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 A 1 / 30
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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.
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30
Calculus
Differential 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).
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30
Calculus
Differential 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
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30
Calculus
Differential 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
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30
Calculus
Differential 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
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30
Calculus
Differential 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
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30
Calculus
Differential 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
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30
Calculus
Differential 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
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30
Calculus
Differential 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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30
Calculus
Differential 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.
√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30
Calculus
Differential 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.
√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30
Calculus
Differential 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.
√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30
Calculus
Differential 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.
√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30
Calculus
Differential 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.
√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30
Calculus
Differential 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.
√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30
Calculus
Differential 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.
√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30
Calculus
Centroid and Moment of Inertia
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30
Calculus
Centroid and Moment of Inertia
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30
Calculus
Centroid and Moment of Inertia
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30
Calculus
Centroid and Moment of Inertia
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30
Calculus
Centroid and Moment of Inertia
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30
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
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis29 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30
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.
Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30