Chapter 13, Problem 5P

(a)

To determine

The heating degree hours that were experienced on a particular day from $t=0$ to $t=24$, that day temperature was modeled by the function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$.

Answer to Problem 5P

On a particular day when temperature was modeled by the function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ the heating degree hours were experienced is 1625.28 heating degree-hours.

Explanation of Solution

Given:

The formula for heating degree-hours is,

$\text{Heating}\text{degree-hours}=\text{temperature}\times \text{time}$

The temperature was modeled by the function is,

$D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$

Where, t was measured hours since midnight.

Calculation:

The area under the function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ is equal to the heating degree hours that were experienced on a particular day from $t=0$ to $t=24$.

Calculate the area under the function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ by below formula,

$A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(x_k\right)}\Delta x$ (1)

$\Delta x$ can be calculated by below formula,

$\Delta x=\frac{b-a}{n}$ (2)

Formula to calculate the value of $x_k$ is,

$x_k=a+k\Delta x$ (3)

Substitute 0 for a and 24 for b in equation (2) to find the value of $\Delta x$,

$\begin{array}{c}\Delta x=\frac{24-0}{n}\\ =\frac{24}{n}\end{array}$

Substitute 0 for a and $\frac{24}{n}$ for $\Delta x$ in equation (3) to find the value of $x_k$,

$\begin{array}{c}{x}_k=0+k\cdot \frac{24}{n}\\ =\frac{24k}{n}\end{array}$

Find the value of $f\left(x_k\right)$ in function $D\left(t\right)$

$\begin{array}{c}f\left(x_k\right)=61+\frac{6}{5}{x}_k-\frac{1}{25}{x}_k{}^2\\ =61+\frac{6}{5}\cdot \left(\frac{24}{n}\right)-\frac{1}{25}\cdot {\left(\frac{24}{n}\right)}^2\\ =61+\frac{144k}{5n}-\frac{576k^2}{25n^2}\end{array}$

Substitute $\frac{24}{n}$ for $\Delta x$, $\frac{24k}{n}$ for $x_k$ and $61+\frac{144k}{5n}-\frac{576k^2}{25n^2}$ for $f\left(x_k\right)$ in equation (1) to find the value of area,

$\begin{array}{c}A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left[61+\frac{144k}{5n}-\frac{576k^2}{25n^2}\right]}\left(\frac{24}{n}\right)\\ =\underset{n\to \infty }{\lim }\left({\displaystyle \sum _{k=1}^n\left(\frac{1464}{n}+\frac{3456k}{5n^2}-\frac{13824k^2}{25n^3}\right)}\right)\\ =\underset{n\to \infty }{\lim }\frac{1464}{n}{\displaystyle \sum _{k=1}^{n}1}+\underset{n\to \infty }{\lim }\frac{3456}{5n^2}{\displaystyle \sum _{k=1}^{n}k}-\underset{n\to \infty }{\lim }\frac{13824}{25n^3}{\displaystyle \sum _{k=1}^n{k}^2}\end{array}$

The value of, ${\displaystyle \sum _{k=1}^{n}1}=n$, ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$ and ${\displaystyle \sum _{k=1}^n{k}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}$

Substitute n for ${\displaystyle \sum _{k=1}^{n}1}$, $\frac{n\left(n+1\right)}{2}$ for ${\displaystyle \sum _{k=1}^{n}k}$, $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ for ${\displaystyle \sum _{k=1}^n{k}^2}$ in the above value of A,

$\begin{array}{c}A=\underset{n\to \infty }{\lim }\frac{1464}{n}\cdot n+\underset{n\to \infty }{\lim }\frac{3456}{5n^2}\cdot \frac{n\left(n+1\right)}{2}-\underset{n\to \infty }{\lim }\frac{13824}{25n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =1464+\frac{3456}{10}\underset{n\to \infty }{\lim }\frac{n^2+n}{n^2}-\frac{13824}{150}\underset{n\to \infty }{\lim }\frac{2n^3+3n^2+n}{n^3}\\ =1464+\frac{3456}{10}\underset{n\to \infty }{\lim }\left(\frac{1}{n}+1\right)-\frac{13824}{150}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)\end{array}$

Substitute $\infty$ for n in above values of W because the $n=\infty$ is limit,

$\begin{array}{c}A=1464+\frac{3456}{10}-\frac{13824}{150}\left(2\right)\\ =1625.28{}^{°}\text{hr}\end{array}$

Thus, on a particular day when temperature was modeled by the function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ the heating degree hours were experienced is 1625.28 heating degree-hours.

(b)

To determine

To find: The maximum temperature that was on the day when temperature was modeled by the function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$.

Answer to Problem 5P

The maximum temperature on that day is ${70}^{\circ }\text{F}$.

Explanation of Solution

Given:

The function of temperature is,

$D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$

Calculation:

The maximum temperature occurs when the graph of the function is at its vertex which is,

$t=\frac{-b}{2a}$

Substitute $\frac{6}{5}$ for b and $-\frac{1}{25}$ for a in above formula,

$\begin{array}{c}t=\frac{-\frac{6}{5}}{2\cdot \left(-\frac{1}{25}\right)}\\ =15\end{array}$

Substitute 15 fro x in function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$,

$\begin{array}{c}D\left(15\right)=-\frac{1}{25}\cdot {\left(15\right)}^2+\frac{6}{5}\cdot \left(15\right)+61\\ ={70}^{\circ }\text{F}\end{array}$

The maximum temperature on that day is ${70}^{\circ }\text{F}$.

(c)

To determine

The heating degree hours that were experienced on a particular day from $t=0$ to $t=24$, that day temperature was modeled by the function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$.

Answer to Problem 5P

On a particular day when temperature was modeled by the function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$ the heating degree hours were experienced is 1488 heating degree-hours..

Explanation of Solution

Given:

The formula for heating degree-hours is,

$\text{Heating degree-hours}=\text{temperature}\times \text{time}$

The temperature was modeled by the function is,

$E\left(t\right)=50+5t-\frac{1}{4}{t}^2$

Where, t was measured hours since midnight.

Calculation:

The area under the function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$ is equal to the heating degree hours that were experienced on a particular day from $t=0$ to $t=24$.

Calculate the area under the function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$ by below formula,

$A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(x_k\right)}\Delta x$ (4)

$\Delta x$ can be calculated by below formula,

$\Delta x=\frac{b-a}{n}$ (5)

Formula to calculate the value of $x_k$ is,

$x_k=a+k\Delta x$ (6)

Substitute 0 for a and 24 for b in equation (5) to find the value of $\Delta x$,

$\begin{array}{c}\Delta x=\frac{24-0}{n}\\ =\frac{24}{n}\end{array}$

Substitute 0 for a and $\frac{24}{n}$ for $\Delta x$ in equation (6) to find the value of $x_k$,

$\begin{array}{c}{x}_k=0+k\cdot \frac{24}{n}\\ =\frac{24k}{n}\end{array}$

Find the value of $f\left(x_k\right)$ in function $D\left(t\right)$

$\begin{array}{c}f\left(x_k\right)=50+5x_k-\frac{1}{4}{\left(x_k\right)}^2\\ =50+5\cdot \left(\frac{24}{n}\right)-\frac{1}{4}\cdot {\left(\frac{24}{n}\right)}^2\\ =50+\frac{120k}{n}-\frac{144k^2}{n^2}\end{array}$

Substitute $\frac{24}{n}$ for $\Delta x$, $\frac{24k}{n}$ for $x_k$ and $50+\frac{120k}{n}-\frac{144k^2}{n^2}$ for $f\left(x_k\right)$ in equation (4) to find the value of area,

$\begin{array}{c}A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left[50+\frac{120k}{n}-\frac{144k^2}{n^2}\right]}\left(\frac{24}{n}\right)\\ =\underset{n\to \infty }{\lim }\left({\displaystyle \sum _{k=1}^n\left(\frac{1200}{n}+\frac{2880k}{n^2}-\frac{3456k^2}{25n^3}\right)}\right)\\ =\underset{n\to \infty }{\lim }\frac{1200}{n}{\displaystyle \sum _{k=1}^{n}1}+\underset{n\to \infty }{\lim }\frac{2880}{n^2}{\displaystyle \sum _{k=1}^{n}k}-\underset{n\to \infty }{\lim }\frac{3456}{n^3}{\displaystyle \sum _{k=1}^n{k}^2}\end{array}$

The value of, ${\displaystyle \sum _{k=1}^{n}1}=n$, ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$ and ${\displaystyle \sum _{k=1}^n{k}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}$

Substitute n for ${\displaystyle \sum _{k=1}^{n}1}$, $\frac{n\left(n+1\right)}{2}$ for ${\displaystyle \sum _{k=1}^{n}k}$, $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ for ${\displaystyle \sum _{k=1}^n{k}^2}$ in the above value of A,

$\begin{array}{c}A=\underset{n\to \infty }{\lim }\frac{1200}{n}\cdot n+\underset{n\to \infty }{\lim }\frac{2880}{n^2}\cdot \frac{n\left(n+1\right)}{2}-\underset{n\to \infty }{\lim }\frac{3456}{n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =1200+\frac{2880}{2}\underset{n\to \infty }{\lim }\frac{n^2+n}{n^2}-\frac{3456}{6}\underset{n\to \infty }{\lim }\frac{2n^3+3n^2+n}{n^3}\\ =1200+\frac{2880}{2}\underset{n\to \infty }{\lim }\left(\frac{1}{n}+1\right)-\frac{3456}{6}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)\end{array}$

Substitute $\infty$ for n in above values of W because the $n=\infty$ is limit,

$\begin{array}{c}A=1200+\frac{2880}{2}-\frac{3456}{6}\cdot \left(2\right)\\ =1488\end{array}$

Thus, on a particular day when temperature was modeled by the function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$ the heating degree hours were experienced is 1488 heating degree-hours.

(d)

To determine

To find: The maximum temperature that was on the day when temperature was modeled by the function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$

Answer to Problem 5P

The maximum temperature on that day is ${75}^{\circ }\text{F}$.

Explanation of Solution

Given:

The function of temperature is,

$E\left(t\right)=50+5t-\frac{1}{4}{t}^2$

Calculation:

The maximum temperature occurs when the graph of the function is at its vertex which is,

$t=\frac{-b}{2a}$

Substitute 5 for b and $-\frac{1}{4}$ for a in above formula,

$\begin{array}{c}t=\frac{-5}{2\cdot \left(-\frac{1}{4}\right)}\\ =10\end{array}$

Substitute 10 fro x in function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$,

$\begin{array}{c}E\left(t\right)=50+5t-\frac{1}{4}{t}^2\\ =50+5\cdot 10-\frac{1}{4}\cdot {\left(10\right)}^2\\ =50+50-25\\ ={75}^{\circ }\text{F}\end{array}$

Thus, the maximum temperature on that day is ${75}^{\circ }\text{F}$.

(e)

To determine

The hotter day between the day with temperature $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ and the day with temperature $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$.

Answer to Problem 5P

The day with temperature $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ is hotter.

Explanation of Solution

From part (a), when temperature was modeled by the function $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ the heating degree hours were experienced is 1625.28 heating degree-hours.

From part (c), when temperature was modeled by the function $E\left(t\right)=50+5t-\frac{1}{4}{t}^2$ the heating degree hours were experienced is 1488 heating degree-hours.

So, the heating degree hours of the day of part(a) is more than the heating degree hours of the day of part (b).

Thus, the day with temperature $D\left(t\right)=61+\frac{6}{5}t-\frac{1}{25}{t}^2$ is hotter.