Intermediate Algebra (11th edition)

SECTION 7.6 Variation 411

Let the weight and the distance from the center of Earth, for some

constant k.

wvaries inversely as the square of d.

At the apogee, the astronaut weighs 57 lb, and the distance from the center of Earth is

6700 mi. Use these values to find k.

Let and.

Solve for k.

Substitute to find the weight at the perigee.

w = Use a calculator. NOW TRY

571670022

1409022

L153 lb

k= 571670022 and d= 4090

k= 571670022

57 = w= 57 d= 6700

k

1670022

w =

k

d^2

w= d=

Space shuttle

at perigee

Space shuttle

at apogee

d 2 d 1

Earth

FIGURE 10

OBJECTIVE 4 Solve joint variation problems.If one variable varies directly

as the productof several other variables (perhaps raised to powers), the first variable

is said to vary jointlyas the others.

Joint Variation

yvaries jointly asxandzif there exists a real number ksuch that

ykxz.

Solving a Joint Variation Problem

The interest on a loan or an investment is given by the formula Here, for a

given principal p, the interest earned, I, varies jointly as the interest rate rand the

time tthe principal is left earning interest. If an investment earns $100 interest at 5% for

2 yr, how much interest will the same principal earn at 4.5% for 3 yr?

We use the formula where pis the constant of variation because it is the

same for both investments.

Let , , and.

Divide by 0.1. Rewrite.

Now we find Iwhen and

Here,. Let and.

The interest will be $135. NOW TRY

I= 10001 0.045 2132 = 135 p= 1000 r=0.045 t= 3

p= 1000,r= 0.045, t=3.

p= 1000

100 = 0.1p

100 = p 1 0.05 2122 I= 100 r=0.05 t= 2

I= prt

I=prt,

I= prt.

EXAMPLE 6

NOW TRY
EXERCISE 5
The weight of an object above
Earth varies inversely as the
square of its distance from the
center of Earth. If an object
weighs 150 lb on the surface
of Earth, and the radius of
Earth is about 3960 mi, how
much does it weigh when
it is 1000 mi above Earth’s
surface?

NOW TRY ANSWERS

5. about 96 lb 6.600 ft^3

NOW TRY
EXERCISE 6
The volume of a right pyramid
varies jointly as the height and
the area of the base. If the vol-
ume is when the area of
the base is and the height
is 10 ft, find the volume when
the area of the base is
and the height is 20 ft.

90 ft^2

30 ft^2

100 ft^3