Category Archives: Engineering Mathematics Blog

Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 1 — Special Second-Ordered Differential Equations

Find the general solution of the differential equation

\frac{d}{dx}\left(\frac{dy}{dx}\right)=6x+3

Solution:

\begin{align*}
\frac{d}{dx}\left(\frac{dy}{dx}\right) & =6x+3  \\\  \\
let\:u & =\frac{dy}{dx} \\\ \\
\frac{du}{dx} & =6x+3 \\\ \\
Integrate,\\
\int \frac{du}{dx} & =\int (6x+3)dx \\\ \\
\int \frac{du}{dx} & =6\int xdx+3\int dx \\\ \\
u & =\frac{6x^2}{2}+3x+C_1 \\\ \\
u & =3x^2+3x+C_1 \\\ \\
Substitute, \\
\frac{dy}{dx} & =3x^2+3x+C_1 \\\ \\
dy & =\left(3x^2+3x+C_1\right)dx \\\ \\
Integrate,\\
\int dy & =\int (3x^2+3x+C_1)dx \\\ \\
\int dy & =3\int x^2dx+3\int xdx+C_1\int dx \\\ \\
y & =\frac{3x^3}{3}+\frac{3x^2}{2}+C_1x+C_2 \\\ \\
Simplify, \\
y & =x^3+\frac{3x^2}{2}+C_1x+C_2 \\
\end{align*}
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College Physics by Openstax Chapter 3 Problem 7

(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R’=A−B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R”=B−A=−R’ ). Show that this is the case.

Solution:

Refer to this problem.

Part A

Consider Figure 3-7A

Figure 3-7A

First, we solve for the value of α using a simple geometry. 

\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ

We can solve for R using the cosine law.

\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
R & = \sqrt{\left( 12.0\ \text{m} \right)^2+\left( 20.0\ \text{m} \right)^2-2\left(  12.0\ \text{m} \right)\left( 20.0\ \text{m} \right) \cos 110^\circ} \\
R & =  26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Before we can solve for θ, we need to solve for 𝛽 first using the sine law.

\begin{align*}
\frac{\sin \beta}{B} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{B \  \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{B \  \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{20.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 44.9290^\circ
\end{align*}

Now, we can solve for θ. 

\begin{align*}
\theta & = \left( 90^\circ + 20^\circ \right)-\beta \\
\theta & = 110^\circ - 44.9290^\circ \\
\theta & = 65.071 \\
\theta & = 65.1 ^\circ
\end{align*}

Therefore, the compass reading is

65.1^\circ, \text{North of East} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

Part B

Refer to Figure 3-7B

Figure 3-7B

First, we solve for the value of α using a simple geometry. 

\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ

We can solve for R using the cosine law. 

\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
R & = \sqrt{\left( 12.0\ \text{m} \right)^2+\left( 20.0\ \text{m} \right)^2-2\left(  12.0\ \text{m} \right)\left( 20.0\ \text{m} \right) \cos 110^\circ} \\
R & =  26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Before we can solve for θ, we need to solve for 𝛽 first using the sine law. 

\begin{align*}
\frac{\sin \beta}{A} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{A \  \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{A \  \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{12.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 25.0708^\circ
\end{align*}

Now, we can solve for θ. 

\begin{align*}
\theta & = 40^\circ + \beta \\
\theta & = 40^\circ +  25.0708^\circ\\
\theta & = 65.0708^\circ\\
\theta & = 65.1 ^\circ
\end{align*}

Therefore, the compass reading is 

65.1^\circ, \text{South of West} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

This is consistent with Part A because (A-B) = -(B-A).

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College Physics by Openstax Chapter 3 Problem 6

Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)

Solution:

Consider Figure 3-6A below with B drawn first before A.

Figure 3-6A

Compute for the value of angle β by adding 20° and the complement of 40°. This is by simple geometry. 

\begin{align*}
\beta & = 20^\circ +\left( 90^\circ -40^\circ \right) \\
\beta & = 70^\circ \\
\end{align*}

Solve for the magnitude of R using cosine law. 

\begin{align*}
R^2 & = A^2 +B^2 - 2 A B \cos \beta \\
R & = \sqrt{A^2 +B^2 - 2 A B \cos \beta} \\
R & = \sqrt{\left( 12.0 \ \text{m} \right)^2+\left( 20.0 \ \text{m} \right)^2-2\left( 12.0 \ \text{m} \right)\left( 20.0 \ \text{m} \right) \cos 70^\circ} \\
R & = 19.4892 \ \text{m}\\
R & = 19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

To solve for θ, we need to solve angle α first. This can be done using the sine law.

\begin{align*}
\frac{\sin \alpha}{A}  & = \frac{\sin \beta }{R} \\
\frac{\sin \alpha}{12.0 \ \text{m}}  & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\
\sin \alpha & = \frac{12.0 \ \text{m}\ \sin 70^\circ}{19.4892 \ \text{m}} \\
\alpha & = \sin ^{-1}  \left(  \frac{12.0 \ \text{m}\ \sin 70^\circ}{19.4892 \ \text{m}} \right) \\
\alpha & = 35.3516 ^ \circ
\end{align*}

Finally, we can solve for θ.

\begin{align*}
\theta &  = 40^\circ - 35.3516^\circ  \\
& = 4.6484 ^\circ \\
& = 4.65 ^\circ
\end{align*}

Therefore, the compass reading is

4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}

This is the same with Problem 5.

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College Physics by Openstax Chapter 3 Problem 5

Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.54, then this problem finds their sum R=A+B.)

Figure 3.54

Solution:

Consider Figure 3.5A shown below.

Figure 3.5A

Before we can use cosine law to solve for the magnitude of R, we need to solve for the interior angle 𝛽 first. The value of 𝛽 can be calculated by inspecting the figure and use simple knowledge on geometry. It is equal to the sum of 20° and the complement of 40°. That is

\beta = 20^\circ +\left( 90^\circ -40^\circ  \right) = 70^\circ

We can use cosine law to solve for R.

\begin{align*}
R^2 & =A^2+B^2 -2AB \cos \beta \\
R^2 & = \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ \\
R & = \sqrt{ \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ} \\
R  & =19.4892 \ \text{m} \\
R & =19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

We can solve for α using sine law.

\begin{align*}
\frac{\sin \alpha}{B} & = \frac{\sin \beta}{R} \\
\frac{\sin \alpha}{20.0\ \text{m}} & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\
\sin \alpha & = \frac{20.0 \ \sin 70^\circ }{19.4892} \\
\alpha & = \sin ^{-1}  \left(  \frac{20.0 \ \sin 70^\circ }{19.4892}  \right) \\
\alpha & = 74.6488 ^\circ 
\end{align*}

Then we solve for the value of θ by subtracting 70° from α.

\theta=74.6488 ^\circ -70 ^\circ = 4.65^\circ

Therefore, the compass reading is

4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}
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College Physics by Openstax Chapter 3 Problem 4

Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements \vec{A} and \vec{B} , as in Figure 3.53, then this problem asks you to find their sum \vec{R}=\vec{A}+\vec{B} .)

Figure 3.53

Solution:

Figure 3.4A

Consider Figure 3.54A.

The resultant of the two vectors \vec{A} and \vec{B} is labeled \vec{R}. This \vec{R} is directed \theta ^{\circ} from the x-axis.

We shall use the right triangle formed to solve for the unknowns.

Solve for the magnitude of the resultant.

\begin{align*}
R & = \sqrt{A^2 +B^2} \\
R & = \sqrt{\left(18.0 \ \text{m}  \right)^2+\left( 25.0 \ \text{m} \right)^2} \\
R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Solve for the value of \theta .

\begin{align*}
\theta & = \arctan \left( \frac{B}{A} \right) \\
\theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\
\theta & = 54.2^\circ 
\end{align*}

We need the complementary angle for the compass angle. 

\begin{align*}
90^\circ -54.2^\circ =35.8^\circ 
\end{align*}

Therefore, the compass angle reading is

\begin{align*}
35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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College Physics by Openstax Chapter 3 Problem 3

Find the north and east components of the displacement for the hikers shown in Figure 3.50.

Figure 3.50

Solution:

Refer to Figure 3-3-A for the north and east components of the displacement s of the hikers.

Figure 3-3-A

Considering the right triangle formed. The north component is computed as

\begin{align*}
\text{s}_{\text{north}} & = \left( 5.00 \ \text{km}  \right)\sin 40^\circ  \\
\text{s}_{\text{north}} & = 3.21 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

Using the same right triangle, the east component is computed as follows. 

\begin{align*}
\text{s}_{\text{east}} & = \left( 5.00 \ \text{km}  \right)\cos 40^\circ  \\
\text{s}_{\text{east}} & = 3.83 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics by Openstax Chapter 3 Problem 2

Find the following for path B in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

\begin{align*}

\text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\
\text{d} & = 1 200\ \text{m}  \ \qquad \  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part B

The magnitude of the displacement is 

\begin{align*}
\text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\
\text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\
\text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\
\text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The direction is

\begin{align*}
\theta & = \arctan \left( \frac{s_y}{s_x} \right) \\
\theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\
\theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics by Openstax Chapter 2 Problem 64

(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t=2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Figure 2.64

Solution:

Part A

To find the slope at t=2.5 s, we need the position values at t= 0 s and t=5 s. When t = 0 \ \text{s}, x = 0 \ \text{m}, and when t = 5 \ \text{s}, x = 17.5 \ \text{m}. The velocity at t=2.5 s is

\begin{align*}

\text{velocity} & =\text{slope} \\
\text{v} & =\frac{\Delta x}{\Delta t} \\
\text{v} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{v} & = \frac{17.5\ \text{m}-0\ \text{m}}{5 \ \text{s}-0\ \text{s}} \\
\text{v} & =3.5 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\

\end{align*}

Part B

When t = 10 \ \text{s}, x=2.5 \ \text{m}. Considering the points at t=5 s and t=10 s, the slope at 7.5 s is

\begin{align*}

\text{velocity} & =\text{slope} \\
\text{v} & =\frac{\Delta x}{\Delta t} \\
\text{v} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{v} & = \frac{2.5\ \text{m}-17.5\ \text{m}}{10 \ \text{s}-5\ \text{s}} \\
\text{v} & =-3.0 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\

\end{align*}
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College Physics by Openstax Chapter 2 Problem 63

Construct the position graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

Figure 2.18

Solution:

The figure with the corresponding examples are shown on this page: https://openstax.org/books/college-physics/pages/2-4-acceleration#import-auto-id2590556

The position-vs-time graph of the train’s motion is also graphed in the first figure here: https://openstax.org/apps/archive/20210713.205645/resources/a697c43432cdf2d09a02df47d2b746283b841fcd

(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) The velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.

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