cosu x x v x x

' 1, sinu x v x x

' 'u x v x dx u x v x u x v x dx cos sin sin sin

sin co

cos

s

x xdx x x xdx x x

x x C

C

x

x

uv vdu

Integration by Parts (76)

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx?xxe dx

Try writing as .'xxe uv

, ' xu x x v x e

' 1, xu x v x e

' 'u x v x dx u x v x u x v x dx

x x x x xxe dx xe e d e Cx x e

2 2cos si 2 sinnx xdx x x x xdx , ' sinu x x v x x

' 1, cosu x v x x

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx

2 cos ?x xdx

Try writing the two factors of the integrand as '.uv

2 , ' cosu x x v x x ' 2 , sinu x x v x x

' 'u x v x dx u x v x u x v x dx

Integrating by Parts More Than Once

2 sin 2 c2 cos ocos s 2sinx x xdx xdx x x x Cx 2 2 sin 2 cos 2sincos x xx xdx x x x C

(76)

3 33 3

1 11 1

ln ln 3l

3

n 3

3ln 3 3 1 ln 3 2

xdx x x dx x

ln

' 1

' 1/

u x x

v x

u x x

v x x

Integration by Parts applies to definite integrals:

' 'b b

b

aa a

u x v x dx u x v x u x v x dx

3

1

ln ?xdx ????

' 1 is also used to find the antiderivative of inverse trig functions.v

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx

Try writing the two factors of the integrand as '.uv

' 'u x v x dx u x v x u x v x dx

(76)

Calculator

cos co sis nx x xe xdx e x e xdx sin con ssix x xe e x e xdxxdx

cos cos sin cosx x x xe xdx e x e x e xdx

sin

'

' cos

x

x

u x x

v x e

u x x

v x e

cos

'

' sin

x

x

u x x

v x e

u x x

v x e

cos ?xe xdx

1 1co2 cos cos s sinsin cos

2 2x x x x x xe xdx e x e x e xd Cx e x e x

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx

Try writing the two factors of the integrand as '.uv

' 'u x v x dx u x v x u x v x dx

Solve for cosxe xdxGoing in circles???

(76)

Shortcuts to Finding Taylor Series Write a Series for… (77)There are several methods for generating new Taylor series from known ones. First of all, we can differentiate and integrate Taylor series term by term within its interval of convergence, by Theorem 2 of Section 11.6. We can also multiply two Taylor series or substitute one Taylor series into another (we omit the proofs of these facts).

Find the Maclaurin series for f (x) = x2ex.

2 3 4 5

0

1

Standard Maclaurin Series for :

2! 3! 4! 5! !

nx

n

x

x x x x xe

n

e

x

4 5 6 7

2 3 4 52

2 3

2

2 12! 3! 4! 5!

2! 3! 4! 5! 2 !

x

n

n

x x x

x x x xx e x x

x xx x

n

Maclaurin Series Converges to forf x f x

2 3 4

0

2 3 41

1

1 3 5 7

0

12

1 1

1 2 3 4

1

2 1 3

1

5 7

1

ln 1

t

1

1 and 1

1 anan

1

1

d

a

n n

n

nn

n

n n

n

x x x x x

x x x xx

n

x x x xx

n

x

x x

x

x

x

x

x

x

2 3

0

1 1 2( ) 1

2! 3! 1a n

nn

a a a a ax a x x xx

This expansion should be memorized!

(78)The Taylor Series PRGM, cleared Derek's RAM?

Maclaurin S Converges terie o s r fof xf x

2 3 4 5

0

2 1 3 5 7

0

2 2 4 6

0

sin

cos

1 ! 2! 3! 4! 5!

1 2 1 ! 3! 5! 7

!

1 1 2 ! 2! 4

All

All

! 6

!

xn

n

nn

n

nn

n

x x x x xx

n

x x x xx

n

x x x x

n

e x

xx

x

2 3 4

0

1

1

All

1 1

n

n

x x x x xx

x

x

These 3 expansions should be memorized!

The sum of the reciprocal powers n−p is called a p-series.

THEOREM 3 Convergence of p-Series The infinite series

converges if p > 1 and diverges otherwise.

Here are two examples of p-series:

2 1/30 faster than .n n

Convergence of a Geometric Series (79)