Author Archives: Engineering Math

$Elementary Differential Equations de la Fuente, Feliciano and Uy$

Separation of Variables| Elementary Differential Equations|dela Fuente, Feliciano, and Uy|Example 2

Find the complete solution of the following differential equation:

$L\:dI+RI\:dt=0\:\left(L\:and\:R\:are\:constants\right).$

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$Elementary Differential Equations de la Fuente, Feliciano and Uy$

Separation of Variables| Elementary Differential Equations|dela Fuente, Feliciano, and Uy|Problem 1

Find the solution of the following differential equation:

$\frac{dy}{dx}=\frac{4x+xy^2}{y-x^2y}$

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College Physics by Openstax Chapter 2 Problem 38

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s² for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Solution:

We are given the following: $v_0=11.5 \ \text{m/s}$ ; $a=0.500 \ \text{m/s}^2$ ; and $\Delta t=7.00 \ \text{s}$ .

Part A

To solve for the final velocity, we are going to use the formula

v_f=v_0+at

Substituting the given values:

\begin{align*} v_f &=v_0+at\\ v_f&=11.5\ \text{m/s}+\left( 0.500\ \text{m/s}^2 \right)\left( 7.00\ \text{s} \right)\\ v_f&=15.0\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let $t_{const}$ be the time it takes to reach the finish line without accelerating:

\begin{align*} t_{const}&=\frac{x}{v_0}\\ t_{const}&=\frac{300\ \text{m}}{11.5\ \text{m/s}}\\ t_{const}&=26.1\ \text{m/s} \end{align*}

Now let $d$ be the distance traveled during the 7 seconds of acceleration. We know $t=7.00 \ \text{s}$ so

\begin{align*} d&=v_0t+\frac{1}{2}at^2\\ d&=\left( 11.5\ \text{m/s} \right)\left( 7.00\ \text{s} \right)+\frac{1}{2}\left( 0.500\ \text{m/s} ^2\right)\left( 7.00\ \text{s} \right)^2\\ d&=92.8\ \text{m} \end{align*}

Let $t'$ be the time it will take the rider at the constant final velocity to complete the race:

\begin{align*} t'&=\frac{x-d}{v}\\ t'&=\frac{300\ \text{m}-92.8\ \text{m}}{15.0\ \text{m/s}}\\ t'&=13.8\ \text{s} \end{align*}

So the total time $T$ it will take the accelerating rider to reach the finish line is

\begin{align*} T&=t+t'\\ T&=7.00\ \text{s}+13.8\ \text{s}\\ T&=20.8\ \text{s} \end{align*}

Finally, let $T^{*}$ be the time saved. So

\begin{align*} T^{*}&=26.1\ \text{s}-20.8\ \text{s}\\ T^{*}&={\color{DarkGreen} 5.3\ \text{s}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For rider 2, we are given the following values: $\Delta x'=295 \ \text{m}$ ; $v'=11.8 \ \text{m/s}$

Let $t_2$ be the time it takes for rider 2 to reach the finish line.

We are going to use the formula

t_2=\frac{\Delta x'}{v'}

Substituting the given values:

\begin{align*} t_2 & =\frac{x'}{v'} \\ t_2 & =\frac{295\:\text{m}}{11.8\:\text{m/s}} \\ t_2 & =25.0\:\text{s} \end{align*}

The time difference is

\begin{align*} \text{time difference} & =t_2-T \\ \text{time difference} & =25.0\:\text{s}-20.817\:\text{s} \\ \text{time difference} & =4.2\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, he finishes 4.2 s after the winner.

When the other racer reaches the finish line, he has been traveling at 11.8 m/s for 4.2 seconds, so the other racer finishes

\begin{align*} \Delta x & =\left(11.8\:\text{m/s}\right)\left(4.2\:\text{s}\right) \\ \Delta x & =49.56\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

behind the other racer.

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Successive Wins: Challenging Problem in Probability

To encourage Elmer’s promising tennis career, his father offers him a prize if he wins (at least) two tennis sets in a row in a three-set series to be played with his father and the club champion alternately: father-champion-father or champion-father-champion, according to Elmer’s choice. The champion is a better player than Elmer’s father. Which series should Elmer choose?

Solution:

Since the champion plays better than the father, it seems reasonable that fewer sets should be played with the champion. On the other hand, the middle set is the key one, because Elmer cannot have two wins in a row without winning the middle one. Let C stand for the champion, F for father, and W and L for win and loss by Elmer. Let f be the probability of Elmer’s winning any set from his father, c the corresponding probability of winning from the champion. The table shows only possible prize-winning sequences together with their probabilities, given independence between sets, for the two choices.

Set with:

Father First

F	C	F	Probability
W	W	W	fcf
W	W	L	fc(1-f)
L	W	W	(1-f)cf
Total			fc(2-f)

Champion First

C	F	C	Probability
W	W	W	cfc
W	W	L	cf(1-c)
L	W	W	(1-c)fc
Total			fc(2-c)

Since Elmer is more likely to best his father than to beat the champion, f is larger than c, and 2-f is smaller than 2-c, and so Elmer should choose CFC. For example, for f=0.8, c=0.4, the chance of winning the prize with FCF is 0.384, that for CFC is 0.512. Thus, the importance of winning the middle game outweighs the disadvantage of playing the champion twice.

Many of us have a tendency to suppose that the higher the expected number of successes, the higher the probability of winning a prize, and often this supposition is useful. But occasionally a problem has special conditions that destroy this reasoning by analogy. In our problem, the expected number of wins under CFC is 2c+f, which is less than the expected number of wins for FCF, 2f+c. In our example with f=0.8 and c=0.4, these means are 1.6 and 2.0 in that order. This opposition of answers gives the problem flavor.

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The Sock Drawer: Challenging Problem in Probability

A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2.

a) How small can the number of socks in the drawer be?

b) How small if the number of black socks is even?

Solution:

Just to set the pattern, let us do a numerical example first. Suppose there were 5 red and 2 black socks; then the probability of the first sock’s being red would be 5/(5+2). If the first were red, the probability of the second’s being red would be 4/(4+2), because one red sock has already been removed. The product of these two numbers is the probability that both socks are red:

\frac{5}{5+2}\times \frac{4}{4+2}=\frac{5\left( 4 \right)}{7\left( 6 \right)}=\frac{10}{21}

This result is close to 1/2, but we need exactly 1/2. Now let us go at the problem algebraically.

Let there be $r$ red and $b$ black socks. The probability of the first sock’s being red is $\frac{r}{r+b}$ ; and if the first sock is red, the probability of the second’s being red now that a red has been removed is $\frac{r-1}{r+b-1}$ . Then we require the probability that both are red to be $\frac{1}{2}$ , or

\frac{r}{r+b}\times \frac{\:r-1}{r+b-1}=\frac{1}{2}

One could just start with $b=1$ and try successive values of $r$ , then go to $b=2$ and try again, and so on. That would get the answer quickly. Or we could play along with a little more mathematics. Notice that

\frac{r}{r+b}>\frac{\:r-1}{r+b-1}

Therefore, we can create the inequalities

\left(\frac{r}{r+b}\right)^2>\frac{1}{2}>\left(\frac{\:r-1}{r+b-1}\right)^2

Taking the square roots, we have, for $r>1$ .

\frac{r}{r+b}>\frac{1}{\sqrt{2}}>\frac{\:r-1}{r+b-1}

From the first inequality we get

r>\frac{1}{\sqrt{2}}\left( r+b \right)

\begin{align*} r & >\frac{1}{\sqrt{2}-1}b \\ \\ r & > \left( \sqrt{2}+1 \right)b \end{align*}

From the second we get

\left( \sqrt{2}+1 \right)b>r-1

or all told

\left(\sqrt{2}+1\right)b+1>r>\left(\sqrt{2}+1\right)b

For $b=1$ , $r$ must be greater than 2.414 and less than 3.414, and so the candidate is $r=3$ . For $r=3, \ b=1$ , we get

P\left(2\:\text{red socks}\right)=\frac{3}{4}\cdot \frac{2}{3}=\frac{1}{2}

And so, the smallest number of socks is 4.

Beyond this, we investigate even values of $b$ .

$b$	$r$ is between	eligible $r$	$P\left(2 \ \text{red socks}\right)$
2	5.8, 4.8	5	$\frac{5\left( 4 \right)}{7\left( 6 \right)} \neq \frac{1}{2}$
4	10.7, 9.7	10	$\frac{10\left( 9 \right)}{14\left( 13 \right)} \neq \frac{1}{2}$
6	15.5, 14.5	15	$\frac{15\left( 14 \right)}{21\left( 20 \right)} = \frac{1}{2}$

And so, 21 is the smallest number of socks when b is even. If we were to go on and ask for further values of r and b so that the probability of two red socks is 1/2, we would be wise to appreciate that this is a problem in the theory of numbers. It happens to lead to a famous result in Diophantine Analysis obtained from Pell’s equation. Try r = 85, b = 35.

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College Physics by Openstax Chapter 2 Problem 37

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

(c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

Solution:

We are given the following: $v_0=0\ \text{m/s}$ ; $v_f=145 \ \text{m/s}$ ; and $\Delta t=4.45 \ \text{sec}$ .

Part A

To compute for the average acceleration $a$ , we are going to use the formula

a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_0}{\Delta t}

Substituting the given values, we have

\begin{align*} a & =\frac{v_f-v_0}{\Delta t} \\ a & =\frac{145\:\text{m/s}-0\:\text{m/s}}{4.45\:\text{s}} \\ a & =32.6\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following: $a=32.6 \ \text{m/s}^2$ ; $v_0=0 \ \text{m/s}$ ; and $\Delta x=402 \ \text{m}$ .

Since we do not have any information on time, we are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

To compute for the final velocity, we have

v_f=\sqrt{\left(v_0\right)^2+2a\Delta \:x}

Substituting the given values:

\begin{align*} v_f & =\sqrt{\left(v_0\right)^2+2a\Delta \:x} \\ v_f & =\sqrt{\left(0\:\text{m/s}\right)^2+2\left(32.6\:\text{m/s}^2\right)\left(402\:\text{m}\right)} \\ v_f & =162\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The final velocity is greater than that used to find the average acceleration because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6 m/s² during the last few meters, but substantially less, and the final velocity would be less than $162\:\text{m/s}$ .

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 1

What is the hydrologic cycle? What are the pathways that precipitation falling onto the land surface of the Earth is dispersed to the hydrologic cycle?

Solution:

The hydrologic cycle is a continuous process in which water is evaporated from water surfaces and the oceans, moves inland as moist are masses, and produces precipitation if the correct vertical lifting conditions exist.

A portion of precipitation (rainfall) is retained in the soil near where it falls and returns to the atmosphere via evaporation (the conversion of water vapor from a water surface) and transpiration (the loss of water vapor through plant tissue and leaves). Combined loss is called evapotranspiration and is a maximum value if the water supply in the soil moisture conditions and soil may reenter channels layer as interflow or may percolate to recharge the shallow groundwater. The remaining portion of the precipitation becomes overland flow or direct runoff which flows generally in a downgradient direction to accumulate in local streams that then flow into rivers.

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 1

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PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)$

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SOLUTION:

A straight substitution of $x=4$ leads to the indeterminate form $\frac{0}{0}$ which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

\begin{align*} \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)& =\lim\limits_{x\to 4}\left(\frac{\left(x-4\right)\left(x^2+4x+16\right)}{\left(x+4\right)\left(x-4\right)}\right)\\ \\ & =\lim\limits_{x\to 4}\left(\frac{x^2+4x+16}{x+4}\right)\\ \\ & =\frac{\left(4\right)^2+4\left(4\right)+16}{4+4}\\ \\ & =\frac{48}{8}\\ \\ & =6 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}

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Brain Teaser #2

If Susan is 10, Arabella is 20, Jim and Neal are both 5, and Richard is 10, how much is Jennifer by the same system?

College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Solution:

Part A

We are given the following: $v_0=22.0\:\text{m/s}$ ; $a=-0.150\:\text{m/s}^2$ ; and $\Delta x=210\:\text{m}$

We are required to solve for time, $t$ . We are going to use the formula

\Delta x=v_0t+\frac{1}{2}at^2

Substituting the given values, we have

\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ 210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2 \end{align*}

If we simplify and rearrange the terms into a general quadratic equation, we have

0.075t^2-0.22t+210=0

Solve for $t$ using the quadratic formula. We are given $a=0.075;\:b=-22t;\:c=210$ .

\begin{align*} t & =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ t& =\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}\\ \end{align*}

There are two values of $t$ that can satisfy the quadratic equation.

t=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}

Discard $t=283.46 \ \text{s}$ as it can be seen from the problem that this is not a feasible solution. So, we have

t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

We have the same given values from part a. We are going to solve $v_f$ in the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

So we have

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\ v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

We are given the same values as in the previous parts, except that for the value of $\Delta x$ since we should incorporate the length of the train. For the distance, $\Delta x$ , we have

\begin{align*} \Delta x & =210\:\text{m}+130\:\text{m} \\ \Delta x & = 340 \ \text{m} \end{align*}

To solve for time $t$ , we are going to use the formula

\Delta x=v_ot+\frac{1}{2}at^2

If we rearrange the formula into a general quadratic equation and solve for $t$ using the quadratic formula, we come up with

t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

Substituting the given values:

\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\ t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\ t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

We have the same given values. We are going to solve for $v_f$ in the equation

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\ v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Engineering Mathematics and Sciences

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Author Archives: Engineering Math

Separation of Variables| Elementary Differential Equations|dela Fuente, Feliciano, and Uy|Example 2

Separation of Variables| Elementary Differential Equations|dela Fuente, Feliciano, and Uy|Problem 1

Find the solution of the following differential equation:

$\frac{dy}{dx}=\frac{4x+xy^2}{y-x^2y}$

College Physics by Openstax Chapter 2 Problem 38

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s² for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Successive Wins: Challenging Problem in Probability

Solution:

The Sock Drawer: Challenging Problem in Probability

A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2.

a) How small can the number of socks in the drawer be?

b) How small if the number of black socks is even?

Solution:

College Physics by Openstax Chapter 2 Problem 37

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 1

What is the hydrologic cycle? What are the pathways that precipitation falling onto the land surface of the Earth is dispersed to the hydrologic cycle?

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 1

PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)$

Brain Teaser #2

If Susan is 10, Arabella is 20, Jim and Neal are both 5, and Richard is 10, how much is Jennifer by the same system?

College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Find the solution of the following differential equation:

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s2 for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Solution:

A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2.

a) How small can the number of socks in the drawer be?

b) How small if the number of black socks is even?

Solution:

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

What is the hydrologic cycle? What are the pathways that precipitation falling onto the land surface of the Earth is dispersed to the hydrologic cycle?

PROBLEM:

Evaluate lim⁡x→4(x3−64x2−16)\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)x→4lim​(x2−16x3−64​)

If Susan is 10, Arabella is 20, Jim and Neal are both 5, and Richard is 10, how much is Jennifer by the same system?

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s² for 7.00 s.

Evaluate $\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)$

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.