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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\%

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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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 20

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PROBLEM:

If \displaystyle f\left(x\right)=\sqrt{x}, find \displaystyle \lim\limits_{x\to 0}\left(\frac{f\left(9+x\right)-f\left(9\right)}{x}\right).

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SOLUTION:

\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(9+x\right)-f\left(9\right)}{x}\right)=\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-\sqrt{9}}{x}\right)

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

\begin{align*}
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-3}{x}\cdot \displaystyle \frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{9+x-9}{x\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{x}{x\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{1}{\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\left(\displaystyle \frac{1}{\left(\sqrt{9+0}+3\right)}\right)\\
\\
& =\displaystyle \frac{1}{6} \ \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 19

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PROBLEM:

If f\left(x\right)=\sqrt{x}, find \displaystyle \lim\limits_{x\to 4}\left(\frac{f\left(x\right)-f\left(4\right)}{x-4}\right)

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SOLUTION:

\displaystyle \lim\limits_{x\to 4}\left(\displaystyle \frac{f\left(x\right)-f\left(4\right)}{x-4}\right)=\lim\limits_{x\to 4}\left(\displaystyle \frac{\sqrt{x}-\sqrt{4}}{x-4}\right)

Direct substitution of x=4 gives the indeterminate form \frac{0}{0}. Therefore, we proceed by rationalizing the numerator.

\begin{align*}
& =\lim\limits_{x\to 4}\left(\displaystyle \frac{\sqrt{x}-2}{x-4}\right)\cdot \displaystyle \frac{\sqrt{x}+2}{\sqrt{x}+2} \\
\\
& =\lim\limits_{x\to 4}\left(\displaystyle \frac{x-4}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right) \\
\\
& =\lim\limits_{x\to 4}\left(\displaystyle \frac{1}{\sqrt{x}+2}\right) \\
\\
& =\left(\displaystyle \frac{1}{\sqrt{4}+2}\right) \\
\\
& =\displaystyle \frac{1}{4} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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