Q. 34 - Precalculus Enhanced with Graphing Utilities

Question

Effect of Elevation on Weight If an object weighs m pounds at sea level, then its weight W (in pounds) at a height of h miles above sea level is given approximately by 

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

(a) If Amy weighs 120 pounds at sea level, how much will she weigh on Pike’s Peak, which is 14,110 feet above sea level? 

(b) Use a graphing utility to graph the function W = W(h). Use m = 120 pounds. 

(c) Create a TABLE with TblStart = 0 and $∆Tbl$ = 0.5 to see how the weight W varies as h changes from 0 to 5 miles. 

(d) At what height will Amy weigh 119.95 pounds? 

(e) Does your answer to part (d) seem reasonable? Explain. 

Step-by-Step Solution

Answer

(a) 119.8 pounds

(b)

(c) 

(d) 0.84 mile

(e) Yes

Part (a )Step 1. Given information

We have been given that if an object weighs m pounds at sea level, then its weight W (in pounds) at a height of h miles above sea level is given approximately by  

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

We have to find that if Amy weighs 120 pounds at sea level, how much will she weigh on Pike’s Peak, which is 14,110 feet above sea level.

Part (a) Step 2. Convert height into miles

We know 

$$\begin{gathered} 1 \mathrm{mile}=5280 \mathrm{feet} \\ 1 \mathrm{feet} =\frac{1}{5280} \mathrm{mile} \\ \mathrm{Therefore}, 14110 \mathrm{feet} =\frac{1}{5280}\times 14110 \mathrm{mile} \\ =2.672 \mathrm{miles} \end{gathered}$$

Part (a) Step 3. Put h = 2 . 672 in the given expression

$$\begin{gathered} W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2 \\ W(2.672)=120{\left(\frac{4000}{4000+2.672}\right)}^2 \\ W(2.672)=119.8 pounds \end{gathered}$$

Part (b) Step 1. Graph the given function

The function $W\left(h\right)=120{\left(\frac{4000}{4000+h}\right)}^2$ is graphed as:

Part (c) Step 1. Create a TABLE

h	W(h)
0	120
0.5	119.97
1	119.94
1.5	119.91
2	119. 88
2.5	119. 85
3	119. 82
3.5	119. 79
4	119. 76
4.5	119. 73
5	119. 70

Part (d) Step 1. Put W ( h ) = 119 . 95 in the given formula and solve for h

$$\begin{gathered} W\left(h\right)=m{\left(\frac{4000}{4000+h}\right)}^2 \\ 119.95=120{\left(\frac{4000}{4000+h}\right)}^2 \\ \frac{119.95}{120}={\left(\frac{4000}{4000+h}\right)}^2 \\ \sqrt{\frac{119.95}{120}}=\frac{4000}{4000+h} \\ 0.99979=\frac{4000}{4000+h} \\ h=\frac{4000}{0.99979}-4000 \\ h=0.84 mile \end{gathered}$$

Part (e) Step 1. Compare the values obtained from part (c) and part (d)

From part (c), we can see that in the weight range of $119.97-119.94$ pounds the height range is $0.5-1.0$ m.

Therefore, our answer to part (d) is reasonable.