Notes: Direct and Inverse Variation
Direct and Inverse Variation: Direct variation, like a proportion, has a direct relationship. The formula
y = mx shows us that the relationship is linear. We say that y varies directly as x and m is the constant
of variation. Since y is directly proportional to x, you could use a proportion to solve the equation
instead of the method below.
1. If the stretch on a spring varies directly as the load it supports, and a load of 8 kg stretches a
spring 9.6 cm, what load would stretch the spring 6 cm? Use your equation from b.
a. If x is the load in kg and y the stretch in the spring find the constant of variation, m.
S = mL
9.6 = m(8)
M = 1.2 cm/kg
b. Write the equation y = mx by replacing m with the constant of variation you found in a.
This represents the relationship between x and y.
S = 1.2L
c. What load would stretch the spring 6 cm? Use your equation from b.
6 = 1.2L
L = 5 kg
2. If y varies directly as x and y = 15 when x = 24, find x when y = 25.
y = mx
15 = m(24)
M = 0.624
Y =0.625x
25 = 0.625x
X = 40
3. If a is directly proportional to b³, and a = 10 when b = 2, find a when b = 4.
a = m(b3)
10 = m(23)
m = 1.25
a = 1.25b3
a = 1.25(43)
a = 80
4. The amount of hydrogen produced when sodium is added to water varies directly as the amount
of sodium added. If 92 g of sodium produces 4 g of hydrogen, find the amount of sodium
needed to produce 10 g of hydrogen.
H = ms
4 = m(92)
m = 1/23
h = (1/23)s
10 = (1/23)s
230 grams of sodium
5. The electrical resistance in ohms (𝛺) of a wire varies directly as its length. If a wire 110 cm long
has a resistance of 7.5 Ω, what length of wire will have a resistance of 12Ω?
Ω = mL
7.5 = m(110)
M = (3/44)L
12 = (3/44)(L)
176 cm
Inverse variation is represented by xy = k or y = k/x, where y varies inversely as x. We also say that y is
inversely proportional to x. The constant “k” is the constant of variation. You cannot use proportions
to solve these problems.
1. If y is inversely proportional to x and y = 6 when x = 5, find x when y is 12.
a. Substitute the 6 and 5 in for y and x to find k, the constant.
y = k/x
6 = k/5
k =30
b. Write the equation that represents the relationship by replacing k in y = k/x with you
answer from a. This equation represents the inverse relationship.
y = 30/x
c. Use this equation to find x when y is 12.
12 = 30/x
X = 2.5
2. If y varies inversely as q, and y = 22 when q = 6, find q when y = 15.
y = k/q
22 = k/6
k = 132
y = 132/q
15 = 132/q
q = 8.8
3. The time required to travel a given distance is inversely proportional to the speed of travel. If a
trip can be made in 3.6 hours at a speed of 70 km/h, how long will it take to make the same trip
at 90km/h?
T = k/s
3.6 = k/70
k = 252
t = 252/s
t = 252/90
2.8 hours
Direct, Inverse, and Joint Variation:
We will now use k as the constant, whether the relationship is direct, inverse because, as we consider
joint variation, we will combine inverse and direct variations in a single problem.

Note that how you read the problem is important. The order of the words tells you the order to
write the problem.

When there is a direct relationship you multiply, and when there is an inverse relationship, you
divide.

For example, if it reads “z varies jointly as x and the square root of y” the equation is
z = kx√𝑦
Note this is only direct variation, which it will be unless the word inverse is used
Also note that you have to include k, the constant.
However, if it read “r varies jointly as s and t and inversely as u” your equation would be
r=

𝑘𝑠𝑡
.
𝑢
You will still use the process of finding the constant “k” by using the complete relationship given
and the writing your equation including k. You then use this equation to solve the incomplete
relationship.
1. If z varies jointly as x and the square root of y, and z = 6 when x = 3 and y =16, find z when x = 7
and y = 4.
Z = kx(√𝑦)
6 = k(3)(4)
k=0.5
Z = 0.5x(√𝑦)
Z = 0.5(7)(2)
Z= 7
2. If r varies jointly as s and t and inversely as u and r = 18 when s = 2, t = 3, and u =4, find s when r
= 6, t = 2, and u = 4.
r = kst/ur
18 = k(2)(3)/4
k = 12
r = 12st/ur
6 = 12(s)(2)/4
s=1
3. The electrical resistance of a wire varies directly as its length and inversely as the square of its
diameter. One hundred meters of wire with diameter 6 mm has a resistance of 12 ohms (𝛺 ).
Eighty meters of a second wire of the same material has a resistance of 15 Ω. Find the diameter
of the second wire.
Ω = kL/d2
12 = k(100)/36
k = 4.32
Ω = 4.32L/d2
15 = 4.32(80)/d2
d = 4.8 mm
4. For cones of equal volume, the height is inversely proportional to the square of the radius of
the base. If the radius of a cone is 10 cm and its height is 15 cm, find the radius of a second
cone of equal volume with a height of 60 cm.
H = k/r2
15 = k/100
K = 1500
H = 1500/r2
60 = 1500/r2
r = 5 cm
5. If c is inversely proportional to d, and c = 2 when d = 3.6, find c when d = 4.5. Find d when c
= 6.4.
C = k/d
2 = k/3.6
k = 7.2
c = 7.2/d
6.4 = 7.2/d
d = 1.125
6. Suppose r varies jointly as t and s and inversely as the square of v. If t = 3, s = 18, and v = 5,
when r =3.78, then find r when t = 4, s= 12, and v = 4.
r = kts/v2
3.78 = k(3)(18)/25
k= 1.75
r = 1.75ts/v2
r = 1.75(4)(12)/16
r = 5.25
7. The surface area of a cylinder varies jointly as the radius and the sum of the radius and the
height. A cylinder with height 8 cm and radius 4 cm has a surface area of 96π cm². Find the
surface area of a cylinder with radius 3 cm and height 10 cm. (note: don’t use the volume
formula; just use the relationships as they are expressed here).
S = kr(r+h)
96π = k(4)(12)
k = 2π
S = 2π(r)(r+h)
S = 2π(3)(13)
S = 78 cm3