Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.
Inverse variation states that whenever the product of corresponding values of two quantities is a constant, then one quantity varies inversely as the other. It is an equation that states that the product of two variables is equal to a constant. In symbols, xy = k or y = k/x. It means that y is inversely proportional to x, or x is inversely proportional to y.
From the word inverse, the opposite happens in an inverse variation. An increase in one quantity brings about a corresponding decrease in the other quantity and vice versa. Consider a rectangle with a width (w) of 2 cm and a length (l) of 18 cm. Retaining its area, 36 cm3 constant, what happens when the width increased to 3 cm or 4 cm?
The table below shows the relationship between the length and the width of the rectangle, given a constant area.
We can see that the product of the length and width in any four pairs is equal to 36. The value 36 is the variation constant. If we let the first, second, third, and fourth set of variables be (l1, w1), (l2, w2), (l3, w3), and (l4, w4), respectively, we can come up with the following equations.
l1 w1 = (2) (18)
l1 w1 = 36
l2 w2 = (3) (12)
l2 w2 = 36
l3, w3 = (4) (9)
l3, w3 = 36
l4 w4 = (5) (7.2)
l4 w4 = 36
Unlike that of direct variation where the ratio of two variables is equal to a constant, for an inverse variation, the product of the two variables is equal to a constant. Hence, l1 w1 = l2 w2 = l3 w3 = l4 w4 = 36.
In general, if we let w1 and w2 represent the widths in cm and l1 and l2 the corresponding lengths in cm, then by the multiplication property of equality, we can come up with the following equation. Use the shown equation if at least three terms are known. Hence, w1/w2 = l1/l2
Graph of an Inverse Variation
What does the graph of an inverse variation look like? Using the example mentioned earlier about the rectangle whose area value is 36 square centimeters, we can construct the graph illustrating the inverse variation given the length and width. The graph is neither a straight line nor a parabola. However, it is a half of a two-part curve known as the hyperbola. The graph is not a straight line since lw = k is not a linear equation.
Example 1: Finding the Constant Of Inverse Variation
Find the variation constant and the variation equation where y varies inversely as x, given y=24 and x=0.3.
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We know that one of the solutions is y=k/x.
24 = k / 0.3
k = 24 (0.3)
k = 7.2
The variation constant is 7.2; therefore, the equation of variation is y=7.2/x.
Example 2: Work Problem Applying Inverse Variation
The time t required to finish a specific job varies inversely as the number of person p who work on the job (assume that they do the same amount of work). If 15 persons are required to finish painting a house in 5 hours, how long would it take three persons to finish the same job?
Find the variation constant and the equation of variation. The variation constant, k = 75, and the equation of variation is the given equation shown below.
t = k/p
5 = k/15
k = 75
t = 75/p
We write the proportion as shown below. Substitute the given values in the proportion, solve the following by Multiplication Property of Equality.
t1 / t2 = p2 / p1
5 / t2 = 3 / 15
3t2 = 75
t2 = 25
t = k/p
t = 75/3
t = 25
Therefore, if three people will work on the same job, it will take them 25 hours to finish it.
Example 3: Pressure Problem Using Inverse Variation
If the temperature remains constant, the pressure of an enclosed gas is inversely proportional to the volume. Absolute gas pressure within a spherical balloon of a radius of 9 inches is 20 lb/in2. If the radius of the balloon increases to 12 inches, approximate the new pressure of the gas. Sketch a graph of the relationship between the pressure and the volume.
If we denote the pressure by P (in lb/in2) and the volume by B (in in3), then since P is inversely proportional to V, the equation will be as shown below, which is for some constant of proportionality k.
P = k/V
We find the constant of proportionality k in the first step. Since the volume V of a sphere of radius r is V = 4/3ℼ(r)3, the initial volume of the balloon is V = 4/3ℼ(9)3 = 972ℼ in3. Considering P = 20 when V=972ℼ, this leads to the following:
20 = k/972ℼ
K = 19440ℼ
Substituting k = 19440ℼ into P = k/V, we find that the pressure corresponding to any volume V is the equation shown below.
P = 19440ℼ / V
If the new radius of the balloon is 12 inches, then find the new volume.
V = 4/3ℼ(12)3
V = 2304ℼ in3
Substituting this number for V in the formula obtained in the third step gives us the following.
P = 19440ℼ / 2304ℼ
P = 135 / 16
P = 8.4375
Thus, the pressure decreases to approximately 8.4 lb/in2 when the radius increases to 12 inches.
Example 4: Fulcrum Problem in Inverse Variation
The distance from the center d of a seesaw varies inversely with the weight of the child w. Jordan, who weighs 100 lbs, sits 4 feet from the fulcrum. How far from the fulcrum must Aaron sit to balance Bongbong if he weighs 80 lbs?
Find the variation constant and the equation of variation. Then substitute the obtained variation constant to the second equation to solve the distance Aaron must sit to balance Jordan.
d = k/w
4 = k/100
k = 400
d = 400/w
d = 400/80
d = 5 feet
Finally, express the functional equation as the equation shown below. Use direct substitution and analyze the equation algebraically.
d1/d2 = w2/w1
4/d2 = 80/100
d2 = 400/80
d2 = 5
Aaron must sit 5 feet away from the fulcrum to balance Jordan.
Example 5: Pressure Problem About Inverse Variation
The pressure (P) varies inversely as the volume (V) for a given gas of constant temperature. If P = 6 when V = 24, find P when V = 36.
Recall that in an inverse variation, the value of one variable increases as the value of the other decreases, and vice versa. Apply this principle to solve the given problem. Using P1V1 = k:
k = P1V1
k = (6) (24)
k = 144
P2V2 = k
P2 = k / V2
P2 = 144/36
P2 = 4
Using P1/P2 = V2/V1:
6 / P2 = 36 / 24
P2 = 4
The value of pressure given the given gas of constant temperature is 4.
Example 6: Work-Related Problem in Inverse Variation
Find the number of days ten workers can complete a job if five workers can complete the same job in 7 days.
The number of days (y) to complete a certain job can vary inversely as the number of workers (x) assigned to do the job. Assuming that each works at the same rate, then y = k/x or yx = k. Using y1x1 = k:
y1x1 = k
k = (7) (5)
k = 35
y2x2 = k
y2 = k/x2
y2 = 35 / 10
y2 = 3.5
Using y1x1 = k and y2x2 = k:
y1x1 = y2x2
(7) (5) = (y2) (10)
y2 = 3.5
Ten workers can complete the job in 3.5 days.
Example 7: Force of Attraction Problem About Inverse Variation
The force of attraction (F) between two opposite electrical charges varies inversely as the square of the distance (d) between them. If F = 18 when d = 10, find F when d = 15.
Convert the given mathematical statement into an equation. Let F be the force of attraction between two opposite electrical charges, q1 and q2, and variable d be the distance between these electrical charges. Since the given problem requires no values for the two electrical charges, consider them as constants. It means that for both conditions given, the numerical values for q1 and q2 are the same. First, solve the value of the constant of variation.
F = k q1q2 / d2
F = k / d2
18 = k / (10)2
k = 1800
Finally, solve for the value of F using the obtained constant of variation k.
F = (1800) / (15)2
F = 8
The value of F when d = 15 and k = 1800 is 8.
Example 8: Work Problem About Inverse Variation
The number of hours required to complete a specific job varies inversely as the number of machines used to do the work, assuming that they operate at the same rate. If four machines can complete a particular job in 10 hours, how long will it take six machines to complete the same job?
Express the mathematical statement into equations. Always take note to consider the constant of variation k. Let T = number of hours required to complete a particular job and m be the number of machines used to do the work.
T = k / m
10 = k / 4
k = 40
T = 40 / 6
T = 6.67 hours
It will take 6.67 hours for six machines to complete the same job.
Example 9: Inverse Variation Graph Problem
Using the graph shown below, answer the following questions.
- If the weight is 18 grams, what is the distance?
- If the distance is 12 cm, what is the weight?
- If 24/6 = w2/2, w2 = ?
- If d1/2 = 24/16, d1 = ?
Solution and Answer
- Using the graph, if w = 18 grams, the corresponding value of d is 2 cm.
- Using the graph, if d = 12 cm, the corresponding value of w = 3 grams.
- If 24/6 = w2/2, then by multiplication property of inequality, w2 = 8 grams.
- If d1/2 = 24/16, then by the multiplication property of inequality, d1 = 3 cm
Example 10: Expressing Mathematical Statements to Equations
Express the following as an equation in which k represents the constant of proportionality.
- The electric current (I) varies inversely as the resistance (R) at a constant voltage.
- The intensity of illumination (I) on a book varies inversely as the square of the distance (d) from the source of light.
- The temperature (T) at which water boils varies inversely as the number of feet (h) above sea level.
- Atmospheric pressure (P) varies inversely as the altitude (h).
Solution and Answer
For all questions, follow the pattern for inverse variation formula y = k/x.
- Let letter I be the electric current, V be the constant voltage, and R be the resistance. Therefore the converted mathematical equation is I=kV/R.
- Variable I is the intensity of illumination and d as the distance from the source of light. Therefore, the mathematical statement is equivalent to I=k/d2.
- Let T be the temperature at which water boils, and h be the height in feet above sea level. Therefore, the inverse variation equation is T = K/h.
- Let P be the atmospheric pressure and h be the altitude. Therefore, the resulting math equation is P = k/h.
This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.
© 2021 Ray