Chapter 7, Problem 55EP

a.

To determine

To find: How many people have entered the park by $5.00$ P.M. $\left(t=17\right)$ .

Answer to Problem 55EP

The number of people entered the park is $6004$ .

Explanation of Solution

Given information:

  $\begin{array}{l}E\left(t\right)=\frac{15600}{t^2-24t+160}\\ L\left(t\right)=\frac{9890}{t^2-38t+370}\end{array}$

Calculation:

Given that the rate at which people entered the amusement park is given by,

  $E\left(t\right)=\frac{15600}{t^2-24t+160}$

Integrating the given function,

  $\begin{array}{c}\text{Number of people}={\displaystyle \underset{9}{\overset{17}{\int }}E\left(t\right)}\\ ={\displaystyle \underset{9}{\overset{17}{\int }}\frac{15600}{t^2-24t+160}}\\ =15600{\displaystyle \underset{9}{\overset{17}{\int }}\frac{1}{t^2-24t+144+16}dt}\\ =15600{\displaystyle \underset{9}{\overset{17}{\int }}\frac{1}{{\left(t-12\right)}^2+4^2}dt}\end{array}$

By substitution method,

  $\begin{array}{c}u=t-12\\ du=dt\end{array}$

To find the new boundaries:

At $x=17$ the upper boundary is,

  $\begin{array}{c}u=17-12\\ =5\end{array}$

At $x=9$ the lower boundary is,

  $\begin{array}{c}u=9-12\\ =-3\end{array}$

Now,

  $\begin{array}{c}\text{Number of people}=15600{\displaystyle \underset{9}{\overset{17}{\int }}\frac{1}{{\left(t-12\right)}^2+4^2}dt}\\ =15600{\displaystyle \underset{-3}{\overset{5}{\int }}\frac{1}{u^2+16}}dt\\ =15600{\displaystyle \underset{-3}{\overset{5}{\int }}\frac{1}{16\left(\frac{u^2}{16}+1\right)}}dt\\ =975{\displaystyle \underset{-3}{\overset{5}{\int }}\frac{1}{\left({\left(\frac{u}{4}\right)}^2+1\right)}}dt\end{array}$

Using substitution method,

  $\begin{array}{c}s=\frac{u}{4}\\ ds=\frac{u}{4}du\end{array}$

To find the new boundaries:

At $u=5$ the upper boundary is,

  $s=\frac{5}{4}$

At $u=-3$ the lower boundary is,

  $s=-\frac{3}{4}$

Now,

  $\begin{array}{c}\text{Number of people}=975{\displaystyle \underset{-3}{\overset{5}{\int }}\frac{1}{\left({\left(\frac{u}{4}\right)}^2+1\right)}}\frac{du}{4}\\ =3900{\displaystyle \underset{\frac{-3}{4}}{\overset{\frac{5}{4}}{\int }}\frac{1}{\left({\left(s\right)}^2+1\right)}}ds\end{array}$

Applying fundamental theorem of calculus,

  ${\displaystyle \int \frac{1}{1+x^2}=\arctan x+C}$

Now,

  $\begin{array}{c}\text{Number of people}=3900{\displaystyle \underset{\frac{-3}{4}}{\overset{\frac{5}{4}}{\int }}\frac{1}{\left({\left(s\right)}^2+1\right)}}ds\\ =3900{\left.\left(\arctan s\right)\right|}_{\frac{-3}{4}}^{\frac{5}{4}}\\ =3900\left(\arctan \frac{5}{4}-\arctan \frac{-3}{4}\right)\\ =3900\left(0.896-\left(-0.6435\right)\right)\\ =3900\left(1.5296\right)\\ \approx 6004.27\end{array}$

Therefore, the number of people entered the park by $5.00$ P.Mis $6004$ .

b.

To determine

To find: How many dollars are collected from admissions to the park on the same day.

Answer to Problem 55EP

The dollars collected from the admissions to the park is $\$104048$ .

Explanation of Solution

Given information:

  $\begin{array}{l}E\left(t\right)=\frac{15600}{t^2-24t+160}\\ L\left(t\right)=\frac{9890}{t^2-38t+370}\end{array}$

Calculation:

Given that the rate at which people entered the amusement park is given by,

  $E\left(t\right)=\frac{15600}{t^2-24t+160}$

Integrating the given function,

  $\begin{array}{c}{\text{Number of people}}_2={\displaystyle \underset{17}{\overset{23}{\int }}E\left(t\right)}\\ ={\displaystyle \underset{17}{\overset{23}{\int }}\frac{15600}{t^2-24t+160}}\\ =15600{\displaystyle \underset{17}{\overset{23}{\int }}\frac{1}{t^2-24t+144+16}dt}\\ =15600{\displaystyle \underset{17}{\overset{23}{\int }}\frac{1}{{\left(t-12\right)}^2+4^2}dt}\end{array}$

By substitution method,

  $\begin{array}{c}u=t-12\\ du=dt\end{array}$

To find the new boundaries:

At $x=23$ the upper boundary is,

  $\begin{array}{c}u=23-12\\ =11\end{array}$

At $x=17$ the lower boundary is,

  $\begin{array}{c}u=17-12\\ =5\end{array}$

Now,

  $\begin{array}{c}{\text{Number of people}}_2=15600{\displaystyle \underset{17}{\overset{23}{\int }}\frac{1}{{\left(t-12\right)}^2+4^2}dt}\\ =15600{\displaystyle \underset{5}{\overset{11}{\int }}\frac{1}{u^2+16}}dt\\ =15600{\displaystyle \underset{5}{\overset{11}{\int }}\frac{1}{16\left(\frac{u^2}{16}+1\right)}}dt\\ =975{\displaystyle \underset{5}{\overset{11}{\int }}\frac{1}{\left({\left(\frac{u}{4}\right)}^2+1\right)}}dt\end{array}$

Using substitution method,

  $\begin{array}{c}s=\frac{u}{4}\\ ds=\frac{u}{4}du\end{array}$

To find the new boundaries:

At $u=11$ the upper boundary is,

  $s=\frac{11}{4}$

At $u=5$ the lower boundary is,

  $s=\frac{5}{4}$

Now,

  $\begin{array}{c}{\text{Number of people}}_2=975{\displaystyle \underset{5}{\overset{11}{\int }}\frac{1}{\left({\left(\frac{u}{4}\right)}^2+1\right)}}\frac{du}{4}\\ =3900{\displaystyle \underset{\frac{5}{4}}{\overset{\frac{11}{4}}{\int }}\frac{1}{\left({\left(s\right)}^2+1\right)}}ds\end{array}$

Applying fundamental theorem of calculus,

  ${\displaystyle \int \frac{1}{1+x^2}=\arctan x+C}$

Now,

  $\begin{array}{c}{\text{Number of people}}_2=3900{\displaystyle \underset{\frac{5}{4}}{\overset{\frac{11}{4}}{\int }}\frac{1}{\left({\left(s\right)}^2+1\right)}}ds\\ =3900{\left.\left(\arctan s\right)\right|}_{\frac{5}{4}}^{\frac{11}{4}}\\ =3900\left(\arctan \frac{11}{4}-\arctan \frac{5}{4}\right)\\ =3900\left(1.222-\left(-0.896\right)\right)\\ =3900\left(0.326\right)\\ \approx 1271.28\end{array}$

The total cost is calculated by,

  $\begin{array}{c}\text{Total cost=15}\left({\text{Number of people}}_{\text{1}}\right)\text{+11}\left({\text{Number of people}}_{\text{2}}\right)\\ =15\left(6004.27\right)+11\left(1271.28\right)\\ =90064.05+13984.08\\ =104048.11\\ \approx \$104048\end{array}$

Therefore, the dollars collected from the admissions to the park is $\$104048$ .

c.

To determine

To find: The value of $H\left(17\right)$ and explain the meaning of $H\left(17\right)$ and $H\text{'}\left(17\right)$ .

Answer to Problem 55EP

The value of $H\left(17\right)$ is $3725$ .

Explanation of Solution

Given information:

  $\begin{array}{l}E\left(t\right)=\frac{15600}{t^2-24t+160}\\ L\left(t\right)=\frac{9890}{t^2-38t+370}\end{array}$

Calculation:

It is known that,

  $\begin{array}{c}H\left(t\right)={\displaystyle {\int }_9^t\left(E\left(t\right)-L\left(t\right)\right)dt}\\ H\text{'}\left(t\right)=E\left(t\right)-L\left(t\right)\end{array}$

At $t=17$ ,

  $\begin{array}{c}H\text{'}\left(t\right)=E\left(t\right)-L\left(t\right)\\ H\text{'}\left(t\right)=\frac{15600}{{\left(17\right)}^2-24\left(17\right)+160}-\frac{9890}{{\left(17\right)}^2-38\left(17\right)+370}\\ =380.488-760.769\\ =-380.28\\ \approx -380\end{array}$

According to this, the number of persons in the park at $5.00\text{ P}\text{.M}$ is $H\left(17\right)=3725$ , and $H\text{'}\left(17\right)$ represents the rate at which that number is changing.

In this instance, the population is now dropping at a rate of roughly $380$ people each hour $t=17$ .

d.

To determine

To find: The value of time $t$ when the number of people in the park is a maximum.

Answer to Problem 55EP

The value of time $t$ is $15.795$ .

Explanation of Solution

Given information:

  $\begin{array}{l}E\left(t\right)=\frac{15600}{t^2-24t+160}\\ L\left(t\right)=\frac{9890}{t^2-38t+370}\end{array}$

Calculation:

When the rate of change equals zero, as shown in the equation below, the number of persons in the park is at its highest.

  $\begin{array}{c}H\text{'}\left(t\right)=E\left(t\right)-L\left(t\right)\\ 0=\frac{15600}{{\left(t\right)}^2-24\left(t\right)+160}-\frac{9890}{{\left(t\right)}^2-38\left(t\right)+370}\end{array}$

Multiply the equation by its least common denominator (LCD). For the equation above,

  $\begin{array}{c}LCD=\left[{(t)}^2-24(t)+160\right]\left[{(t)}^2-38(t)+370\right]\\ 0=\frac{15600}{{(t)}^2-24(t)+160}-\frac{9890}{{(t)}^2-38(t)+370}\\ =\left[\frac{15600}{{(t)}^2-24(t)+160}-\frac{9890}{{(t)}^2-38(t)+370}\right]\left[{(t)}^2-24(t)+160\right]\left[{(t)}^2-38(t)+370\right]\\ =\frac{15600}{t^2-24t+160}\left(t^2-24t+160\right)\left(t^2-38t+370\right)-\frac{9890}{t^2-38t+370}\left(t^2-24t+160\right)\left(t^2-38t+370\right)\\ =15600\left(t^2-38t+370\right)-9890\left(t^2-24t+160\right)\\ =15600t^2-592800t+5772000-9890t^2+237360-1582400\\ =5710t^2-355440t+4189600\end{array}$

Using quadratic formula, to solve for $t$ .

The standard form is given by,

  $a{x}^2+bx+c=0$

The quadratic formula is given by,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Dividing on both sides by $5710$ to get,

  $\begin{array}{c}0=5710t^2-355440t+418960\\ \frac{0}{5710}=\frac{5710t^2}{5710}-\frac{355440t}{5710}+\frac{4189600}{5710}\\ 0=t^2-\frac{35544t}{571}+\frac{418960}{571}\\ t_{1,2}=\frac{-\left(-\frac{3554}{571}\right)\pm \sqrt{{\left(-\frac{35544}{571}\right)}^2-4\cdot 1\cdot \frac{418960}{571}}}{2\cdot 1}\\ t_{1,2}=\frac{\left(\frac{25541}{571}\right)\pm \sqrt{\frac{{35544}^2}{526011}-\frac{1675811}{571}}}{2-1}\\ t=46.45;15.795\end{array}$

The period when the number of persons is at its peak is $t=15.795$ because the interval for $t$ is $9\le t\le 23$ .

Therefore, the value of time $t$ when the number of people in the park is a maximumis $15.795$ .