Post on 15-Feb-2016
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L’Hopital’s Rule (62)3
42
8Use L'Hopital's Rule to evaluate lim .2 20x
xx x
2
32
3 12lim 613 74 2 4x
xx
2
2
4Evaluate lim .sinx
xx
2
2 4limcos
4x
xx
2
/ 2
cosEvaluate lim .1 sinx
xx
/ 2 / 2
2cos sinlim lim 2sinco
2sx x
x x xx
Note that the quotient
2cos sincosx xx
is still indeterminate at x = π/2. We removed this indeterminacy by cancelling the factor − cos x.
Always check forindeterminate form.
00
32
Show that converges and compute its value.dxx
Step 1. Integrate over a finite interval [2, R].
22 2 2
22
23
1 1 1 12 2 2 4
1 18 2
RR
R
dx x x Rx R
Step 2. Compute the limit as R → ∞.
3 22
18
1 1lim lim8 2
R
R R
dxx R
Improper Integrals (63)
1
Determine whether converges.dxx
11
n n0 l ll nR
R
dx xx
RR
1
lim lim lnR R
R
dx Rx
Improper Integrals (63)
9
0
?dxx
1f 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 ydt
0
0
11 /kt
ydy y Aky y t Cdt A e C y A
1.2
thus, 1.2 & 44so,
1 /t
k A
ye C
0
0
13
yCy A
C
1.2
41 3 ty
e
Logistics Growth (64)
0.40.4 1000 500 1500 or
1 9 500 1000 9t
t ee
0.4 ln 990.
5.493 s4
yearte t
0.4
0.4 & 1000
0.4 11000
10001 /t
k AdP Pydt
ye C
0
00.4
10001 9
19 tyyC
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 Cdt A e C y A
kty Cey A
Solve for t
`
Same formula for anypoint in the plane.
(64)
2
12
17 1 2 50 5
A Bx x xx xx
Finding the Constants
2 ?7 10dx
x x
2 52 5
2 5 2 5
1 5 2
x x A Bx xx x x x
A x B x
12 1 33
15 1 33
x A A
x B B
2
1 17 10 3 2 3 5dx dx dx
x x x x
1 1ln 2 ln 53 3
x 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 10dx
x x
Partial Fractions (65)
Calculator:APPS; A+ Calc; Calc Pack II; Partial Fractions
2
1 / 3 1 / 3 1 17 10 2 5 3 2 3 5dx 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 2x dx dx dxdx
x x x x x x
2
1 22
81 2 8 22A B Cx x
xxx x x
2 2 ?
1 2 8 2x dx
x x x
11 3 6 36
4 18 3 6 1
12 6 3 126
x A A
x B B
x C C
1 ln 42
1 1ln 1 ln 26 6
x xx C
Multiply by thedenominator ofour integrand.
Let a # whichdrops a parameter.
x
Partial Fractions (65)
2 2 ?
1 2 8 2x dx
x x x
2
1 22
81 2 8 22A B Cx x
xxx x x
2 2 2 8 2 1 2 1 2 8x A x x B x x C x x
11 3 6 36
4 18 3 6 1
12 6 3 126
x A A
x B B
x C C
2 2 1 1
1 2 8 2 6 1 2 8 6 2x dx dx dxdx
x x x x x x
1 1 1ln 1 l 1 1ln 1 ln 26 6
1n 2 l22 6
n 2 86
xdux x C x xu
C
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 2x dx
x x x
2 2 1 / 6 1 1 / 6
1 2 8 2 1 2 8 2
1 16 1 2 8 6 2
x dxx x x x x x
dx dx dxx 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
' ' , 'dr 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' , , ' 1,04 2 2 2
r r
tangent line is parametrized by
2 2At , ,4 4 2 2
t 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
32
t
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 ' 'ds x t y tdt
1/23' 2, '2
x t y t t
944
2.5 m/mindsdt
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 2
t dt t t c t c t c
210 2,1 1, 2, 12
s t dt t t 21, and 0 2, 12, 1
21 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 .
ta 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
1 4sin 8 sin2
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 .2
K f d
2 1sin 1 cos 2
2
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
212
K r d
/3 /3
0 0
/3 /3
0 0
22
00
21 1 1 cos62 2 2
1 1 cos
sin 3
1
64 4
1 1cos sin12 24 12 24 2
d d
d d
udu u
2 1 cos 6sin 32
6 6u du d
Area Inside the Polar Curve (72)
sin cos// cos sin
dr rdy dy d ddrdx 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.2
r
2 22 02 22cos2 sin cos
2cos2 cos sin 2 22 01
2 2
rrr
r
2 212 2
y x
2cos2
2sin2
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; 6Y 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
00.10.20.30.4
ttttt
0
1
2
3
4
000.010.030.06
yyyyy
kt ky , sin cosk k k kF t y t y 1 ,k k k ky y hF t y
sin 0 cos 0 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
Here's some good advice:(76)
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 cos2 2
x 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 xen
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 xn
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 xxn
x x x xxn
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 xxn
x x x xxn
x x x xn
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 seriesconverges 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)