Chapter 11.1, Problem 11.5P

A group of hikers uses a GPS while doing a 40-mile trek in Colorado. A curve fit to the data shows that their altitude can be approximated by the function y(t) = 0.12t⁵ − 6.75t⁴ + 135t³ − 1120t² + 3200t + 9070, where y and t are expressed in feet and hours, respectively. During the 18-hour hike, determine (a) the maximum altitude that the hikers reach, (b) the total feet they ascend, (c) the total feet they descend. Hint: You will need to use a calculator or computer to solve for the roots of a fourth-order polynomial.

To determine

The maximum altitude $\left(y_{\max }\right)$ that the hikers reach.

Answer to Problem 11.5P

The maximum altitude $\left(y_{\max }\right)$ that the hikers reach is $\underset{\_}{11,980\text{ft}}$.

Explanation of Solution

Given information:

The curve fit equation with function of time is $y\left(t\right)=0.12t^5-6.75t^4+135t^3-1120t^2+3200t+9070$.

Calculation:

Write the relation for the motion of car:

$y\left(t\right)=0.12t^5-6.75t^4+135t^3-1120t^2+3200t+9070$ (1).

Here, position of car is x and time is t.

Differentiate Equation (1) with respect to time.

$\begin{array}{l}\frac{dx}{dt}=0.6t^4-27t^4+405t^3-2240t^2+3200\\ 0=0.6t^4-27t^4+405t^3-2240t^2+3200\end{array}$

Solve the above equation to find the roots.

The roots values are $2.151\text{hours}$, $8.606\text{hours}$, $14.884\text{hours}$ and $19.36\text{hours}$.

Calculate the hike $y\left(2.151\right)$:

Substitute $2.151\text{hours}$ for t in Equation (1).

$\begin{array}{c}y\left(2.151\right)=\left\{\begin{array}{c}0.12{\left(2.151\right)}^5-6.75{\left(2.151\right)}^4+135{\left(2.151\right)}^3\\ -1120{\left(2.151\right)}^2+3,200\left(2.151\right)+9,070\end{array}\right\}\\ =5.756-144.499+1,343.554-5,182.017+6,883.2+9,070\\ =11,976.044\text{ft}\approx 11,976\text{ft}\end{array}$

Calculate the hike $y\left(8.606\right)$:

Substitute $8.606\text{hours}$ for t in Equation (1).

$\begin{array}{c}y\left(8.606\right)=\left\{\begin{array}{c}0.12{\left(8.606\right)}^5-6.75{\left(8.606\right)}^4+135{\left(8.606\right)}^3\\ -1120{\left(8.606\right)}^2+3200\left(8.606\right)+9070\end{array}\right\}\\ =\left(\begin{array}{l}5,664.844-37,026.1998+86,047.408\\ -82,950.824+27,539.2+9,070\end{array}\right)\\ =8,344.428\text{ft}\approx \text{8,344}\text{ft}\end{array}$

Calculate the hike $y\left(14.884\right)$

Substitute $14.884\text{hours}$ for t in equation (1).

$\begin{array}{c}y\left(14.884\right)=\left\{\begin{array}{c}0.12{\left(14.884\right)}^5-6.75{\left(14.884\right)}^4+135{\left(14.884\right)}^3\\ -1,120{\left(14.884\right)}^2+3,200\left(14.884\right)+9,070\end{array}\right\}\\ =\left(\begin{array}{l}87,655.577-331,270.237+445,136.035\\ -248,117.471+47,628.8+9,070\end{array}\right)\\ =10,102.704\text{ft}\approx \text{10,103}\text{ft}\end{array}$

Neglect the root 19.36 because the value is greater than 18 hours.

Calculate the hike $y\left(18\right)$ using equation (1).

Substitute $18$ for $t$ in equation (1).

$\begin{array}{c}y\left(18\right)=\left\{\begin{array}{c}0.12{\left(18\right)}^5-6.75{\left(18\right)}^4+135{\left(18\right)}^3\\ -1,120{\left(18\right)}^2+3,200\left(18\right)+9,070\end{array}\right\}\\ =226,748.16-708,588+787,320-362,880+57,600+9,070\\ =9,270.16\text{ft}\approx \text{9,270}\text{ft}\end{array}$

Therefore, the maximum altitude $\left(y_{\max }\right)$ that the hikers reach is $\underset{\_}{11,976\text{ft}}$.

To determine

The total feet they ascend.

Answer to Problem 11.5P

The total feet they ascend is $\underset{\_}{4,665\text{ft}}$.

Explanation of Solution

Given information:

The function of time is $y\left(t\right)=0.12t^5-6.75t^4+135t^3-1120t^2+3200t+9070$.

Calculation:

Calculate the hike with respective time (t) as in Table 1.

Time (t)	Hike (ft)
0	9070
1	11278.4
2	11965.8
3	11717.4
4	10984.9
5	10101.3
6	9295.12
7	8705.09
8	8394.16
9	8364.13
10	8570
11	8934.37
12	9361.84
13	9753.41
14	10020.9
15	10101.3
16	9971.12
17	9661.09
18	9270.16

Plot the graph for time versus hike as in Figure (1).

Calculate the hike $y\left(0\right)$:

Substitute $0\text{hours}$ for t in Equation (1).

$\begin{array}{c}y\left(0\right)=\left\{\begin{array}{c}0.12{\left(0\right)}^5-6.75{\left(0\right)}^4+135{\left(0\right)}^3\\ -1,120{\left(0\right)}^2+3,200\left(0\right)+9,070\end{array}\right\}\\ =0-0+0-0+3,200+9,070\\ =9,070\text{ft}\end{array}$

Calculate the total feet they ascend using the relation:

$\text{Total}\text{feet}\text{ascend}\text{=}\left(y\left(2.151\right)-y\left(0\right)\right)+\left(y\left(14.884\right)-y\left(8.606\right)\right)$

Substitute $9070\text{ft}$ for $y\left(0\right)$, $\text{11,976}\text{ft}$ for $y\left(2.151\right)$, $\text{8,340}\text{ft}$ for $y\left(8.606\right)$, and $\text{10,103}\text{ft}$ for $y\left(14.884\right)$.

$\begin{array}{c}\text{Total}\text{feet}\text{ascend}=\left(11,976-9,070\right)+\left(10,103-8,344\right)\\ =2,906+1,759\\ =4,665\text{ft}\end{array}$

Therefore, the total feet they ascend is $\underset{\_}{4,665\text{ft}}$.

To determine

The total feet they descend.

Answer to Problem 11.5P

The total feet they descend is $\underset{\_}{4,469\text{ft}}$.

Explanation of Solution

Given information:

The function of time is $y\left(t\right)=0.12t^5-6.75t^4+135t^3-1120t^2+3200t+9070$.

Calculation:

Calculate the total feet they descend using the relation:

$\text{Total}\text{feet}\text{ascend}\text{=}\left(y\left(2.151\right)-y\left(8.606\right)\right)+\left(y\left(14.884\right)-y\left(18\right)\right)$

Substitute $9,270\text{ft}$ for $y\left(18\right)$, $\text{11,976}\text{ft}$ for $y\left(2.151\right)$, $\text{8,340}\text{ft}$ for $y\left(8.606\right)$, and $\text{10,103}\text{ft}$ for $y\left(14.884\right)$.

$\begin{array}{c}\text{Total}\text{feet}\text{descend}=\left(11,976-8,340\right)+\left(10,103-9,270\right)\\ =3,636+833\\ =4,469\text{ft}\end{array}$

Therefore, the total feet they descend is $\underset{\_}{4,469\text{ft}}$.