Chapter 2.2, Problem 34AYU

(a)

To determine

To find: the weight on Pike’s Peak.

Answer to Problem 34AYU

The weight on Pike’s Peak is 5.85 pounds.

Explanation of Solution

Given information:

If an object weights $m$ pounds at sea level, then its weight $W$ at a height of $h$ miles above sea level is given approximately by

  $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$

If Amy weights 120 pounds at sea level, and Pike’s Peak which is 14110 feet above sea level.

Calculation:

Substitute $m=120$ pounds and $h=14110$ feet in function $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$ ,

  $\begin{array}{c}W\left(14110\right)=\left(120\right){\left(\frac{4000}{4000+14110}\right)}^2\\ =\left(120\right){\left(\frac{4000}{18110}\right)}^2\\ \approx 5.85\end{array}$

Therefore the weight on Pike’s Peak is 5.85 pounds.

(b)

To determine

To sketch: a graph of the function $W=W\left(h\right)$ using graphing utility.

Explanation of Solution

Given information:

If an object weights $m$ pounds at sea level, then its weight $W$ at a height of $h$ miles above sea level is given approximately by

  $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$

If Amy weights 120 pounds at sea level, and Pike’s Peak which is 14110 feet above sea level.

Calculation:

To sketch the graph of function $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$ in graphing calculator input the function

  $Y_1=120{\left(\frac{4000}{4000+h}\right)}^2$

Set windows as $\left[X_{\min }=0,X_{\max }=20000,X_{scl}=2000\right]$ by $\left[Y_{\min }=0,Y_{\max }=150,Y_{scl}=25\right]$ .

Press the Graph key to obtain the graph. The graph of function $Y_1$ appears as shown below:

  

(c)

To determine

To create: a table with $\text{TblStart}=\text{0}$ and $\Delta \text{Tbl}=0.5$ to see how the weight $W$ varies as $h$ changes from 0 to 5 miles.

Answer to Problem 34AYU

The above table we can see that the as height change from 0 to 5 miles weight change from 120 to 119.7 or difference is 0.3.

Explanation of Solution

Given information:

If an object weights $m$ pounds at sea level, then its weight $W$ at a height of $h$ miles above sea level is given approximately by

  $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$

If Amy weights 120 pounds at sea level, and Pike’s Peak which is 14110 feet above sea level.

Calculation:

Table with $\text{TblStart}=\text{0}$ and $\Delta \text{Tbl}=0.5$ is appears as below:

  

From the above table we can see that the as height change from 0 to 5 miles weight change from 120 to 119.7 or difference is 0.3.

(d)

To determine

To find: the height that A will weight 119.95 pounds.

Answer to Problem 34AYU

At height 0.84 A weights 119.95 pounds.

Explanation of Solution

Given information:

If an object weights $m$ pounds at sea level, then its weight $W$ at a height of $h$ miles above sea level is given approximately by

  $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$

If Amy weights 120 pounds at sea level, and Pike’s Peak which is 14110 feet above sea level.

Calculation:

Substitute $m=120$ pounds and $W=119.95$ feet in function $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$ ,

  $\begin{array}{c}119.95=\left(120\right){\left(\frac{4000}{4000+h}\right)}^2\\ \frac{119.95}{120}={\left(\frac{4000}{4000+h}\right)}^2\\ \sqrt{\frac{119.95}{120}}=\frac{4000}{4000+h}\\ \frac{4000}{4000+h}=0.99979\\ 1+\frac{h}{4000}=\frac{1}{0.99979}\\ h=4000\left(\frac{1}{0.99979}-1\right)\\ h=8.4\end{array}$

Therefore at height 0.84 A weights 119.95 pounds.

(e)

To determine

Whether the answer to part (d) seem reasonable.

Answer to Problem 34AYU

Explanation of Solution

Given information:

If an object weights $m$ pounds at sea level, then its weight $W$ at a height of $h$ miles above sea level is given approximately by

  $W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2$

If Amy weights 120 pounds at sea level, and Pike’s Peak which is 14110 feet above sea level.

Calculation:

Yes. From the table we can see that 119.95 are between 0.5 and 1. Therefore 0.84 is reasonable.