19. By observing a significantly smaller object orbiting a
large celestial body, astronomers can easily determine
the mass of the larger. Appealing to Kepler’s third law
of planetary motion, we know the mass of the large
body varies directly with the cube of the mean
distance to the smaller and inversely with the square
of its orbital period. Write the variation equation.
Using the mean Earth/Sun distance of km
and the Earth’s orbital period of 1 yr, the mass of the
Sun has been calculated to be kg.
Given the orbital period of Mars is 1.88 yr, find its
mean distance from the Sun.
1.98892 ⫻ 10
30
1.496 ⫻ 10
8
20. The maximum load that can be supported by a
rectangular beam varies jointly with its width and
its height squared and inversely with its length. If a
beam 10 ft long, 3 in. wide, and 4 in. high can
support 624 lb, how many pounds could a 12-ft-long
beam with the same dimensions support?
2–91 Calculator Exploration and Discovery 277
College Algebra G&M—
CALCULATOR EXPLORATION AND DISCOVERY
Studying Joint Variations
Although a graphing calculator is limited to displaying the relationship between only two variables (for the most part),
it has a feature that enables us to see how these two are related with respect to a third. Consider the variation equation
from Example 8 in Section 2.6: . If we want to investigate the relationship between fuel consumption and
velocity, we can have the calculator display multiple versions of the relationship simultaneously for different values of d.
This is accomplished using the “{” and “}” symbols, which are functions to the parentheses. When the calculator
sees values between these grouping symbols and separated by commas, it is programmed to use each value independently
of the others, graphing or evaluating the relation for each value in the set. We illustrate by graphing the relationship
for three different values of d. Enter the equation on the screen as which
tells the calculator to graph the equations , , and on the same grid.Y
1
⫽ 0.051302X
2
Y
1
⫽ 0.051202X
2
Y
1
⫽ 0.051102X
2
Y
1
⫽ 0.05510, 20, 306X
2
,
Y=
f ⫽ 0.05dv
2
2nd
F ⫽ 0.05dv
2
Note that since d is constant, each graph is a parabola. Set the viewing window
using the values given in Example 8 as a guide. The result is the graph shown in
Figure 2.97, where we can study the relationship between these three variables
using the up and down arrows. From our work with the toolbox functions
and transformations, we know the widest parabola used the coefficient “10,” while
the narrowest parabola used the coefficient “30.” As shown, the graph tells us that
at a speed of 15 nautical miles per hour , it will take 112.5 barrels of fuel
to travel 10 mi (the first number in the list). After pressing the key, the cursor
jumps to the second curve, which shows values of and . This
means at 15 nautical miles per hour, it would take 225 barrels of fuel to travel 20
mi. Use these ideas to complete the following exercises:
Exercise 1: The comparison of distance covered versus fuel consumption at different speeds also makes an interesting
study. This time velocities are constant values and the distance varies. On the screen, enter .
What family of equations results? Use the up/down arrow keys for (a distance of 15 mi) to find how many
barrels of fuel it takes to travel 15 mi at 10 mph, 15 mi at 20 mph, and 15 mi at 30 mph. Comment on what you notice.
Exercise 2: The maximum safe load S for a wooden horizontal plank supported at both ends varies jointly with the
width W of the beam, the square of its thickness T, and inversely with its length L. A plank 10 ft long, 12 in. wide, and
1 in. thick will safely support 450 lb. Find the value of k and write the variation equation, then use the equation to
explore:
a. Safe load versus thickness for a constant width and given lengths (quadratic function). Use in. and {8, 12, 16}
for L.
b. Safe load versus length for a constant width and given thickness (reciprocal functional). Use in. and
for thickness.
5
1
4
,
1
2
,
3
4
6
w ⫽ 8
w ⫽ 8
x ⫽ 15
Y
1
⫽ 0.05x 510, 20, 306
2
Y=
Y ⫽ 225X ⫽ 15
1X ⫽ 152
30
800
0
0
Figure 2.97
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