Fundamentals of Engineering Calculus, Differential Equations

Fundamentals of EngineeringCalculus, 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 Analysis1 / 30

Overview

Overview

Agenda:

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

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.


Overview

Overview

Agenda:



Quiz (30 minutes)

Topics:






Overview

Overview

Agenda:



Quiz (30 minutes)

Topics:






Overview

Overview

Agenda:



Quiz (30 minutes)

Topics:






Overview

Overview

Agenda:



Quiz (30 minutes)

Topics:






Overview

Overview

Agenda:



Quiz (30 minutes)

Topics:






Overview

Overview

Agenda:



Quiz (30 minutes)

Topics:






Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:

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

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.

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

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus



Solution:

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

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.



Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus



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

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.



Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus




f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.



Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus




f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.



Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus




f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.



Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus




f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.



Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus




f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.



Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.


Calculus


Problem 2: Find dydx if y = xx.

Solution:

Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).


Calculus



Solution:

Apply logarithm and then use implicit differentiation.


dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).


Calculus



Solution:Apply logarithm and then use implicit differentiation.


dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).


Calculus




Differentiate ln y = x ln x w.r.t x.

1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).


Calculus





dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).


Calculus





dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).


Calculus





dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).


Calculus


Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:

Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

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


Calculus



Solution:

Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).


2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.



Calculus



Solution:Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).


2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.



Calculus





2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.



Calculus





2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.



Calculus





2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.



Calculus





2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.



Calculus





2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.



Calculus


Problem 4: Evaluate the following limit.

limx→∞

xe−x.

Solution:

Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.

Recall: L’Hopital’s 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



limx→∞

xe−x.

Solution:

Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.



Calculus



limx→∞

xe−x.

Solution:Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.



Calculus



limx→∞

xe−x.


So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.



Calculus



limx→∞

xe−x.


So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.



Calculus


Problem 5: Find the partial derivative ∂m∂c if

m =m0√

1− v2/c2.

Solution:

Notation: if z = f (x, y) then ∂z∂x = ∂f

∂x = fx.

Treat other variables as constants and differentiate w.r.t. indicatedvariable.∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32


Calculus



m =m0√

1− v2/c2.

Solution:


∂x = fx.


=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32


Calculus



m =m0√

1− v2/c2.

Solution:


∂x = fx.


=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32


Calculus



m =m0√

1− v2/c2.

Solution:


∂x = fx.

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

∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32


Calculus



m =m0√

1− v2/c2.

Solution:


∂x = fx.


=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32


Calculus



m =m0√

1− v2/c2.

Solution:


∂x = fx.


=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32


Calculus


Problem 6: What is the slope of the curve y =5− 3x5 + 3x

when it crosses the

positive x-axis?

Solution:

The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5

3 . The slope is given by dydx .

Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.


Calculus



when it crosses the

positive x-axis?

Solution:

The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5


Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.


Calculus



when it crosses the

positive x-axis?

Solution:The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5


Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.


Calculus



when it crosses the

positive x-axis?



Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.


Calculus



when it crosses the

positive x-axis?



Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.


Calculus



when it crosses the

positive x-axis?



Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.


Calculus


Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of 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 whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:

Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

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

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v 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 = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)


Calculus



Solution:

Costs must be converted to a per mile basis.







Calculus



Solution:Costs must be converted to a per mile basis.







Calculus










Calculus










Calculus










Calculus










Calculus

Integral Calculus

Problem 8: Calculate the definite integral∫ 2

0xex2

dx

Solution:

Use u-substitution: u = x2 so du = 2x dx.

Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!∫ 2

0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.


Calculus

Integral Calculus


0xex2

dx

Solution:



0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.


Calculus

Integral Calculus


0xex2

dx

Solution:



0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.


Calculus

Integral Calculus


0xex2

dx

Solution:


Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!

∫ 2

0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.


Calculus

Integral Calculus


0xex2

dx

Solution:



0xex2

dx =12

∫ 4

0eu du.

∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.


Calculus

Integral Calculus


0xex2

dx

Solution:



0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.


Calculus

Integral Calculus

Problem 9: Determine the indefinite integral∫

x2e−x dx

Solution:

Integration by parts:∫

u dv = u v−∫

v du.

u = x2, du = 2x dx, dv = e−x dx, v = −e−x,∫x2e−x dx = −x2e−x + 2

∫xe−x dx.

u = x, du = dx, dv = e−x dx, v = −e−x dx,∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C

Finally, (incorporating the 2 again)∫x2e−x dx = −x2e−x − 2xe−x − 2e−x + C.


Calculus

Integral Calculus


x2e−x dx

Solution:


u dv = u v−∫

v du.


∫xe−x dx.





Calculus

Integral Calculus


x2e−x dx

Solution:


u dv = u v−∫

v du.


∫xe−x dx.





Calculus

Integral Calculus


x2e−x dx

Solution:


u dv = u v−∫

v du.


∫xe−x dx.





Calculus

Integral Calculus


x2e−x dx

Solution:


u dv = u v−∫

v du.


∫xe−x dx.





Calculus

Integral Calculus


x2e−x dx

Solution:


u dv = u v−∫

v du.


∫xe−x dx.





Calculus

Integral Calculus

Problem 10: Find the volume of revolution from x = 1 to x = 3 whenf (x) = x− 1 is rotated about the y-axis.

Solution:

Volume =∫ b

a 2πxf (x) dx. (y-axis)

Volume =∫ b

a π(f (x))2 dx. (x-axis)

Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.


Calculus

Integral Calculus


Solution:

Volume =∫ b


Volume =∫ b


Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.


Calculus

Integral Calculus


Solution:

Volume =∫ b


Volume =∫ b


Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.


Calculus

Integral Calculus


Solution:

Volume =∫ b


Volume =∫ b


Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.


Calculus

Integral Calculus


Solution:

Volume =∫ b


Volume =∫ b


Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.


Calculus

Integral Calculus

Problem 11: Find the area between the curves y = x and y =√

x.

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

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.


Calculus

Integral Calculus


x.

Solution:


x ≥ x.


x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.


Calculus

Integral Calculus


x.

Solution:


x ≥ x.


x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.


Calculus

Integral Calculus


x.

Solution:


x ≥ x.


x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.


Calculus

Integral Calculus


x.

Solution:


x ≥ x.


x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.


Calculus

Integral Calculus

Problem 12: Find the indefinite integral∫ 1

x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.

Use cover-up method or common denominator:

1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.

Equate terms: A + B = 0 (x terms) and A = 1 (constant terms) soB = −1.

Finally we integrate:∫1

x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.


Calculus

Integral Calculus


x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.


1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.



x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.


Calculus

Integral Calculus


x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.


1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.



x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.


Calculus

Integral Calculus


x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.


1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.



x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.


Calculus

Integral Calculus


x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.


1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.



x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.


Calculus

Integral Calculus


x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.


1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.



x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.


Calculus

Centroid and Moment of Inertia

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

Solution:

xc = A−1∫

x dA and yc = A−1∫

y dA where A = Area.

A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.


Calculus



Solution:

xc = A−1∫



A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.


Calculus



Solution:

xc = A−1∫



A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.


Calculus



Solution:

xc = A−1∫



A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.


Calculus



Solution:

xc = A−1∫



A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.


Calculus


Problem 14: Find the moment of inertia of a rectangle with side lengths a andb about its bottom edge (side length a).

Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.

Imagining the bottom edge as the x-axis we want Ix:

Ix =

∫ b

0y2(a− 0) dy =

ab3

3.

Note: We set this up using dy because points on horizontal slices havethe same distance from the axis of rotation. Points on vertical slices (aswith dx) would not.


Calculus



Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.


Ix =

∫ b

0y2(a− 0) dy =

ab3

3.



Calculus



Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.


Ix =

∫ b

0y2(a− 0) dy =

ab3

3.



Calculus



Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.


Ix =

∫ b

0y2(a− 0) dy =

ab3

3.



Calculus



Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.


Ix =

∫ b

0y2(a− 0) dy =

ab3

3.



Calculus

Vector Calculus

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

Solution:

Gradient is∇f where∇ = ∂∂x~i + ∂

∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.

The gradient is a vector valued function and is the direction of maximumincrease of f .

Directional derivative in direction of unit vector ~u: D~uf = ∇f ·~u.


Calculus

Vector Calculus


Solution:


∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.




Calculus

Vector Calculus


Solution:


∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.




Calculus

Vector Calculus


Solution:


∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.




Calculus

Vector Calculus


Solution:


∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.




Calculus

Vector Calculus


Solution:


∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.




Calculus

Vector Calculus

Problem 16: Let F(x, y, z) = xy~i + y2z~j + xz~k. Find DivF and curlF.

Solution:

If F(x, y, z) = f (x, y, z)~i + g(x, y, z)~j + h(x, y, z)~k then

DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.

The curl of F is given by∇× F,

curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

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

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.


Calculus

Vector Calculus


Solution:


DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.


curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

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

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.


Calculus

Vector Calculus


Solution:


DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.


curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

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

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.


Calculus

Vector Calculus


Solution:


DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.


curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

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

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.


Calculus

Vector Calculus

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

Solution:

The Laplacian of φ is ∇2φ = ∂2φ∂x2 + ∂2φ

∂y2 + ∂2φ∂z2 .

φx = yz, φy = xz, and φz = xy.

φxx = 0, φyy = 0, and φzz = 0, so∇2φ(x, y, z) = 0.

Functions for which∇2φ = 0 are called potential functions.


Calculus

Vector Calculus


Solution:


∂y2 + ∂2φ∂z2 .





Calculus

Vector Calculus


Solution:


∂y2 + ∂2φ∂z2 .





Calculus

Vector Calculus


Solution:


∂y2 + ∂2φ∂z2 .





Calculus

Vector Calculus


Solution:


∂y2 + ∂2φ∂z2 .





Calculus

Vector Calculus


Solution:


∂y2 + ∂2φ∂z2 .





Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curvex = t2, y = sin (2πt) at t = 2.

Solution:

Chain Rule: dydt = dy

dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.

The tangent line is thus y = π2 (x− 4).


Calculus

Vector Calculus


Solution:


dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.



Calculus

Vector Calculus


Solution:


dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.



Calculus

Vector Calculus


Solution:


dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.



Calculus

Vector Calculus


Solution:


dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.



Calculus

Vector Calculus


Solution:


dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.



Differential Equations and Transforms

Differential Equations

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

Solution:

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

The [homogeneous or complementary] solution is y(x) = Aex + Be2x

Then y′(x) = Aex + 2Be2x and

y(0) = 1 = A + B y′(0) = 0 = A + 2B.

B = −1 and A = 2 yield the particular solution: y(x) = 2ex − e2x.





Solution:




y(0) = 1 = A + B y′(0) = 0 = A + 2B.






Solution:




y(0) = 1 = A + B y′(0) = 0 = A + 2B.






Solution:




y(0) = 1 = A + B y′(0) = 0 = A + 2B.






Solution:




y(0) = 1 = A + B y′(0) = 0 = A + 2B.






Solution:




y(0) = 1 = A + B y′(0) = 0 = A + 2B.





Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

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

Substitute:

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

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

A = 1, B = −4, and C = 6.





Solution:



Substitute:

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


A = 1, B = −4, and C = 6.





Solution:



Substitute:

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


A = 1, B = −4, and C = 6.





Solution:



Substitute:

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


A = 1, B = −4, and C = 6.





Solution:



Substitute:

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


A = 1, B = −4, and C = 6.





Solution:



Substitute:

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


A = 1, B = −4, and C = 6.





Solution:



Substitute:

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


A = 1, B = −4, and C = 6.




Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:

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).

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

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).





Solution:

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


Substitute:









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


Substitute:











Substitute:











Substitute:











Substitute:











Substitute:











Substitute:







Fourier Series

Problem 22: Find the Fourier Series of f (x) = x, 0 ≤ x ≤ 1, as a 1-periodicfunction.

Solution:

Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.

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

bn = 2∫ 1

0x sin 2πnx dx = (Integrate by parts) = − 1

πn.



Fourier Series


Solution:

Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.


bn = 2∫ 1


πn.



Fourier Series


Solution:Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.


bn = 2∫ 1


πn.



Fourier Series



f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.


bn = 2∫ 1


πn.



Fourier Series



f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.


bn = 2∫ 1


πn.



Laplace Transform

Problem 23: Find the Laplace transform of f (t) = eat.

Solution:

Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(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


Solution:

Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)


L{f ′(t)} = sF(s)− y(0).




Laplace Transform


Solution:Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)


L{f ′(t)} = sF(s)− y(0).




Laplace Transform



F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)


L{f ′(t)} = sF(s)− y(0).




Laplace Transform



F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)


L{f ′(t)} = sF(s)− y(0).




Laplace Transform

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

Solution:

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

So Y(s) = y(0)/(s + 2) = 2/(s + 2).

We find y(t) by recognizing the form of Y(s) using known Laplacetransforms:

y(t) = 2L−1{

1s + 2

}= 2e−2t.



Laplace Transform


Solution:

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

So Y(s) = y(0)/(s + 2) = 2/(s + 2).


y(t) = 2L−1{

1s + 2

}= 2e−2t.



Laplace Transform


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

So Y(s) = y(0)/(s + 2) = 2/(s + 2).


y(t) = 2L−1{

1s + 2

}= 2e−2t.



Laplace Transform



So Y(s) = y(0)/(s + 2) = 2/(s + 2).


y(t) = 2L−1{

1s + 2

}= 2e−2t.



Laplace Transform



So Y(s) = y(0)/(s + 2) = 2/(s + 2).


y(t) = 2L−1{

1s + 2

}= 2e−2t.



Laplace Transform

Problem 25: Find the inverse Laplace transform of Y(s) = s+4s2+4 .

Solution:

Consulting a table, the relevant transforms are:

L{cosωt} =s

s2 + ω2 L{sinωt} =ω

s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).

Often we must use partial fractions to first factor an expression intorecognizable pieces.



Laplace Transform


Solution:

Consulting a table, the relevant transforms are:

L{cosωt} =s


s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).




Laplace Transform


Solution:Consulting a table, the relevant transforms are:

L{cosωt} =s


s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).




Laplace Transform



L{cosωt} =s


s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).




Laplace Transform



L{cosωt} =s


s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).




Laplace Transform



L{cosωt} =s


s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).




Separable ODEs

Problem 26: Solve the initial value problem: (1 + x2) dydx = 3xy, y(0) = 1.

Solution:

This differential equation is separable: (integrate RHS using u = 1 + x2)∫1y

dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.

Exponentiation leads to y = A(1 + x2)32 and since y(0) = 1, we have

A = 1.



Separable ODEs


Solution:


dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.


A = 1.



Separable ODEs


Solution:


dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.


A = 1.



Separable ODEs


Solution:


dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.


A = 1.



Euler’s Method

Problem 27: Use Euler’s Method to predict y(0.5) if dydx = 1 + xy, y(0) = 0,

and ∆x = 0.25.

Solution:

Euler’s Method: yk+1 = yk + dydx

∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.

Interpret y2 ≈ y(x0 + 2∆x) = y(0.5) so y(0.5) ≈ 0.516.



Euler’s Method


and ∆x = 0.25.

Solution:


∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.




Euler’s Method


and ∆x = 0.25.

Solution:


∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.




Euler’s Method


and ∆x = 0.25.

Solution:


∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.




Euler’s Method


and ∆x = 0.25.

Solution:


∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.




Euler’s Method


and ∆x = 0.25.

Solution:


∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.



Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:

The bisection method starts with points, L0 and R0 for 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).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 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.


Numerical Analysis

Bisection Method


Solution:

The bisection method starts with points, L0 and R0 for 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).






Numerical Analysis

Bisection Method


Solution:The bisection method starts with points, L0 and R0 for 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).






Numerical Analysis

Bisection Method








Numerical Analysis

Bisection Method








Numerical Analysis

Bisection Method








Numerical Analysis

Bisection Method








Documents

Fundamentals of Engineering Calculus, Differential Equations