Tag Archives: Feliciano and Uy

Header Photos for Elementary Differential Equations Chapter 9 Special Second Ordered Equations

Chapter 9: Special Second-Ordered Equations

Supplementary Problems

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

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College Physics by Openstax Chapter 2 Problem 65

A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t=5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

A graph of v(t) is shown for a world-class track sprinter in a 100-m race.
Figure 2.67

Solution:

Part A

To find the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

v_{\text{ave}}=6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

a=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 5

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \infty }\left(\frac{8x-5}{\sqrt{4x^2+3}}\right)

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SOLUTION:

Divide by the highest denominator power

\begin{align*}

\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8x-5}{\sqrt{4x^2+3}}\right) & =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8x-5}{\sqrt{4x^2+3}}\cdot \displaystyle \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\right) \\
\\

& =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{8x}{x}-\displaystyle \frac{5}{x}}{\sqrt{\displaystyle \frac{4x^2}{x^2}+\displaystyle \frac{3}{x^2}}}\right)\\
\\
& =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8-\displaystyle \frac{5}{x}}{\displaystyle \sqrt{4+\displaystyle \frac{3}{x^2}}}\right) \\

\\
& =\displaystyle \frac{8-0}{\sqrt{4+0}} \\
\\
& =\displaystyle \frac{8}{2} \\
\\
& =\displaystyle 4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 4

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \infty }\left(\frac{x^3+x+2}{x^2-1}\right)

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Solution:

Divide by the highest denominator power

\begin{align*}

\lim\limits_{x\to \infty }\left(\displaystyle \frac{x^3+x+2}{x^2-1}\right) & =\lim\limits_{x\to \infty }\left(\displaystyle \frac{x^3+x+2}{x^2-1}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\
\\
&=\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{x^3}{x^3}+\displaystyle \frac{x}{x^3}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{x^2}{x^3}-\displaystyle \frac{1}{x^3}}\right)\\
\\
&=\lim\limits_{x\to \infty }\left(\displaystyle \frac{1+\displaystyle \frac{1}{x^2}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{1}{x}-\displaystyle \frac{1}{x^3}}\right)\\
\\
&=\displaystyle \frac{1+0+0}{0-0}\\
\\
&=\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 3

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\right)

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Solution:

Divide by the highest denominator power

\begin{align*}
\lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\right) & =\lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\cdot \frac{\displaystyle\frac{1}{x^2}}{\displaystyle\frac{1}{x^2}}\right) \\
\\
&=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4x}{x^2}+\displaystyle\frac{5}{x^2}}{\displaystyle\frac{x^2}{x^2}+\displaystyle\frac{1}{x^2}}\right)\\
\\
&=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4}{x}+\displaystyle\frac{5}{x^2}}{1+\displaystyle\frac{1}{x^2}}\right) \\
\\
&=\displaystyle\frac{0+0}{1+0} \\
\\
&=0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 2

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right)

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Solution:

Divide by the highest denominator power.

\begin{align*}

\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right) & =\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\cdot \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right)\\
\\
&=\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3x^2}{x^3}+\displaystyle \frac{x}{x^3}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{x^3}{x^3}+\frac{8x}{x^3}+\frac{1}{x^3}}\right)\\
\\
& =\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3}{x}+\frac{1}{x^2}+\displaystyle \frac{2}{x^3}}{1+\displaystyle \frac{8}{x^2}+\frac{1}{x^3}}\right)\\
\\
&=\frac{0+0+0}{1+0+0} \\
\\
&=0\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 1

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \infty }\left( \frac{6x^3+4x^2+5}{8x^3+7x-3}\right)

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Solution:

Divide by the highest denominator power.

\begin{align*}
\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\right)&=\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\
\\
& =\displaystyle \lim\limits_{x\to \infty \:}\left(\displaystyle \frac{6+\displaystyle \frac{4}{x}+\displaystyle \frac{5}{x^3}}{8+\displaystyle \frac{7}{x^2}-\frac{3}{x^3}}\right)\\
\\
& =\displaystyle \frac{\lim\limits_{x\to \infty \:}\left(6+\displaystyle \frac{4}{x}+\frac{5}{x^3}\right)}{\lim\limits_{x\to \infty \:}\left(8+\displaystyle\frac{7}{x^2}-\displaystyle \frac{3}{x^3}\right)}\\
\\
& =\displaystyle \frac{6+0+0}{8+0-0}\\
\\
& =\displaystyle \frac{6}{8}\\
\\
& =\displaystyle \frac{3}{4} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 22

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PROBLEM:

If  \displaystyle f\left(x\right)=x^2-2x+3, find \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(x+2\right)-f\left(2\right)}{x}\right).

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SOLUTION:

\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(x+2\right)-f\left(2\right)}{x}\right)&=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(2^2-2\cdot 2+3\right)}{x}\right)\\
\\
& =\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(3\right)}{x}\right)\\
\\
&=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)\right)}{x}\right)\\
\\
\end{align*}

Direct substitution of x=0 gives the indeterminate form \frac{0}{0}. Therefore, we proceed by factoring the numerator.

\begin{align*}
& =\displaystyle \lim\limits_{x\to 0}\displaystyle \frac{\left(x+2\right)\left(x+2-2\right)}{x}\\
\\
& =\lim\limits_{x\to 0}\frac{\left(x+2\right)\left(x\right)}{x} \\
\\
& =\lim\limits_{x\to 0}\left(x+2\right) \\
\\
& =0+2 \\
\\
& =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 21

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PROBLEM:

If \displaystyle f\left(x\right)=x^2-2x+3, find \displaystyle \lim\limits_{x\to 2}\left(\frac{f\left(x\right)-f\left(2\right)}{x-2}\right).

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SOLUTION:

\begin{align*}
\displaystyle \lim\limits_{x\to 2}\left(\displaystyle \frac{f\left(x\right)-f\left(2\right)}{x-2}\right) & =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-\left(2^2-2\cdot 2+3\right)}{x-2}\right) \\
\\
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-3}{x-2}\right)\\
\\
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{x^2-2x}{x-2}\right)\\
\\

\end{align*}

Direct substitution of x=2 gives the indeterminate form \frac{0}{0}. Therefore, we proceed by factoring the numerator.

\begin{align*}
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{x\left(x-2\right)}{x-2}\right) \\
\\
& =\lim\limits_{x\to 2}\left(x\right) \\
\\
& =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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