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Direct Variation

$y = kx$  or  $\frac{y}{x} = k$.

EXAMPLE 1  Solving a Direct Variation Problem

The volume of blood $B$ in a person's body varies directly as their weight $W$.  A person weighing 160 pounds has approximately 5 quarts of blood.  Find $B$ for a person weighing 200 pounds.

# B = k*W
# 5 = k*160

k = 5/160
k

B(W) = k*W
B

B(200)

_.n()

# OR -
# since B/W = k,
# B/200 = 5/160.

B = 200*5/160
B

Direct Variation with Powers

$y$ varies directly as the nth power of $x$:     $y = kx^n$     or     $\frac{y}{x^n} = k$

EXAMPLE 2  Solving a Direct Variation Problem

The distance $s$ that a body falls varies directly as the square of the time $t$ of the fall.  If you fall 64 feet in 2 seconds, how far will you fall in 4.5 seconds?

# s = k*t^2
# 64 = k*2^2

k = 64/2^2
k

s(t) = k*t^2

s(4.5)

# OR -
# s/t^2 = k
# 64/2^2 = s/4.5^2

s = 64/2^2 * 4.5^2
s

Inverse Variation

$y$ varies inversely as $x$:  $y = \frac{k}{x}$     OR     $yx = k$.

EXAMPLE 3  Solving an Inverse Variation Problem

Pressure $P$ varies inversely as volume $V$.  If $P$ is 12 pounds per square inch in a container where $V$ is 8 cubic inches, what is the new $P$ when $V$ expands to 22 cubic inches?

# P = k/V
# 12 = k/8

k = 12*8
k

P(V) = k/V
P

P(22)

_.n()

# OR
# 12*8 = 22*P

P = 12*8/22
P

EXAMPLE 4  Solving a Combined Variation Problem

Suppose that the number of sales $S$ vary directly as amount spent in advertising $A$ and inversely as price $P$.  It turns out that monthly sales are 12,000 when $60,000 is spent on advertising and price is set at $40.

4a  Write an equation that describes this situation.

# S = k*A/P
# 12000 = k*60000/40

k = 12000*40/60000
k

S(A, P) = k*A/P
S

4b  Determine sales if advertising is increased to \$70,000.

S(70000, 40)

EXAMPLE 5  Modeling Joint and Combined Variation

Centrifugal force $C$ of a body moving in a circle varies jointly with the radius $r$ and the body's mass $m$, and inversely with the square of the time $t$ it takes for one full revolution.

If a 6-gram body moving in a circle with radius 100 centimeters at a rate of 1 revolution in 2 seconds has a force of 6000 dynes, find the force of an 18-gram body moving at the rate of 1 revolution in 3 seconds.

# C = k*r*m/t^2
# 6000 = k*100*6/2^2

k = 6000/100/6*2^2
k

C(m, r, t) = 40*r*m/t^2

C(18, 100, 3)

# OR
# 6000*2^2/(100*6) = C*3^2/(100*18)

C = 6000*2^2/(100*6)*(100*18)/3^2
C