Category Archives: Engineering Mathematics Blog

College Physics by Openstax Chapter 3 Problem 18

You drive 7.50 km in a straight line in a direction 15º east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

Solution:

Part A

Consider the illustration shown.

Let DE be the east component of the distance, and DN be the north component of the distance. 

\begin{align*}
D_E & = 7.50 \  \sin 15^\circ  \\
D_E & = 1.9411 \ \text{km} \\
D_E & = 1.94 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
\begin{align*}
D_N & = 7.50 \  \cos 15^\circ  \\
D_N & = 7.2444\ \text{km} \\
D_N & = 7.24 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part B

It can be obviously seen from the figure below that you still arrive at the same point if the east and north legs are reversed in order.

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College Physics by Openstax Chapter 3 Problem 17

Repeat Exercise 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

Figure 3.58 The two displacements A and B add to give a total displacement R having magnitude R and direction Î¸.

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

\begin{align*}
R & = \sqrt{A^2+B^2} \\
& = \sqrt{\left( 18.0 \text{m} \right)^2+\left( 25.0 \text{m} \right)^2} \\
& =30.806  \text{m} \\
& \approx 30.8  \text{m}  \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle. 

\begin{align*}
\theta & = \arctan \left( \frac{B}{A} \right)  \\
& = \arctan \left( \frac{25.0 \text{m}}{18.0 \text{m}} \right) \\
& = 54.246 ^\circ \\
& \approx  54.2 ^ \circ \\
\end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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College Physics by Openstax Chapter 3 Problem 16

Solve the following problem using analytical techniques: 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 A and B, as in Figure 3.58, then this problem asks you to find their sum R=A+B.)

Figure 3.58 The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

\begin{align*}
R & = \sqrt{A^2+B^2} \\
& = \sqrt{\left( 18.0\ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\
& =30.806 \ \text{m} \\
& \approx 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle. 

\begin{align*}
\theta & = \arctan\left( \frac{B}{A} \right) \\
& = \arctan\left( \frac{25.0\ \text{m}}{18.0\ \text{m}} \right) \\
& = 54.246 ^\circ \\
& \approx  54.2 ^ \circ 
\end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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College Physics by Openstax Chapter 3 Problem 15

Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.57.

Figure 3.57

Solution:

Consider the following figure.

Using the right triangle formed, we can solve for the east component and the north component. Let SE be the east component and SN be the north component of S.

\begin{align*}
S_E & = S \cos 45^\circ \\
& = \left( 123 \ \text{km} \right) \cos45^\circ \\
& = 86.974 \ \text{km} \\
& \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
\begin{align*}
S_N & = S \sin 45^\circ \\
& = \left( 123 \ \text{km} \right) \sin 45^\circ \\
& = 86.974 \ \text{km} \\
& \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the east and north components are equal at about 87.0 km.

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College Physics by Openstax Chapter 3 Problem 14

Find the following for path D in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

Figure 3.56 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

Looking at path D, we can see that it moves 2 blocks downward, 6 blocks to the right, 4 blocks upward, and 1 block to the left. Thus, the total distance of path D is

\begin{align*}
\text{distance} & = \left( 2\times 120\ \text{m} \right)+\left( 6\times 120\ \text{m} \right)+\left( 4\times 120\ \text{m} \right)+\left( 1\times 120\ \text{m} \right) \\
& = 1\ 560 \ \text{m} \\
& = 1.56 \times 10^{3} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part B

Looking at the initial and final position of path D, the final position is 5 blocks to the right or 600 meters to the right of the initial position, and 2 blocks or 240 meters upward from the initial position. Refer to the figure below.

Using the right triangle, we can solve for the displacement using the Pythagorean Theorem. 

\begin{align*}
\text{displacement} & = \sqrt{\left( 600\ \text{m} \right)^2+\left( 240\ \text{m} \right)^2} \\
& = 646.2198 \ \text{m} \\
& \approx 646 \  \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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Elementary Differential Equations by Dela Fuente, Feliciano and Uy Physical Application 2: Exponential Growth and Decay

A certain radioactive material follows the law of exponential change and has a half life of 38 hours. Find how long it takes for 90% of the radioactivity to be dissipated.

Solution:

Use the formula:

S=Ce^{-kt}

First, find the constant of proportionality. In the problem, after 38 hours, half of the radioactivity has been dissipated and a half has been retained. So we can assume that S = 0.5So when t = 38 hrs and C = So.

\left(0.5\right)So=\left(So\right)e^{-k\left(38\right)}

And then solve for k:

k=-0.018241

And then substitute k to the formula:

S=Ce^{-0.018241\left(t\right)}

Now we can solve for the time(t). According to the problem, 90% of the radioactivity is dissipated, so 10% is retained. So we can assume that S = 0.1So and change C = So.

\left(0.1\right)So=\left(So\right)e^{-0.018241\left(t\right)}

And then solve for time(t):

t\:=\:126.23\:hrs
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Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 10 Problem 3 — Applications of Ordinary First-Ordered Differential Equations

A tank contains 400 liters of brine. Twelve liters of brine, each containing 2.5 N of dissolved salt, enter the tank per minute, and the mixture, assumed uniform leaves at the rate of 8 liters per min. If the concentration is to be 2 N/litre at the end of one hour, how many newtons of salt should there be present in the tank originally?

Solution:

Consider the following illustration

\frac{dS}{dt}=\left(\frac{dS}{dt}\right)_{en}-\left(\frac{dS}{dt}\right)_{es}

Using,

V_{brine}+\left(rate\:of\:brine\:out\right)t\\\frac{dS}{dt}=\frac{8L}{M}\left(\frac{S}{400+4t}\right)\\=\frac{8S}{\left(400+4t\right)}\\\frac{dS}{dt}=\frac{2S}{\left(100+t\right)}

Using the general solution:

\frac{dS}{dt}=30-\frac{2S}{\left(100+t\right)}\\\frac{dS}{dt}+\frac{2S}{\left(100+t\right)}=30

To solve we will use First Order Linear Differential Equation (FOLDE) where:

P_{\left(t\right)}=\frac{2}{\left(100+t\right)}\:,\:Q_{\left(t\right)}=30

Solve for the integrating factor using the formula:

\sigma =e^{\int \:P_{\left(t\right)}dt}

Apply,

\sigma =e^{\int \:\frac{2}{100+t}dt}\\\sigma =e^{2ln\left(100+t\right)}\\\sigma \:=e^{ln\left(100+t\right)^2}\\\sigma \:=\left(100+t\right)^2

Substitute the given value to the formula:

S\sigma =\int \:\sigma Q\left(t\right)dt+C

Apply,

S\left(100+t\right)^2=\int \:\left(100+t\right)^230dt+C\\S\left(100+t\right)^2=30\int \:\left(100+t\right)^2dt+C\\S\left(100+t\right)^2=30\:\frac{\left(100+t\right)^{^3}}{3}dt+C\\S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C\rightarrow eqn.1

Evaluate C; @t=1hr

Convert 1hr to minutes, where 1hr is simply 60 minutes.

S\left(100+60\right)\:2\:=10\left(100+60\right)\:^3\\C=\frac{2N}{L}\:;\:C=\frac{S}{\left(400+4t\right)}

Get the value of S using the equation:

\:C=\frac{S}{\left(400+4t\right)}

Isolate S,

S=C\left(400+4t\right);\:C=2,\:t=60\\S=2\left(400+4\left(60\right)\right)\\S=1280N

Get the value of C using Eqn.1

S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C; S=1280 , t=60\\
1280\left(100+60\right)^2=10\left(100+60\right)^{^3}+C\\32768000=40960000+C\\32768000-40960000=C\\ C=-8192000

With the presence of the value of C we will now have our working equation:

S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000

Using the given working equation, solve for the value of S @ t=0

S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000; t=0\\S\left(100+0\right)^2=10\left(100+0\right)^{^3}-8192000\\\frac{S\left(1000\right)^2}{1000}=\frac{1808000}{1000}\\S=180.8 N
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 10 Problem 12 — Applications of Ordinary First-Ordered Differential Equations

A bacterial population follows the law of exponential growth. If between noon and 2 p.m. the population triples, at what time should the population become 100 times what it was at noon? At 10 a.m. what percentage was present?

SOLUTION:

First, we denote

P as the population of bacteria at anytime

Po as the original bacterial population

t = 0 (12 noon)

t = 2 (2 p.m.)

Let us determine the given and the required

GIVEN:

@12nn to 2p.m.; P= 3Po

REQUIRED:

  1. what time should the population become 100 times
  2. at noon
  3. percentage at 10 a.m.

Using the formula of Applications of Ordinary First-Ordered Differential Equations under Exponential Growth or Decay

\frac{dP}{dt}=kP \\
\int \:\frac{dP}{P}=\int \:kdt\\
e^{ln\:P}\:=\:e^{kt\:+\:C}\\
P=\:Ce^{kt}\:\:\:\:\:(Eq.1)\\
@t=0; P=P_o\\
P_o=Ce^{kt}\\
P_o=Ce^{k\left(0\right)}\\
P_o = C

Substituting to Eq.1., we get

P=\:P_{o\:}e^{kt}\:\:\:\:\:\:\:(Eq.2)

Then from the given condition, from 12 noon to 2 p.m., the population triples (using Eq.2), we will solve for the value of k

@t= 2\:;\:P= 3P_o\\
P=\:P_{o\:}e^{kt}\\
3P_{o\:}=\:P_{o\:}e^{k\left(2\right)}\\
k=0.54931

We will then come up with the working equation (WE), this will help us solve the required problems

P_{\:}=\:P_{o\:}e^{\left(0.54931\right)t}

1.) what time should the population become 100 times

Using WE,

t=?\:\:;\:\:P=100P_o\\
P_{\:}=\:P_{o\:}e^{\left(0.54931\right)t}\\
100P_{o\:}=\:P_{o\:}e^{\left(0.54931\right)t}\\
t=8.38\: hrs.\\
t= 8:22:48\: p.m. \; or\:8:23\:p.m.

2.) at noon 

P=P_o

3.) percentage at 10 a.m. 

@10 a.m.\:\:;\:\:t=-2\\
P_{\:}=\:P_{o\:}e^{\left(0.549\right)\left(-2\right)}\\
P_{\:}=\:P_{o\:}\left(0.33333\right)\\
\%=\frac{P}{P_o}{(100)}=\frac{P_o\left(0.33333\right)}{P_o}{(100)}\\
\%=\:33.33\%

Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 10 Problem 2 — Applications of Ordinary First-Ordered Differential Equations

Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.

Solution:

Plot points on the curve, 

A(x_{1},y_{1})

We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.

y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)

As the normal intersects the x-axis, y = 0

Substituting to the previous equation, we get

\begin{align*}
-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right)\\
-my_{1}&=-1\left(x-x_{1}\right)\\
-my_{1}&=-x+x_{1}\\
x&=x_{1}+my_{1}
\end{align*}

By using distance formula, from the origin (0,0), to point (x1+y1) = from intersection to (x1+y1) 

\begin{align*}
\sqrt{\left(x_{1}-0\right)^2+\left(y_{1}-0\right)^2}&=\sqrt{\left(x_{1}+my_{1}-x_{1}\right)^2+\left(0-y_{1}\right)^2}\\
\sqrt{x_{1}^2+y_{1}^2}&=\sqrt{m^2y_{1}^2+y_{1}^2}\\
x_{1}^2+y_{1}^2&=m^2y_{1}^2+y_{1}^2\\
x_{1}^2&=m^2y_{1}^2\\
x_{1}&=m_{1}y_{1}\:\:\:\:;m=\frac{dy}{dx}\\
x_{1}&=\frac{dy}{dx}y_{1}
\end{align*}

Change x1 and y1 to x and y,

\begin{align*}
x&=y\frac{dy}{dx}\\
xdx&=ydy
\end{align*}

By integrating,

\begin{align*}

\int \:ydx&=\int \:xdy\\
\frac{y^2}{2}&=\frac{x^2}{2}+C\\
y^2&={x^2}+2C\\
\end{align*}

We get,

y^2-x^2-=2C
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 1 – Special Second-Ordered Differential Equations

Find the General Solution

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

Solution:

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

 solve\; the\; equation\; using\; case\; 1,\\

let\; u=\frac{dy}{dx} \\
\int \:\frac{du}{dx}\:=\:\int \:\left(6x\:+\:3\right)\\

using\;  separation\; of\; variable\; divide\;   both\;  sides\;  by\;  dx,\\

\int \:du\:=\:\int \:\left(6x\:+\:3\right)dx\\

by\; integrating\; using\; the\; sum\; rule:\\
we\; get,\\

u=3x^2\:+\:3x\:+\:C_1\\

substitute\; the\; value\; of\; u=\frac{dy}{dx} \\

\frac{dy}{dx}=3x^2\:+\:3x\:+\:C_1\\

using\;  separation\; of\; variable:\\

\int \:dy = \int \:\left(3x^2+3x\:+\:C_1\right)dx\\

apply\; the\; sum\; rule:\\

\int \:dy=\int \:3x^2dx+\int  \:3xdx+ \int \:C_1dx\\

y=x^3+\frac{3x^2}{2}+C_1x+C_2\\
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