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Fundamentals of Engineering

Calculus, Differential Equations & Transforms, and

Numerical Analysis

Brody Dylan Johnson

St. Louis University

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical

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Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)


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Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)Group work with more problems (30 minutes)


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Overview

Overview

Agenda:


Quiz (30 minutes)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical2 / 30
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Overview

Overview

Agenda:


Quiz (30 minutes)

Topics:Calculus: Differential Calculus, Integral Calculus, Centroids and

Moments of Inertia, Vector Calculus.

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Overview

Overview

Agenda:


Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids and


Differential Equations and Transforms: Differential Equations, Fourier

Series, Laplace Transforms, Eulers Approximation


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Overview

Overview

Agenda:


Quiz (30 minutes)

Topics:





Numerical Analysis: Root Solving with Bisection Method and Newtons

Method.


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Overview

Overview

Agenda:


Quiz (30 minutes)

Topics:





Numerical Analysis: Root Solving with Bisection Method and Newtons

Method.

Acknowledgement: Many problems are taken from the Hughes-Hallett,

Gleason, McCallum, et al. Calculus textbook.Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical2 / 30

Calculus
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Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values off(x) =x3

3x2

+20on[1, 3].


Calculus
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Calculus


Problem 1: Find the maximum and minimum values off

(x

) =x3

3x2

+20on[1, 3].Solution:

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Calculus

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Calculus


Problem 1: Find the maximum and minimum values off(x

) =x3

3x2

+20

on[1, 3].Solution:

Check endpoints and critical points (f(x) =0).

f

(x) =3x2

6x, f

(x) =0 = x= 0, 2.


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3x2 +20

on[1, 3].Solution:


f

(x) =3x2

6x, f

(x) =0 = x= 0, 2.f(x) =6x 6, f(0)< 0 concave down; local max, f(2)>0 concave up; local min.


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3x2 +20

on[1, 3].Solution:


f

(x) =3x2

6x, f


Compare values: f(1) =16,f(0) =20,f(2) =16,f(3) =20.


Calculus
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3x2 +20

on[1, 3].Solution:


f

(x) =3x2

6x, f



Minima:f(x) =16 atx = 1 orx =2.


Calculus
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3x2 +20

on[1, 3].Solution:


f

(x) =3x2

6x, f



Minima:f(x) =16 atx = 1 orx =2.Maxima:f(x) =20 atx =0 orx =3.


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Problem 2: Find dydx ify =xx.


Calculus
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Problem 2: Find dydx ify =x

x

.

Solution:


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x

.

Solution:

Apply logarithm and then use implicit differentiation.


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x

.

Solution:


Differentiate lny=x lnxw.r.tx.


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x

.

Solution:



1y

dy

dx=lnx+x 1

x(product rule).


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x

.

Solution:



1y

dy

dx=lnx+x 1

x(product rule).

dy

dx=y(lnx+1).


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Problem 2: Find dy

dx ify =xx

.

Solution:



1y

dy

dx=lnx+x 1

x(product rule).

dy

dx=y(lnx+1).

dy

dx =xx(lnx+1).


Calculus

Diff i l C l l
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Problem 3: Find the standard form of the tangent line to the hyperbola

x2

2y2

=8 at the point(4, 2).


Calculus

Diff i l C l l

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x2

2y2

=8 at the point(4, 2).Solution:


Calculus

Diff ti l C l l

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x2

2y2


Tangent Line toy = f(x)at x = a: y=f(a) +f(a)(x a).


Calculus

Diff ti l C l l

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x2

2y2


Tangent Line toy = f(x)at x = a: y=f(a) +f(a)(x a).Use implicit differentiation to find dy

dxsincefis not given explicitly.


Calculus

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x2

2y2




2x 4y dydx

=0


Calculus

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x2

2y2




2x 4y dydx

=0

dy

dx=

2x

4y.

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Calculus


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x2

2y2




2x 4y dydx

=0

dy

dx=

2x

4y.

dy

dx(4,2) =

8

8 = 1.Tangent line: y=2+ (1)(x (4))or y = x 2.


Calculus

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Problem 4: Evaluate the following limit.

limx

xex.


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limx

xex.

Solution:


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limx

xex.

Solution:

Indeterminate form: 0 . Use LHopitals Rule.


Calculus

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limx

xex.

Solution:

Indeterminate form: 0 . Use LHopitals Rule.So,

limx

xex = limx

x

exH= lim

x

1

ex =

1

=0.


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limx

xex.

Solution:

Indeterminate form: 0 . Use LHopitals Rule.So,

limx

xex = limx

x

exH= lim

x

1

ex =

1

=0.

Recall: LHopitals Rule states that the limit of an indeterminate formf(x)/g(x)can be evaluated using the ratio of the derivatives f(x)/g(x).


Calculus

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Problem 5: Find the partial derivative mc

if

m= m01 v2/c2 .


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if

m= m01 v2/c2 .

Solution:

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Calculus


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if

m= m01 v2/c2 .

Solution:

Notation: ifz =f(x,y)then zx = fx =fx.

Treat other variables as constants and differentiate w.r.t. indicated

variable.


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if

m= m01 v2/c2 .

Solution:

Notation: ifz =f(x,y)then zx =

fx =fx.


variable.

m

c

=

c m0

(1 v2/c2)1

2=

1

2

m0

(1 v2/c2)3

2

(

2)(

v2/c3)


Calculus

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if

m= m01 v2/c2 .

Solution:

Notation: ifz =f(x,y)then zx =

fx =fx.


variable.

m

c

=

c m0

(1 v2/c2)1

2=

1

2

m0

(1 v2/c2)3

2

(

2)(

v2/c3)

Simplify. m

c=

m0v2c3(1 v2/c2) 32


Calculus

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Problem 6: What is the slope of the curvey = 5 3x

5+3x

when it crosses the

positivex-axis?


Calculus

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5+3x

when it crosses the

positivex-axis?

Solution:


Calculus

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5+3x

when it crosses the

positivex-axis?

Solution:

The graph crosses thepositive x-axis wheny = 0 andx >0, i.e.,5

3x=0 orx = 53 . The slope is given by

dydx

.


Calculus


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Problem 6: What is the slope of the curvey = 5 3x5

+3x

when it crosses the

positivex-axis?

Solution:



dydx

.

Quotient Rule: d

dx

f

g

=

fg gfg2

.


Calculus

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Problem 6: What is the slope of the curvey = 5 3x5

+3x

when it crosses the

positivex-axis?

Solution:



dydx

.

Quotient Rule: d

dx

f

g

=

fg gfg2

.

dy

dx=

(3)(5+3x) (3)(5 3x)(5+3x)2

.


Calculus

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Problem 6: What is the slope of the curvey = 5 3x5+3x

when it crosses the

positivex-axis?

Solution:



dydx

.

Quotient Rule: d

dx

f

g

=

fg gfg2

.

dy

dx=

(3)(5+3x) (3)(5 3x)(5+3x)2

.

dydx

x= 5

3

=(3)(10) 3(0)102

= 0.3.

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Calculus


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Problem 7: The cost of fuel to propel a boat through the water (dollars per

hour) is proportional to the cube of the speed. A certain ferry boat uses $100

of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At what

speed should it travel so as to minimize the cost per miletraveled?

Solution:


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Solution:

Costs must be converted to aper milebasis.


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Solution:

Costs must be converted to aper milebasis.Fuel Cost: f(v) =Cv3 (dollars/hour) wherev is speed. f(10) =100 soC=0.1.


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Solution:


Fixed Cost: g(v) =675 (dollars/hour) is constant w.r.t. speed.


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Solution:



Total Cost: h(v) =675+0.1v3

(dollars/hour). At speedv it takes 1/vhours to go one mile.


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Solution:



Total Cost: h(v) =675+0.1v3

(dollars/hour). At speedv it takes 1/vhours to go one mile.

Minimize:C(v) =675/v+0.1v2,C(v) = 675/v2 +0.2v= (0.2v3 675)/v2. Critical Pointv= 35(675) =15 miles per hour. (Check thatC(15)> 0.)


Calculus

Integral Calculus
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Problem 8: Calculate the definite integral 2

0xex

2

dx


Calculus

Integral Calculus
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0xex

2

dx

Solution:


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Integral Calculus
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0xe

x2

dx

Solution:

Useu-substitution:u= x2 sodu = 2x dx.


Calculus

Integral Calculus
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0xe

x2

dx

Solution:


Dont forget to change the limits of integration: u(0) =0,u(2) =4!


Calculus

Integral Calculus
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0xe

x2

dx

Solution:


Dont forget to change the limits of integration: u(0) =0,u(2) =4! 20

xex2

dx= 1

2

40

eu du.


Calculus

Integral Calculus
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0xe

x2

dx

Solution:


Dont forget to change the limits of integration: u(0) =0,u(2) =4! 20

xex2

dx= 1

2

40

eu du.

2

0

xex2

dx= 1

2

eu4

0

= e4 1

2

.


Calculus

Integral Calculus 2

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Problem 9: Determine the indefinite integral

x2ex dx


Calculus

Integral Calculus 2

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x2ex dx

Solution:


Calculus

Integral Calculus 2 x

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x2ex dx

Solution:Integration by parts:

u dv= u v v du.


Calculus

Integral Calculus

P bl 9 D i h i d fi i i l

2 x d

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x2ex dx


u dv= u v v du.

u=x2,du = 2x dx,dv = ex dx,v = ex,

x

2

ex

dx= x2

ex

+2

xex

dx.


Calculus

Integral Calculus

P bl 9 D t i th i d fi it i t l

2 x d

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x2e x dx


u dv= u v v du.

u=x2,du = 2x dx,dv = ex dx,v = ex,

x

2

ex

dx= x2

ex

+2

xex

dx.

u=x,du =dx,dv =ex dx,v = ex dx,

xex dx= xex + ex dx= xe

x

ex +C

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Calculus

Integral Calculus

Problem 10: Find the volume of revolution from x = 1 to x = 3 when

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Problem 10: Find the volume of revolution fromx =1 tox = 3 whenf(x) = x 1 is rotated about they-axis.


Calculus

Integral Calculus

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Problem 10: Find the volume of revolution fromx =1 tox = 3 whenf(x) = x 1 is rotated about they-axis.Solution:


Calculus

Integral Calculus

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Problem 10: Find the volume of revolution fromx 1 tox 3 whenf(x) = x 1 is rotated about they-axis.Solution:

Volume=b

a 2xf(x) dx. (y-axis)


Calculus

Integral Calculus

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Problem 10: Find the volume of revolution fromx 1 tox 3 whenf(x) = x 1 is rotated about they-axis.Solution:

Volume=b


Volume=

b

a(f(x))2 dx. (x-axis)


Calculus

Integral Calculus

Problem 10: Find the volume of revolution fromx =1 tox = 3 when
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w

f(x) = x 1 is rotated about they-axis.Solution:

Volume=b


Volume=

b

a(f(x))2 dx. (x-axis)

Here,

Volume=

31

2x(x 1) dx

=2 x3

3x2

2 3

1

=2

26

3 8

2

=

28

3 .


Calculus

Integral Calculus

Problem 11: Find the area between the curves y =xand y =

x.
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y y


Calculus

Integral Calculus


x.
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y y

Solution:


Calculus

Integral Calculus


x.
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Solution:

The curves bound a region over 0 x 1 with x x.


Calculus

Integral Calculus


x.

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Solution:

The curves bound a region over 0 x 1 with x x.The area is given by the integral of

x x.


Calculus

Integral Calculus


x.

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Solution:

The curves bound a region over 0 x 1 with x x.The area is given by the integral of

x x.

Hence,

Area= 1

0

x x dx

=

2

3x

32 1

2x2

1

0

= 2

31

2

= 1

6.


Calculus

Integral Calculus

Problem 12: Find the indefinite integral

1x2+x

dx.
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x +x


Calculus

Integral Calculus


1x2+x

dx.
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+

Solution:


Calculus

Integral Calculus


1x2+x

dx.

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Solution:

Partial fractions: 1

x2 +x=

1

x(x+1)=

A

x+

B

x+1.

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Calculus

Integral Calculus


1x2+x

dx.

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Solution:


x2 +x=

1

x(x+1)=

A

x+

B

x+1.

Use cover-up method or common denominator:

1

x(x+1) = A

x + B

x+1 = A(x+1) +Bx

x(x+1) .

Equate terms: A+B=0 (xterms) andA = 1 (constant terms) soB= 1.


Calculus

Integral Calculus


1x2+x

dx.
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Solution:


x2 +x=

1

x(x+1)=

A

x+

B

x+1.

Use cover-up method or common denominator:

1

x(x+1) = A

x + B

x+1 = A(x+1) +Bx

x(x+1) .

Equate terms: A+B=0 (xterms) andA = 1 (constant terms) soB= 1.

Finally we integrate: 1

x2 +xdx=

1

x 1

x+1 dx= lnx ln (x+1) +C.

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Calculus

Centroid and Moment of Inertia

Problem 13: Find the centroid of the region bounded by 0 x 2,0 y x2

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0 y x .

Solution:xc= A

1

x dAandyc =A1

y dAwhereA = Area.

A= 2

0

x2 dx= 8

3.

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Calculus


Problem 13: Find the centroid of the region bounded by 0 x 2,0 y x2

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0 y x .

Solution:xc= A

1

x dAandyc =A1

y dAwhereA = Area.

A= 2

0

x2 dx= 8

3.

xcoordinate:

xc = 3

8

20

xf(x) dx= 3

8

1

4x4

2

0=

3

2.

ycoordinate:

yc = 3

8

40

y(2y)dy= 38

2y 2

3y

32

40

=1.

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Problem 14: Find the moment of inertia of a rectangle with side lengthsa and
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g g

babout its bottom edge (side length a).



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g g


Solution:



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g g


Solution:

Ix =

y2 dAand Iy=

x2 dA.



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Solution:

Ix =

y2 dAand Iy=

x2 dA.

Imagining the bottom edge as thex-axis we wantIx:

Ix =

b0

y2(a 0) dy= ab3

3 .

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Calculus

Vector Calculus

5 ( ) ( )

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Problem 15: Find the gradient off(x,y,z) =xey +ln (xz).


Vector Calculus

P bl 15 Fi d h di f f ( ) y l ( )
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Solution:


Vector Calculus

P bl 15 Fi d th di t f f ( ) y + l ( )
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Solution:

Gradient is fwhere = xi+

yj+

zk.


Vector Calculus

P bl 15 Fi d th di t f f ( ) y + l ( )
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Solution:


yj+

zk.

So,

f(x,y,z) = (ey

+

1

x )i+xey

j+

1

zk.


Vector Calculus

Problem 15: Find the gradient of f (x y ) xey + ln (x )
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Solution:


yj+

zk.

So,

f(x,y,z) = (ey

+

1

x )i+xey

j+

1

zk.

The gradient is a vector valued function and is the direction of maximum

increase off.


Vector Calculus

Problem 15: Find the gradient of f (x y z) = xey + ln (xz)
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Solution:


yj+

zk.

So,

f(x,y,z) = (ey

+

1

x )i+xey

j+

1

zk.

The gradient is a vector valued function and is the direction of maximum

increase off.

Directional derivative in direction of unit vector u: Duf =

f

u.


Vector Calculus

Problem 16: LetF(x,y,z) =xyi+y2zj+xzk. Find DivFand curlF.
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Calculus

Vector Calculus


Solution:

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IfF(x,y,z) =f(x,y,z)i+g(x,y,z)j+h(x,y,z)kthen

DivF(x,y,z) = F= fx

+g

y+h

z=y+2yz+x.


Calculus

Vector Calculus


Solution:
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IfF(x,y,z) =f(x,y,z)i+g(x,y,z)j+h(x,y,z)kthen

DivF(x,y,z) = F= fx

+g

y+h

z=y+2yz+x.

The curl ofFis given by F,

curlF(x,y,z) =

i j kx

y

z

f(x,y,z) g(x,y,z) h(x,y,z)

=

i j kx

y

z

xy y2z xz

= (0y2)i (z 0)j+ (0x)k= y2i zj xk.


Calculus

Vector Calculus
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Problem 17: Find the Laplacian of=f(x,y,z) =xyz.


Calculus

Vector Calculus
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Solution:


Calculus

Vector Calculus
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Solution:

The Laplacian ofis 2= 2x2

+ 2

y2 +

2z2

.


Calculus

Vector Calculus
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Solution:


+ 2

y2 +

2z2

.

x =yz,y=xz, andz =xy.


Calculus

Vector Calculus
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Solution:


+ 2

y2 +

2z2

.

x =yz,y=xz, andz =xy.xx =0,yy=0, andzz=0, so 2(x,y,z) =0.


Calculus

Vector Calculus
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Solution:


+ 2

y2 +

2z2

.

x =yz,y=xz, andz =xy.xx =0,yy=0, andzz=0, so 2(x,y,z) =0.Functions for which 2=0 are calledpotential functions.


Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curve

x = t2 y = sin (2t) at t = 2
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x= t ,y = sin (2t)att=2.


Calculus

Vector Calculus


x = t2 y = sin (2t) at t = 2
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x= t ,y = sin (2t)att=2.

Solution:

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Calculus

Vector Calculus


x = t2, y = sin (2t) at t = 2.

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x t ,y sin (2t)att 2.

Solution:

Chain Rule: dydt

= dydx dx

dt

x(t) =2t,y(t) =2 cos (2t)


Calculus

Vector Calculus


x= t2,y = sin (2t)att=2.
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, y ( )

Solution:

Chain Rule: dydt

= dydx dx

dt

x(t) =2t,y(t) =2 cos (2t)

Sox(2) =4,y(2) =0, and

dy

dx

t=2

= 2 cos (4)

2(2) =

2.


Calculus

Vector Calculus


x= t2,y = sin (2t)att=2.
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, y ( )

Solution:

Chain Rule: dydt

= dydx dx

dt

x(t) =2t,y(t) =2 cos (2t)

Sox(2) =4,y(2) =0, and

dy

dx

t=2

= 2 cos (4)

2(2) =

2.

The tangent line is thusy =

2(x 4).


Differential Equations and Transforms

Differential Equations

Problem 19: Find the particular solution ofy 3y +2y=0 satisfyingy(0) =1, y(0) =0.
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( ) ( )




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Solution:




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Solution:

Characteristic Equation: r2 3r+2=0 or(r 2)(r 1) =0 sor=1, 2.




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Solution:


The [homogeneous or complementary] solution isy(x) =Aex

+Be2x




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Solution:



+Be2x

Theny(x) =Aex +2Be2x and

y(0) =1=A+B y(0) =0= A+2B.




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Solution:



+Be2x

Theny(x) =Aex +2Be2x and

y(0) =1=A+B y(0) =0= A+2B.

B= 1 andA = 2 yield the particular solution: y(x) =2ex

e2x

.




Problem 20: Find a particular solution ofy +2y +y = x2.
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Solution:





S l i
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Solution:Method of undetermined coefficients. Assume yp(x) =Ax

2 +Bx+C.

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S l ti

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2 +Bx+C.

Differentiate: y(x) =2Ax+B,y(x) =2A.

Substitute:

(2A) +2(2Ax+B) + (Ax2 +Bx+C) =x2.

Equate like terms: x2 forcesA =1,x forces 4A+B = 0 and 1 leads to2A+2B+C=0.





S l ti

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2 +Bx+C.

Differentiate: y(x) =2Ax+B,y(x) =2A.

Substitute:

(2A) +2(2Ax+B) + (Ax2 +Bx+C) =x2.

Equate like terms: x2 forcesA =1,x forces 4A+B = 0 and 1 leads to2A+2B+C=0.

A=1,B = 4, andC=6.




Problem 21: Find the general solution ofy 3y= sin (2x).
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Problem 21: Find the general solution ofy 3y= sin (2x).Solution:
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Method of undetermined coefficients Assume
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Method of undetermined coefficients. Assume

yp(x) =A cos (2x) +B sin (2x).






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yp(x) =A cos (2x) +B sin (2x).

Differentiate: y(x) = 2A sin (2x) +2B cos (2x).






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Method of undetermined coefficients. Assumeyp(x) =A cos (2x) +B sin (2x).

Differentiate: y(x) = 2A sin (2x) +2B cos (2x).Substitute:

(2A sin (2x) +2B cos (2x)) 3(A cos (2x) +B sin (2x)) =sin (2x).






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Equate like terms, sin (2x): 2A 3B= 1 and cos (2x): 2B 3A=0.





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Equate like terms, sin (2x): 2A 3B= 1 and cos (2x): 2B 3A=0.A=2/13,B = 3/13 soyp(x) =2/13 cos (2x) +3/13 sin (2x).





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Equate like terms, sin (2x): 2A 3B= 1 and cos (2x): 2B 3A=0.A=2/13,B = 3/13 soyp(x) =2/13 cos (2x) +3/13 sin (2x).

Char. polynomial isr 3= 0, soyh(x) =Ae3x andy(x) =yh(x) +yp(x).



Fourier SeriesProblem 22: Find the Fourier Series off(x) =x, 0 x 1, as a 1-periodicfunction.
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Solution:
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Solution:

Fourier Series:
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f(x) a02

+

n=1

ancos

2n

Tx+bnsin

2n

Tx

.




Solution:

Fourier Series:
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f(x) a02

+

n=1

ancos

2n

Tx+bnsin

2n

Tx

.

Coefficients:

an= 2

T

T0

f(x) cos2n

Tx dx bn=

2

T

T0

f(x) sin2n

Tx dx.




Solution:

Fourier Series:
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f(x) a02

+

n=1

ancos

2n

Tx+bnsin

2n

Tx

.

Coefficients:

an= 2

T

T0

f(x) cos2n

Tx dx bn=

2

T

T0

f(x) sin2n

Tx dx.

In this case,a0 =1, an=0, n 1, and

bn=2

10

x sin2nx dx=(Integrate by parts)= 1n

.



Laplace TransformProblem 23: Find the Laplace transform off(t) =eat.
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Solution:
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Solution:

Laplace transform:

F(s) =

f (t)est dt
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F(s) =

0

f(t)e dt.




Solution:

Laplace transform:

F(s) =

f (t)est dt
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F(s) =

0

f(t)e dt.

In this case,

F(s) =

0e

(s

a)t dt= 1

(s a) e

(s

a)t

0 = 1

s a . (s> a)




Solution:

Laplace transform:

F(s) =

f (t)est dt
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F(s) =

0

f(t)e dt.

In this case,

F(s) =

0e

(s

a)t dt= 1

(s a) e

(s

a)t

0 = 1

s a . (s> a)

Laplace transforms behave nicely with derivatives:

L{f(t)

}=sF(s)

y(0).

We can use this to solve differential equations.



Laplace Transform

Problem 24: Use Laplace transforms to solve y +2y= 0,y(0) =2.
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Laplace Transform


Solution:
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Laplace Transform


Solution:

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Apply the Laplace transform to the equation: sY(s)y(0) +2Y(s) =0.SoY(s) =y(0)/(s+2) =2/(s+2).



Laplace Transform


Solution:

( ) ( ) ( )

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Apply the Laplace transform to the equation: sY(s)y(0) +2Y(s) =0.SoY(s) =y(0)/(s+2) =2/(s+2).

We findy(t)by recognizing the form ofY(s)using known Laplace

transforms:

y(t) =2L1

1

s+2

=2e2t.



Laplace Transform

Problem 25: Find the inverse Laplace transform ofY(s) = s+4s2+4

.
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Laplace Transform


.

Solution:
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Laplace Transform


.

Solution:

Consulting a table, the relevant transforms are:
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L{cost} = ss2 +2

L{sint} = s2 +2

.



Laplace Transform


.

Solution:

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L{cost} = ss2 +2

L{sint} = s2 +2

.

We have

Y(s) = s+4s2 +4

= s

s2 +4+2

2s2 +4

.



Laplace Transform


.

Solution:

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L{cost} = ss2 +2

L{sint} = s2 +2

.

We have

Y(s) = s+4s2 +4

= s

s2 +4+2 2

s2 +4.

Soy(t) =cos (2t) +2 sin (2t).



Laplace Transform


.

Solution:

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L{cost} = ss2 +2

L{sint} = s2 +2

.

We have

Y(s) = s+4s2 +4

= s

s2 +4+2 2

s2 +4.

Soy(t) =cos (2t) +2 sin (2t).

Often we must use partial fractions to first factor an expression into

recognizable pieces.



Separable ODEs

Problem 26: Solve the initial value problem:(1+x2) dydx

=3xy,y(0) =1.
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Separable ODEs

Problem 26: Solve the initial value problem:(1+x2) dydx

=3xy,y(0) =1.

Solution:
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Eulers Method

Problem 27: Use Eulers Method to predicty(0.5)if dydx

=1+xy,y(0) =0,

andx=0.25.

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Eulers Method


=1+xy,y(0) =0,

andx=0.25.Solution:
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So ut o :



Eulers Method


=1+xy,y(0) =0,

andx=0.25.Solution:
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Eulers Method:yk+1 =yk+ dydx

(xk,yk)

x.



Eulers Method


=1+xy,y(0) =0,

andx=0.25.Solution:
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(xk,yk)

x.

y0=0 is given, soy1 =0+ (1+0(0)) 0.25= 0.25



Eulers Method


=1+xy,y(0) =0,

andx=0.25.Solution:
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(xk,yk)

x.

y0=0 is given, soy1 =0+ (1+0(0)) 0.25= 0.25Now,y2 =0.25+ (1+ (0.25)(0.25)) 0.25=0.515625.



Eulers Method


=1+xy,y(0) =0,

andx=0.25.Solution:
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(xk,yk)

x.

y0=0 is given, soy1 =0+ (1+0(0)) 0.25= 0.25Now,y2 =0.25+ (1+ (0.25)(0.25)) 0.25=0.515625.Interprety2 y(x0+2x) =y(0.5)so y(0.5) 0.516.


Numerical Analysis

Bisection MethodProblem 28: Use two iterations of the bisection method to approximate a root

off(x) =x2 3x 2.
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Numerical Analysis


off(x) =x2 3x 2.Solution:
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Numerical Analysis



The bisection method starts with points, L0andR0for which one off(L0),f(R0)is negative and the other is positive. The intermediate value
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theorem says a root lies in between (assuming fis continuous).

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Numerical Analysis



The bisection method starts with points, L0andR0for which one off(L0),f(R0)is negative and the other is positive. The intermediate valueth t li i b t ( i f i ti )

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Noticef(0) = 2< 0 andf(4) =2 >0, so we can useL0 =0 andR0 =4.First iteration: Checkf(0.5(L0+ R0)) =f(2) = 10, so setL1 =2,R1 =4.


Numerical Analysis



The bisection method starts with points, L0andR0for which one off(L0),f(R0)is negative and the other is positive. The intermediate valueth t li i b t ( i f i ti )
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Second iteration: Checkf(0.5(L1+ R1)) =f(3) = 2, so setL2 =3,R2 =4.


Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a root


The bisection method starts with points, L0andR0for which one off(L0),f(R0)is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous)
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Second iteration: Checkf(0.5(L1+ R1)) =f(3) = 2, so setL2 =3,R2 =4.

At this point our estimate of the root would be x = 3.5. Noticef(3.5) = 0.25 which is much closer.

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