Chapter 6: Proportions and Modeling Using Variation

Solve problems involving joint variation.

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation . For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c , cost, varies jointly with the number of students, n , and the distance, d .

A General Note: Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

For instance, if x  varies directly with both y  and z , we have x  = kyz . If x  varies directly with y  and inversely with z , we have [latex]x=\frac{ky}{z}[/latex]. Notice that we only use one constant in a joint variation equation.

Example 4: Solving Problems Involving Joint Variation

A quantity x  varies directly with the square of y  and inversely with the cube root of z . If x  = 6 when y  = 2 and z  = 8, find x  when y  = 1 and z  = 27.

Begin by writing an equation to show the relationship between the variables.

Substitute x  = 6, y  = 2, and z  = 8 to find the value of the constant k .

Now we can substitute the value of the constant into the equation for the relationship.

To find x  when y  = 1 and z  = 27, we will substitute values for y  and z  into our equation.

x  varies directly with the square of y  and inversely with z . If x  = 40 when y  = 4 and z  = 2, find x  when y  = 10 and z  = 25.

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Joint Variation – Formula, Examples | How to Solve Problems Involving Joint Variation?

Joint Variation definition, rules, methods and formulae are here. Check the joint variation problems and solutions to prepare for the exam. Refer to problems of direct and inverse variations and the relationship between the variables. Know the different type of variations like inverse, direct, combined and joint variation. Go through the below sections to check definition, various properties, example problems, value tables, concepts etc.

Joint Variation – Introduction

Joint Variation refers to the scenario where the value of 1 variable depends on 2 or more and other variables that are held constant. For example, if C varies jointly as A and B, then C = ABX for which constant “X”. The joint variation will be useful to represent interactions of multiple variables at one time.

Most of the situations are complicated than the basic inverse or direct variation model. One or the other variables depends on the multiple other variables. Joint Variation is nothing but the variable depending on 2 or more variables quotient or product. To understand clearly with an example, The amount of busing candidates for each of the school trip varies with the no of candidates attending the distance from the school. The variable c (cost) varies jointly with n (number of students) and d (distance).

Joint Variation problems are very easy once you get the perfection of the lingo. These problems involve simple formulae or relationships which involves one variable which is equal to the “one” term which may be linear (with just an “x” axis), a quadratic equation (like “x²) where more than one variable (like “hr²”), and square root (like “\sqrt{4 – r^2\,}4−r2​”) etc.

Functions of 2 or More Variables

It is very uncommon for the output variable to depend on 2 or more inputs. Most of the familiar formulas describe the several variables functions. For suppose, if the rectangle perimeter depends on the length and width. The cylinder volume depends on its height and radius. The travelled distance depends on the time and speed while travelling. The function notation of the formulas can be written as

P = f(l,w) = 2l + 2w where P is the perimeter and is a function of width and length

V = f(r,h) = Πr²h where V is the volume and is a function of radius and height

d = f(r,t) = rt where d is the distance and is a function of time and rate.

Tables of Values

Just for the single variable functions, we use the tables to describe two-variable functions. The heading of the table shows row and column and it shows the value if two input variables and the complete table shows the values of the output variable.

You can easily make graphs in three dimensions for two-variable functions. Instead of representing graphs, we represent functions by holding two or one variable constants.

Also, Read:

  • What is Variation
  • Practice Test on Ratio and Proportion

How to Solve Joint Variation Problems?

Follow the step by step procedure provided below to solve problems involving Joint Variation and arrive at the solution easily. They are along the lines

Step 1: Write the exact equation. The problems of joint variation can be solved using the equation y =kxz. While dealing with the word problems. you should also consider using variables other than x,y and z. Use the variables which are relevant to the problem being solved. Read the problem carefully and determine the changes in the equation of joint variation such as cubes, squares or square roots.

Step 2: With the help of the information in the problem, you have to find the value of k which is called the constant of proportionality and variation.

Step 3: Rewrite the equation starting with 1 substituting the value of k and found in step 2.

Step 4: Use the equation in step 3 and the information in the problem to answer the question. While solving the word problems, remember including the units in the final answer.

Joint Variation Problems with Solutions

The area of a triangle varies jointly as the base and the height. Area = 12m² when base = 6m and height = 4m. Find base when Area = 36m² and height = 8m?

The area of the triangle is represented with A

The base is represented with b

Height is represented with h

As given in the question,

A = 12m² when B = 6m and H = 4m

We know the equation,

A = kbh where k is the constant value

12 = k(6)(4)

Divide by 24 on both sides, we get

12/24 = k(24)/24

The value of k = 1/2

As the equation is

To find the base of the triangle of A = 36m² and H = 8m

36 = 1/2(b)(8)

Dividing both sides by 4, we get

36/4 = 4b/4

The value of base = 9m

Hence, the base of the triangle when A = 36m² and H = 8m is 9m

Wind resistance varies jointly as an object’s surface velocity and area. If the object travels at 80 miles per hour and has a surface area of 30 square feet which experiences 540 newtons wind resistance. How much fast will the car move with 40 square feet of the surface area in order to experience a wind resistance of 495 newtons?

Let w be the wind resistance

Let s be the object’s surface area

Let v be the object velocity

The object’s surface area = 80 newtons

The wind resistance = 540 newtons

The object velocity = 30

w = ksv where k is the constant

(540) = k (80) (30)

540 = k (2400)

540/2400 = k

The value of k is 9/40

To find the velocity of the car with s = 40, w = 495 newtons and k = 9/40

Substitute the values in the equation

495 = (9/40) (40) v

The velocity of a car is 55mph for which the object’s surface area is 40 and wind resistance is 495 newtons

Hence, the final solution is 55mph

For the given interest, SI (simple interest) varies jointly as principal and time. If 2,500 Rs left in an account for 5 years, then the interest of 625 Rs. How much interest would be earned, if you deposit 7,000 Rs for 9 years?

Let i be the interest

Let p be the principal

Let t be the time

The interest is 625 Rs

The principal is 2500

The time is 5 hours

i = kpt where k is the constant

Substituting the values in the equation,

(625) = k(2500)(5)

625 = k(12,500)

Dividing 12,500 on both the sides

625/12,500 = k (12,500)/12,500

The value of k = 1/20

To find the interest where the deposit is 7000Rs for 9 years, use the equation

i = (1/20) (7000) (9)

i = (350) (9)

Therefore, the interest is 3,150 Rs, if you deposit 7,000 Rs for 9 years

Thus, the final solution is Rs. 3,150

The volume of a pyramid varies jointly as its height and the area of the base. A pyramid with a height of 21 feet and a base with an area of 24 square feet has a volume of 168 cubic feet. Find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet?

Let v be the volume of a pyramid

Let h be the height of a pyramid

Let a be the area of a pyramid

The volume v = 168 cubic feet

The height h = 21 feet

The area a = 24 square feet

V = Kha where K is the constant,

168 = k(21)(24)

168 = k(504)

Divide 504 on both sides

168/504 = k(504)/504

The value of k = 1/3

To find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet

h = 18 feet

a = 42 square feet

V = (1/3) (18) (42)

V = (6) (42)

V = 252 ft³

The volume of the pyramid = 252 ft³ which has a height of 18 feet and a base with an area of 42 square feet

Therefore, the final solution is 252 ft³

The amount of oil used by a ship travelling at a uniform speed varies jointly with the distance and the square of the speed. If the ship uses 200 barrels of oil in travelling 200 miles at 36 miles per hour, determine how many barrels of oil are used when the ship travels 360 miles at 18 miles per hour?

No of barrels of oil = 200

The distance at which the oil is travelling = 200 miles

The distance at which the ship is travelling = 36 miles per hour

A = kds² where k is constant

200 = k.200.(36)²

Dividing both sides by 200

200/200 = k.200.(36)²/200

1 = k.(36)²

The value of k is 1/1296

To find the no of barrels when the ship travels 360 miles at 18 miles per hour

A = 1/1296 * 360 * 18²

Therefore, 90 barrels of oil is used when the ship travels 360 miles at 18 miles per hour

Thus, the final solution is 90 barrels

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  • Joint Variation

If more than two variables are related directly or one variable changes with the change product of two or more variables it is called as joint variation .

If X is in joint variation with Y and Z, it can be symbolically written as X α YZ. If Y is constant also then X is in direct variation with Z. So for joint variation two or more variables are separately in direct variation. So joint variation is similar to direct variation but the variables for joint variation are more than two.

Equation for a joint variation is X = KYZ where K is constant.

One variable quantity is said to vary jointly as a number of other variable quantities, when it varies directly as their product. If the variable A varies directly as the product of the variables B, C and D, i.e., if.A ∝ BCD or A = kBCD (k = constant ), then A varies jointly as B, C and D. 

For solving a problems related to joint variation first we need to build the correct equation by adding a constant and relate the variables. After that we need determine the value of the constant. Then substitute the value of the constant in the equation and by putting the values of variables for required situation we determine the answer.

We know, area of a triangle = ½ × base × altitude. Since ½ is a constant, hence area of a triangle varies jointly as its base and altitude.  A is said to vary directly as B and inversely as C if A ∝ B ∙ \(\frac{1}{C}\)  or A = m ∙ B ∙ \(\frac{1}{C}\)  (m = constant of variation) i.e., if A varies jointly as B and \(\frac{1}{C}\) .

If x men take y days to plough z acres of land, then x varies directly as z and inversely as y.

1.  The variable x is in joint variation with y and z. When the values of y and z are 4 and 6, x is 16. What is the value of x when y = 8 and z =12?

The equation for the given problem of joint variation is

x = Kyz where K is the constant.

For the given data

16 = K  ×  4  ×  6

or, K = \(\frac{4}{6}\) .

So substituting the value of K the equation becomes

x =  \(\frac{4yz}{6}\)

Now for the required condition

x = \(\frac{4 × 8 × 12}{6}\)

   = 64

Hence the value of x will be 64.

2.  A is in joint variation with B and square of C. When A = 144, B = 4 and C = 3. Then what is the value of A when B = 6 and C = 4?

From the given problem equation for the joint variation is

From the given data value of the constant K is

K =   \(\frac{BC^{2}}{A}\)

K = \(\frac{4 × 3^{2}}{144}\)

   =  \(\frac{36}{144}\)

    = \(\frac{1}{4}\) .

Substituting the value of K in the equation

A =  \(\frac{BC^{2}}{4}\)

A = \(\frac{6 × 4^{2}}{4}\)

   = 24

3. The area of a triangle is jointly related to the height and the base of the triangle. If the base is increased 10% and the height is decreased by 10%, what will be the percentage change of the area?

We know the area of triangle is half the product of base and height. So the joint variation equation for area of triangle is A = \(\frac{bh}{2}\)  where A is the area, b is the base and h is the height.

Here \(\frac{1}{2}\)  is the constant for the equation.

Base is increased by 10%, so it will be b x \(\frac{110}{100}\)  = \(\frac{11b}{10}\) .

Height is decreased by 10%, so it will be h x \(\frac{90}{100}\)  = \(\frac{9h}{10}\) .

So the new area after the changes of base and height is

\(\frac{\frac{11b}{10} \times \frac{9h}{10}}{2}\)

= (\(\frac{99}{100}\) )\(\frac{bh}{2}\)  = \(\frac{99}{100}\) A.

So the area of the triangle is decreased by 1%.

4. A rectangle’s length is 6 m and width is 4 m. If length is doubled and width is halved, how much the perimeter will increase or decrease?

Formula for the perimeter of rectangle is P = 2(l + w) where P is perimeter, l is length and w is width.

This is joint variation equation where 2 is constant.

So P = 2(6 + 4) = 20 m

If length is doubled, it will become 2l.

And width is halved, so it will become \(\frac{w}{2}\) .

So the new perimeter will be P = 2(2l +  \(\frac{w}{2}\) ) = 2(2 x 6 +  \(\frac{4}{2}\) ) = 28 m.

So the perimeter will increase by (28 - 20) = 8 m.

●   Variation

  • What is Variation?
  • Direct Variation
  • Inverse or Indirect Variation
  • Theorem of Joint Variation
  • Worked out Examples on Variation
  • Problems on Variation

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1.8: Variation - Constructing and Solving Equations

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Solving Problems involving Direct, Inverse, and Joint variation

Certain relationships occur so frequently in applied situations that they are given special names. Variation equations show how one quantity changes in relation to other quantities. The relationship between the quantities can be described as direct , inverse , or joint variation .

Direct Variation

Many real-world problems encountered in the sciences involve two types of functional relationships. The first type of functional relationship can be explored using the fact that the distance \(s\) in feet an object falls from rest, without regard to air resistance, can be approximated using the following formula:

\(s=16t^{2}\)

Here \(t\) represents the time in seconds the object has been falling. For example, after \(2\) seconds the object will have fallen \(s = 16 ( 2 ) ^ { 2 } = 16 \cdot 4 = 64\) feet.

In this example, we can see that the distance varies over time as the product of a constant \(16\) and the square of the time \(t\). This relationship is described as direct variation  and \(16\) is called the constant of variation or the  constant of proportionality . 

Definition: Direct Variation (\(y=kx\))

Direct variation is a relationship where quantities behave in a like manner. If one increases, so does the other. If one decreases, so does the other.

For two quantities \(x\) and \(y\), this relationship is described as "\(y\) varies directly as \(x\)" or "\(y\) is directly proportional to \(x\)" .

The equation that describes this relationship is \(y=kx\) , where \(k\) is a non-zero constant called the constant of variation or the proportionality constant .

how-to.png

  • Translate the given English statement containing the words varies  or proportional , into a model equation.
  • Substitute a given set of values into the equation and solve for \(k\), the constant of variation.
  • Rewrite the equation obtained in step 1 as a formula with a value for \(k\) found in step 2 defined. Make note of the units used for each variable in the formula.
  • Use the equation from step 3, and another set of values (with one value missing) to solve for the unknown quantity.

Example \(\PageIndex{1}\): Direct Variation

An object’s weight on Earth varies directly to its weight on the Moon. If a man weighs \(180\) pounds on Earth, then he will weigh \(30\) pounds on the Moon. Set up an algebraic equation that expresses the weight on Earth in terms of the weight on the Moon and use it to determine the weight of a woman on the Moon if she weighs \(120\) pounds on Earth.

Step 1. Translate “the weight on Earth varies directly to the weight on the Moon.”    \(E = kM \) 

Step 2. Find \(k\) using "If a man weighs \(180\) pounds on Earth, then he will weigh \(30\) pounds on the Moon."    \(E=180\) pounds, \(M=30\) pounds

\(\begin{array} { cr } E = kM & \text{Model equation} \\180=k \cdot 30\\ \frac { 180 } { 30 } = k  \\ { 6 = k } \end{array}\)

Step 3. The formula is \(E = 6M \), where \(E\) is the weight on Earth in pounds and \(M\) is the weight on the moon in pounds.

Step 4. Answer the question: "determine the weight of a woman on the Moon if she weighs \(120\) pounds on Earth."    \(E=120\) pounds, find M

\(\begin{array} { cll } E = 6M & \text{Formula:} & \text{ \(E\) pounds on Earth}\\  && \text{ \(M\) pounds on the Moon}\\{ 120 = 6 M } \\ { \frac { 120 } { 6 } = M } \\ { 20 = M } \end{array}\)

The woman weighs \(20\) pounds on the Moon.

Indirect Variation

The second functional relationship can be explored using the model that relates the intensity of light \(I\) to the square of the distance from its source \(d\).

\(I = \frac { k } { d ^ { 2 } }\)

Here \(k\) represents some constant. A foot-candle is a measurement of the intensity of light. One foot-candle is defined to be equal to the amount of illumination produced by a standard candle measured one foot away. For example, a \(125\)-Watt fluorescent growing light is advertised to produce \(525\) foot-candles of illumination. This means that at a distance \(d=1\) foot, \(I=525\) foot-candles and we have:

\(\begin{array} { l } { 525 = \frac { k } { ( 1 ) ^ { 2 } } } \\ { 525 = k } \end{array}\)

Using \(k=525\) we can construct a formula which gives the light intensity produced by the bulb:

\(I = \frac { 525 } { d ^ { 2 } }\)

Here \(d\) represents the distance the growing light is from the plants. In the chart above, we can see that the amount of illumination fades quickly as the distance from the plants increases.

This type of relationship is described as an inverse variation . We say that I is inversely proportional  to the square of the distance \(d\), where \(525\) is the constant of proportionality.

Definition: Indirect Variation (\(y=\frac{k}{x}\))

Indirect variation is a relationship between quantities where if one increases, the other decreases.

For two quantities \(x\) and \(y\), this relationship is described as "\(y\) varies indirectly as \(x\)" or "\(y\) is inversely proportional to \(x\)" .

The equation that describes this relationship is \(y=\dfrac{k}{x}\) , where \(k\) is a non-zero constant called the constant of variation or the proportionality constant .

Example \(\PageIndex{2}\): Indirect Variation

The weight of an object varies inversely as the square of its distance from the center of Earth. If an object weighs \(100\) pounds on the surface of Earth (approximately \(4,000\) miles from the center), how much will it weigh at \(1,000\) miles above Earth’s surface?

Step 1. Translate  “\(w\) varies inversely as the square of \(d\)”    \(w = \frac { k } { d ^ { 2 } }\)

Step 2. Find \(k\) using "An object weighs \(100\) pounds on the surface of Earth, approximately \(4,000\) miles from the center".     \(w = 100\) when \(d = 4,000\)

\(\begin{aligned} \color{Cerulean}{( 4,000 ) ^ { 2 }}\color{black}{ \cdot} 100 & =\color{Cerulean}{ ( 4,000 ) ^ { 2 }}\color{black}{ \cdot} \frac { k } { ( 4,000 ) ^ { 2 } } \\ 1,600,000,000 &= k \\ 1.6 \times 10 ^ { 9 } &= k \end{aligned}\)

Step 3. The formula is \(w = \frac { 1.6 \times 10 ^ { 9 } } { d ^ { 2 } }\), where \(w\) is the weight of the object in pounds and \(d\) is the distance of the object from the center of the Earth in miles.

Step 4. Answer the question: "how much will it weigh at \(1,000\) miles above Earth’s surface?"     Since the object is \(1,000\) miles above the surface, the distance of the object from the center of Earth is  \(d = 4,000 + 1,000 = 5,000 \:\:\text{miles}\)

\(\begin{aligned} y & = \frac { 1.6 \times 10 ^ { 9 } } { ( \color{OliveGreen}{5,000}\color{black}{ )} ^ { 2 } } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 25,000,000 } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 2.5 \times 10 ^ { 9 } } \\ & = 0.64 \times 10 ^ { 2 } \\ & = 64 \end{aligned}\)

The object will weigh \(64\) pounds at a distance \(1,000\) miles above the surface of Earth.

Joint Variation

Lastly, we define relationships between multiple variables.

Definition: Joint Variation and Combined Variation

Joint variation is a relationship in which one quantity is proportional to the product of two or more quantities.

Combined variation exists when combinations of direct and/or inverse variation occurs

Example \(\PageIndex{3}\): Joint Variation

Step 1. If we let \(A\) represent the area of an ellipse, then we can use the statement “area varies jointly as \(a\) and \(b\)” to write

Step 2. To find the constant of variation \(k\), use the fact that the area is \(300π\) when \(a=10\) and \(b=30\).

\(\begin{array} { c } { 300 \pi = k ( \color{OliveGreen}{10}\color{black}{ )} (\color{OliveGreen}{ 30}\color{black}{ )} } \\ { 300 \pi = 300 k } \\ { \pi = k } \end{array}\)

Step 3. Therefore, the formula for the area of an ellipse is

\(A=πab\)

The constant of proportionality is \(π\) and the formula for the area of an ellipse is \(A=abπ\).

try-it.png

Given that \(y\) varies directly as the square of \(x\) and inversely with \(z\), where \(y=2\) when \(x=3\) and \(z=27\), find \(y\) when \(x=2\) and \(z=16\).

Joint or Combined Variation

These lessons help Algebra students learn about joint or combined variation.

Related Pages: Proportions Joint Variation Word Problems Direct Variation Inverse Variation More Algebra Lessons

The following figure shows Joint Variation. Scroll down the page for more examples and solutions of Joint and Combine Variations.

Joint Variation

What is Joint Variation or Combined Variation?

Joint Variation or Combined Variation is when one quantity varies directly as the product of at least two other quantities.

For example: y = kxz y varies jointly as x and z, when there is some nonzero constant k

Joint Variation Examples

Example: Suppose y varies jointly as x and z. What is y when x = 2 and z = 3, if y = 20 when x = 4 and z = 3?

Example: z varies jointly with x and y. When x = 3, y = 8, z = 6. Find z, when x = 6 and y = 4.

Joint Variation Application

Example: The energy that an item possesses due to its motion is called kinetic energy. The kinetic energy of an object (which is measured in joules) varies jointly with the mass of the object and the square of its velocity. If the kinetic energy of a 3 kg ball traveling 12 m/s is 216 Joules, how is the mass of a ball that generates 250 Joules of energy when traveling at 10 m/s?

Distinguish between Direct, Inverse and Joint Variation

Example: Determine whether the data in the table is an example of direct, inverse or joint variation. Then, identify the equation that represents the relationship.

Combined Variation

In Algebra, sometimes we have functions that vary in more than one element. When this happens, we say that the functions have joint variation or combined variation. Joint variation is direct variation to more than one variable (for example, d = (r)(t)). With combined variation, we have both direct variation and indirect variation.

How to set up and solve combined variation problems?

Example: Suppose y varies jointly with x and z. When y = 20, x = 6 and z = 10. Find y when x = 8 and z =15.

Lesson on combining direct and inverse or joint and inverse variation

Example: y varies directly as x and inversely as the square of z, and when x = 32, y = 6 and z = 4. Find x when y = 10 and z = 3.

How to solve problems involving joint and combined variation?

If t varies jointly with u and the square of v, and t is 1152 when u is 8 and v is 4, find t when v is 5 and u is 5.

The amount of oil used by a ship traveling at a uniform speed varies jointly with the distance and the square of the speed. If the ship uses 200 barrels of oil in traveling 200 miles at 36 miles per hour, determine how many barrels of oil are used when the ship travels 360 miles at 18 miles per hour.

Designer Dolls found that its number of Dress-Up Dolls sold, N, varies directly with their advertising budget, A, and inversely proportional with the price of each doll, P. When $54,00 was spent on advertising and the price of the doll is $90, then 9,600 units are sold. Determine the number of dolls sold if the amount of advertising budget is increased to $144,000.

Example: y varies jointly as x and z and inversely as w, and y = 3/2, when x = 2, z =3 and w = 4. Find the equation of variation.

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  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Solve direct variation problems.
  • Solve inverse variation problems.
  • Solve problems involving joint variation.

A pre-owned car dealer has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. As she considers the offer, she takes into account the typical price of the dealer's cars, the overall market, and how many she can reasonably expect to sell. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.

Solving Direct Variation Problems

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula e = 0.16 s e = 0.16 s tells us her earnings, e , e , come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 1 .

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other.

Figure 1 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula y = k x n y = k x n is used for direct variation. The value k k is a nonzero constant greater than zero and is called the constant of variation . In this case, k = 0.16 k = 0.16 and n = 1. n = 1. We saw functions like this one when we discussed power functions.

  • Direct Variation

If x and y x and y are related by an equation of the form

then we say that the relationship is direct variation and y y varies directly with, or is proportional to, the n n th power of x . x . In direct variation relationships, there is a nonzero constant ratio k = y x n , k = y x n , where k k is called the constant of variation , which help defines the relationship between the variables.

Given a description of a direct variation problem, solve for an unknown.

  • Identify the input, x , x , and the output, y . y .
  • Determine the constant of variation. You may need to divide y y by the specified power of x x to determine the constant of variation.
  • Use the constant of variation to write an equation for the relationship.
  • Substitute known values into the equation to find the unknown.

Solving a Direct Variation Problem

The quantity y y varies directly with the cube of x . x . If y = 25 y = 25 when x = 2 , x = 2 , find y y when x x is 6.

The general formula for direct variation with a cube is y = k x 3 . y = k x 3 . The constant can be found by dividing y y by the cube of x . x .

Now use the constant to write an equation that represents this relationship.

Substitute x = 6 x = 6 and solve for y . y .

The graph of this equation is a simple cubic, as shown in Figure 2 .

Do the graphs of all direct variation equations look like Example 1 ?

No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through ( 0,0 ) . ( 0,0 ) .

The quantity y y varies directly with the square of x . x . If y = 24 y = 24 when x = 3 , x = 3 , find y y when x x is 4.

Solving Inverse Variation Problems

Water temperature in an ocean varies inversely to the water’s depth. The formula T = 14,000 d T = 14,000 d gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.

If we create Table 2 , we observe that, as the depth increases, the water temperature decreases.

We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations .

For our example, Figure 3 depicts the inverse variation . We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula y = k x y = k x for inverse variation in this case uses k = 14,000. k = 14,000.

  • Inverse Variation

If x x and y y are related by an equation of the form

where k k is a nonzero constant, then we say that y y varies inversely with the n th n th power of x . x . In inversely proportional relationships, or inverse variations , there is a constant multiple k = x n y . k = x n y .

Writing a Formula for an Inversely Proportional Relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.

Recall that multiplying speed by time gives distance. If we let t t represent the drive time in hours, and v v represent the velocity (speed or rate) at which the tourist drives, then v t = distance . v t = distance . Because the distance is fixed at 100 miles, v t = 100 v t = 100 so t = 100/ v . t = 100/ v . Because time is a function of velocity, we can write t ( v ) . t ( v ) .

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. We say that time varies inversely with velocity.

Given a description of an indirect variation problem, solve for an unknown.

  • Determine the constant of variation. You may need to multiply y y by the specified power of x x to determine the constant of variation.

Solving an Inverse Variation Problem

A quantity y y varies inversely with the cube of x . x . If y = 25 y = 25 when x = 2 , x = 2 , find y y when x x is 6.

The general formula for inverse variation with a cube is y = k x 3 . y = k x 3 . The constant can be found by multiplying y y by the cube of x . x .

Now we use the constant to write an equation that represents this relationship.

The graph of this equation is a rational function, as shown in Figure 4 .

A quantity y y varies inversely with the square of x . x . If y = 8 y = 8 when x = 3 , x = 3 , find y y when x x is 4.

Solving Problems Involving Joint Variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation . For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c , c , cost, varies jointly with the number of students, n , n , and the distance, d . d .

Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

For instance, if x x varies directly with both y y and z , z , we have x = k y z . x = k y z . If x x varies directly with y y and inversely with z , z , we have x = k y z . x = k y z . Notice that we only use one constant in a joint variation equation.

A quantity x x varies directly with the square of y y and inversely with the cube root of z . z . If x = 6 x = 6 when y = 2 y = 2 and z = 8 , z = 8 , find x x when y = 1 y = 1 and z = 27. z = 27.

Begin by writing an equation to show the relationship between the variables.

Substitute x = 6 , x = 6 , y = 2 , y = 2 , and z = 8 z = 8 to find the value of the constant k . k .

Now we can substitute the value of the constant into the equation for the relationship.

To find x x when y = 1 y = 1 and z = 27 , z = 27 , we will substitute values for y y and z z into our equation.

A quantity x x varies directly with the square of y y and inversely with z . z . If x = 40 x = 40 when y = 4 y = 4 and z = 2 , z = 2 , find x x when y = 10 y = 10 and z = 25. z = 25.

Access these online resources for additional instruction and practice with direct and inverse variation.

  • Direct and Inverse Variation

5.8 Section Exercises

What is true of the appearance of graphs that reflect a direct variation between two variables?

If two variables vary inversely, what will an equation representing their relationship look like?

Is there a limit to the number of variables that can vary jointly? Explain.

For the following exercises, write an equation describing the relationship of the given variables.

y y varies directly as x x and when x = 6 , y = 12. x = 6 , y = 12.

y y varies directly as the square of x x and when x = 4 , y = 80 .  x = 4 , y = 80 . 

y y varies directly as the square root of x x and when x = 36 , y = 24. x = 36 , y = 24.

y y varies directly as the cube of x x and when x = 36 , y = 24. x = 36 , y = 24.

y y varies directly as the cube root of x x and when x = 27 , y = 15. x = 27 , y = 15.

y y varies directly as the fourth power of x x and when x = 1 , y = 6. x = 1 , y = 6.

y y varies inversely as x x and when x = 4 , y = 2. x = 4 , y = 2.

y y varies inversely as the square of x x and when x = 3 , y = 2. x = 3 , y = 2.

y y varies inversely as the cube of x x and when x = 2 , y = 5. x = 2 , y = 5.

y y varies inversely as the fourth power of x x and when x = 3 , y = 1. x = 3 , y = 1.

y y varies inversely as the square root of x x and when x = 25 , y = 3. x = 25 , y = 3.

y y varies inversely as the cube root of x x and when x = 64 , y = 5. x = 64 , y = 5.

y y varies jointly with x x and z z and when x = 2 x = 2 and z = 3 , y = 36. z = 3 , y = 36.

y y varies jointly as x , z , x , z , and w w and when x = 1 , z = 2 , w = 5 , x = 1 , z = 2 , w = 5 , then y = 100. y = 100.

y y varies jointly as the square of x x and the square of z z and when x = 3 x = 3 and z = 4 , z = 4 , then y = 72. y = 72.

y y varies jointly as x x and the square root of z z and when x = 2 x = 2 and z = 25 , z = 25 , then y = 100. y = 100.

y y varies jointly as the square of x x the cube of z z and the square root of W . W . When x = 1 , z = 2 , x = 1 , z = 2 , and w = 36 , w = 36 , then y = 48. y = 48.

y y varies jointly as x x and z z and inversely as w w . When x = 3 , z = 5 x = 3 , z = 5 , and w = 6 w = 6 , then y = 10. y = 10.

y y varies jointly as the square of x x and the square root of z z and inversely as the cube of w .  w .  When x = 3 , z = 4 , x = 3 , z = 4 , and w = 3 , w = 3 , then y = 6. y = 6.

y y varies jointly as x x and z z and inversely as the square root of w w and the square of t . t . When x = 3 , z = 1 , w = 25 , x = 3 , z = 1 , w = 25 , and t = 2 , t = 2 , then y = 6. y = 6.

For the following exercises, use the given information to find the unknown value.

y y varies directly as x . x . When x = 3 , x = 3 , then y = 12. y = 12. Find y y wneh x = 20. x = 20.

y y varies directly as the square of x . x . When x = 2 , x = 2 , then y = 16. y = 16. Find y y when x = 8. x = 8.

y y varies directly as the cube of x . x . When x = 3 , x = 3 , then y = 5. y = 5. Find y y when x = 4. x = 4.

y y varies directly as the square root of x . x . When x = 16 , x = 16 , then y = 4. y = 4. Find y y when x = 36. x = 36.

y y varies directly as the cube root of x . x . When x = 125 , x = 125 , then y = 15. y = 15. Find y y when x = 1,000. x = 1,000.

y y varies inversely with x . x . When x = 3 , x = 3 , then y = 2. y = 2. Find y y when x = 1. x = 1.

y y varies inversely with the square of x . x . When x = 4 , x = 4 , then y = 3. y = 3. Find y y when x = 2. x = 2.

y y varies inversely with the cube of x . x . When x = 3 , x = 3 , then y = 1. y = 1. Find y y when x = 1. x = 1.

y y varies inversely with the square root of x . x . When x = 64 , x = 64 , then y = 12. y = 12. Find y y when x = 36. x = 36.

y y varies inversely with the cube root of x . x . When x = 27 , x = 27 , then y = 5. y = 5. Find y y when x = 125. x = 125.

y y varies jointly as x and z . x and z . When x = 4 x = 4 and z = 2 , z = 2 , then y = 16. y = 16. Find y y when x = 3 x = 3 and z = 3. z = 3.

y y varies jointly as x , z , and w . x , z , and w . When x = 2 , x = 2 , z = 1 , z = 1 , and w = 12 , w = 12 , then y = 72. y = 72. Find y y when x = 1 , x = 1 , z = 2 , z = 2 , and w = 3. w = 3.

y y varies jointly as x x and the square of z. z. When x = 2 x = 2 and z = 4 , z = 4 , then y = 144. y = 144. Find y y when x = 4 x = 4 and z = 5. z = 5.

y y varies jointly as the square of x x and the square root of z . z . When x = 2 x = 2 and z = 9 , z = 9 , then y = 24. y = 24. Find y y when x = 3 x = 3 and z = 25. z = 25.

y y varies jointly as x x and z z and inversely as w . w . When x = 5 , x = 5 , z = 2 , z = 2 , and w = 20 , w = 20 , then y = 4. y = 4. Find y y when x = 3 x = 3 and z = 8 , z = 8 , and w = 48. w = 48.

y y varies jointly as the square of x x and the cube of z z and inversely as the square root of w .  w .  When x = 2 , x = 2 , z = 2 , z = 2 , and w = 64 , w = 64 , then y = 12. y = 12. Find y y when x = 1 , x = 1 , z = 3 , z = 3 , and w = 4. w = 4.

y y varies jointly as the square of x x and of z z and inversely as the square root of w w and of t . t . When x = 2 , x = 2 , z = 3 , z = 3 , w = 16 , w = 16 , and t = 3 , t = 3 , then y = 1. y = 1. Find y y when x = 3 , x = 3 , z = 2 , z = 2 , w = 36 , w = 36 , and t = 5. t = 5.

For the following exercises, use a calculator to graph the equation implied by the given variation.

y y varies directly with the square of x x and when x = 2 , y = 3. x = 2 , y = 3.

y y varies directly as the cube of x x and when x = 2 , y = 4. x = 2 , y = 4.

y y varies directly as the square root of x x and when x = 36 , y = 2. x = 36 , y = 2.

y y varies inversely with x x and when x = 6 , y = 2. x = 6 , y = 2.

y y varies inversely as the square of x x and when x = 1 , y = 4. x = 1 , y = 4.

For the following exercises, use Kepler’s Law, which states that the square of the time, T , T , required for a planet to orbit the Sun varies directly with the cube of the mean distance, a , a , that the planet is from the Sun.

Using Earth’s time of 1 year and mean distance of 93 million miles, find the equation relating T T and a . a .

Use the result from the previous exercise to determine the time required for Mars to orbit the Sun if its mean distance is 142 million miles.

Using Earth’s distance of 150 million kilometers, find the equation relating T T and a . a .

Use the result from the previous exercise to determine the time required for Venus to orbit the Sun if its mean distance is 108 million kilometers.

Using Earth’s distance of 1 astronomical unit (A.U.), determine the time for Saturn to orbit the Sun if its mean distance is 9.54 A.U.

Real-World Applications

For the following exercises, use the given information to answer the questions.

The distance s s that an object falls varies directly with the square of the time, t , t , of the fall. If an object falls 16 feet in one se c ond, how long for it to fall 144 feet?

The velocity v v of a falling object varies directly to the time, t t , of the fall. If after 2 seconds, the velocity of the object is 64 feet per second, what is the velocity after 5 seconds?

The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 24 inches long and vibrates 128 times per second, what is the length of a string that vibrates 64 times per second?

The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what is the volume when the pressure is 300 millimeters of mercury?

The weight of an object above the surface of Earth varies inversely with the square of the distance from the center of Earth. If a body weighs 50 pounds when it is 3960 miles from Earth’s center, what would it weigh it were 3970 miles from Earth’s center?

The intensity of light measured in foot-candles varies inversely with the square of the distance from the light source. Suppose the intensity of a light bulb is 0.08 foot-candles at a distance of 3 meters. Find the intensity level at 8 meters.

The current in a circuit varies inversely with its resistance measured in ohms. When the current in a circuit is 40 amperes, the resistance is 10 ohms. Find the current if the resistance is 12 ohms.

The force exerted by the wind on a plane surface varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a surface of 65 square feet with a velocity of 30 miles per hour.

The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute (rpm) and the cube of the diameter. If the shaft of a certain material 3 inches in diameter can transmit 45 hp at 100 rpm, what must the diameter be in order to transmit 60 hp at 150 rpm?

The kinetic energy K K of a moving object varies jointly with its mass m m and the square of its velocity v . v . If an object weighing 40 kilograms with a velocity of 15 meters per second has a kinetic energy of 1000 joules, find the kinetic energy if the velocity is increased to 20 meters per second.

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  • Authors: Jay Abramson
  • Publisher/website: OpenStax
  • Book title: College Algebra 2e
  • Publication date: Dec 21, 2021
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
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Direct, Inverse, Joint and Combined Variation

When you start studying algebra, you will also study how two (or more) variables can relate to each other specifically. The cases you’ll study are:

  • Direct Variation , where one variable is a constant multiple of another. For example, the number of dollars I make varies directly (or varies proportionally ) to the number of hours I work. Or, the perimeter of a square varies directly with the length of a side of the square.
  • Inverse or Indirect Variation , where when one of the variables increases, the other one decreases (their product is constant). For example, the temperature in my house varies indirectly (same or inversely ) with the amount of time the air conditioning is running. Or, the number of people I invite to my bowling party varies inversely with the number of games they might get to play (or you can say is proportional to the inverse of ).
  • Joint Variation , where at least two variables are related directly. For example, the area of a triangle is jointly related to both its height and base.
  • Combined Variation , which involves a combination of direct or joint variation, and indirect variation. For example, the average number of phone calls per day between two cities has found to be  jointly proportional  to the populations of the cities, and  inversely proportional  to the square of the distance between the two cities.
  • Partial (Direct) Variation , where two variables are related by a formula, such as the formula for a straight line (with a non-zero $ y$-intercept). For example, the total cost of my phone bill consists of a fixed cost per month, and also a charge per minute.

Note : Just because two variables have a direct relationship, the relationship may not necessarily be a causal relationship (causation) , meaning one variable directly affects the other. There may be another variable that affects both of the variables. For example, there may be a correlation between the number of people buying ice cream and the number of people buying shorts. People buying ice cream do not cause people to buy shorts, but most likely warm weather outside is causing both to happen.

Here is a table for the types of variation we’ll be discussing:

Direct or Proportional Variation

When two variables are related directly, the ratio of their values is always the same. If $ k$, the constant ratio is positive, the variables go up and down in the same direction. (If $ k$ is negative, as one variable goes up, the other goes down; this is still considered a direct variation, but is not seen often in these problems.) Note that $ k\ne 0$.

Think of linear direct variation as a “$ y=mx$” line, where the ratio of $ y$ to $ x$ is the slope ($ m$). With direct variation, the $ y$-intercept is always 0 (zero); this is how it’s defined. Direct variation problems are typically written:    →    $ \boldsymbol {y=kx}$, where $ k$ is the ratio of $ y$ to $ x$ (which is the same as the slope or rate ).

Some problems will ask for that $ k$  value (which is called the constant ratio ,  constant of variation or constant of proportionality  – it’s like a slope!); others will just give you 3 out of the 4 values for $ x$ and $ y$ and you can simply set up a ratio to find the other value. I’m thinking the $ k$ comes from the word “constant” in another language.

Remember the example of making $10 an hour at the mall ($ y=10x$)? This is an example of  direct variation, since the ratio of how much you make to how many hours you work is always constant.

We can also set up direct variation problems in a ratio , as long as we have the same variable in either the top or bottom of the ratio, or on the same side . This will look like the following. Don’t let this scare you; the subscripts just refer to either the first set of variables $ ({{x}_{1}},{{y}_{1}})$, or the second $ ({{x}_{2}},{{y}_{2}})$:   $ \displaystyle \frac{{{{y}_{1}}}}{{{{x}_{1}}}}\,\,=\,\,\frac{{{{y}_{2}}}}{{{{x}_{2}}}}$.

Notes: Partial Variation   (see below), or “varies partly” means that there is an extra fixed constant, so we’ll have an equation like $ y=mx+b$, which is our typical linear equation. Also, I’m assuming in these examples that direct variation is linear ; sometime I see it where it’s not, like in a Direct Square Variation where $ y=k{{x}^{2}}$. There is a word problem example of this here .

Direct Variation Word Problem:

We can solve the following direct variation problem in one of two ways, as shown. We do these methods when we are given any three of the four values for $ x$ and $ y$.

It’s really that easy. Can you see why the proportion method can be the preferred method, unless you are asked to find the $ k$ constant in the formula?

Again, if the problem asks for the  equation that models this situation , it would be “$ y=10x$”.

Here’s another; let’s use the proportion method :

See how similar these types of problems are to the Proportions problems we did earlier?

Direct Square Variation Word Problem:

Again, a Direct Square Variation is when $ y$ is proportional to the square of $ x$, or $ y=k{{x}^{2}}$. Let’s work a word problem with this type of variation and show both the formula and proportion methods:

Inverse or Indirect Variation

Inverse  or Indirect  Variation refers to relationships of two variables that go in the opposite direction (their product is a constant, $ k$). Let’s suppose you are comparing how fast you are driving (average speed) to how fast you get to your school. You might have measured the following speeds and times:

(Note that $ \approx $ means “approximately equal to”).

Do you see how when the $ x$ variable goes up, the $ y$ goes down, and when you multiply the $ x$ with the $ y$, we always get the same number? (Note that this is different than a negative slope, or negative $ k$ value, since with a negative slope, we can’t multiply the $ x$’s and $ y$’s to get the same number).

The formula for inverse or indirect variation is:    →    $ \displaystyle \boldsymbol{y=\frac{k}{x}}$  or  $ \boldsymbol{xy=k}$, where $ k$ is always the same number.

(Note that you could also have an Indirect Square Variation or Inverse Square Variation , like we saw above for a Direct Variation. This would be of the form $ \displaystyle y=\frac{k}{{{{x}^{2}}}}\text{ or }{{x}^{2}}y=k$.)

Here is a sample graph for inverse or indirect variation. This is actually a type of Rational Function  (function with a variable in the denominator) that we will talk about in the Rational Functions, Equations and Inequalities section .

Inverse Variation Word Problem:

We might have a problem like this; we can solve this problem in one of two ways, as shown. We do these methods when we are given any three of the four values for $ x$ and $ y$:

Here’s another; let’s use the product method:

“Work” Inverse Proportion Word Problem:

Here’s a more advanced problem that uses inverse proportions in a  “work” word problem ; we’ll see more “work problems”  here in the  Systems of Linear Equations Section  and  here in the  Rational Functions and Equations section .

Recognizing Direct or Indirect Variation

You might be asked to look at functions  (equations or points that compare $ x$’s to unique $ y$’s   – we’ll discuss later in the Algebraic Functions section) and determine if they are direct, inverse, or neither:

Joint Variation and Combined Variation

Joint variation is just like direct variation, but involves more than one other variable. All the variables are directly proportional, taken one at a time. Let’s set this up like we did with direct variation, find the $ k$, and then solve for $ y$; we need to use the Formula Method:

Joint Variation Word Problem:

We know the equation for the area of a triangle is $ \displaystyle A=\frac{1}{2}bh$ ($ b=$ base and $ h=$ height), so we can think of the area having a joint variation with $ b$ and $ h$, with $ \displaystyle k=\frac{1}{2}$. Let’s do an area problem, where we wouldn’t even have to know the value for $ k$:

Another Joint Variation Word Problem:

Combined Variation

Combined variation involves a combination of direct or joint variation, and indirect variation. Since these equations are a little more complicated, you probably want to plug in all the variables, solve for $ k$, and then solve back to get what’s missing. Let’s try a problem:

Combined Variation Word Problem:

Here’s another; this one looks really tough, but it’s really not that bad if you take it one step at a time:

Combined Variation Word Problem:

One word of caution : I found a variation problem in an SAT book that stated something like this: “ If $ x$ varies inversely with $ y$ and varies directly with $ z$, and if $ y$ and $ z$ are both 12 when $ x=3$, what is the value of $ y+z$ when $ x=5$ ”. I found that I had to solve it setting up two variation equations with two different $ k$ ‘s  (otherwise you can’t really get an answer). So watch the wording of the problems.  🙁 Here is how I did this problem:

Partial Variation

You don’t hear about  Partial Variation  or something being  partly varied  or  part varied very often, but it means that two variables are related by the sum of two or more variables (one of which may be a constant). An example of part variation is the relationship modeled by an equation of a line that doesn’t go through the origin. Here are a few examples:

We’re doing really difficult problems now – but see how, if you know the rules, they really aren’t bad at all?

Learn these rules, and practice, practice, practice!

For Practice : Use the Mathway  widget below to try a Variation  problem. Click on Submit (the blue arrow to the right of the problem) and click on Find the Constant of Variation  to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps , or Click Here , you can register at Mathway for a free trial , and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Introduction to the Graphing Display Calculator (GDC) . I’m proud of you for getting this far!

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Chapter 2: Linear Equations

2.7 Variation Word Problems

Direct variation problems.

There are many mathematical relations that occur in life. For instance, a flat commission salaried salesperson earns a percentage of their sales, where the more they sell equates to the wage they earn. An example of this would be an employee whose wage is 5% of the sales they make. This is a direct or a linear variation, which, in an equation, would look like:

[latex]\text{Wage }(x)=5\%\text{ Commission }(k)\text{ of Sales Completed }(y)[/latex]

[latex]x=ky[/latex]

A historical example of direct variation can be found in the changing measurement of pi, which has been symbolized using the Greek letter π since the mid 18th century. Variations of historical π calculations are Babylonian [latex]\left(\dfrac{25}{8}\right),[/latex] Egyptian [latex]\left(\dfrac{16}{9}\right)^2,[/latex] and Indian [latex]\left(\dfrac{339}{108}\text{ and }10^{\frac{1}{2}}\right).[/latex] In the 5th century, Chinese mathematician Zu Chongzhi calculated the value of π to seven decimal places (3.1415926), representing the most accurate value of π for over 1000 years.

Pi is found by taking any circle and dividing the circumference of the circle by the diameter, which will always give the same value: 3.14159265358979323846264338327950288419716… (42 decimal places). Using an infinite-series exact equation has allowed computers to calculate π to 10 13 decimals.

[latex]\begin{array}{c} \text{Circumference }(c)=\pi \text{ times the diameter }(d) \\ \\ \text{or} \\ \\ c=\pi d \end{array}[/latex]

All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are:

  • [latex]x[/latex] varies directly as [latex]y[/latex]
  • [latex]x[/latex] varies as [latex]y[/latex]
  • [latex]x[/latex] varies directly proportional to [latex]y[/latex]
  • [latex]x[/latex] is proportional to [latex]y[/latex]
  • [latex]x[/latex] varies directly as the square of [latex]y[/latex]
  • [latex]x[/latex] varies as [latex]y[/latex] squared
  • [latex]x[/latex] is proportional to the square of [latex]y[/latex]
  • [latex]x[/latex] varies directly as the cube of [latex]y[/latex]
  • [latex]x[/latex] varies as [latex]y[/latex] cubed
  • [latex]x[/latex] is proportional to the cube of [latex]y[/latex]
  • [latex]x[/latex] varies directly as the square root of [latex]y[/latex]
  • [latex]x[/latex] varies as the root of [latex]y[/latex]
  • [latex]x[/latex] is proportional to the square root of [latex]y[/latex]

Example 2.7.1

Find the variation equation described as follows:

The surface area of a square surface [latex](A)[/latex] is directly proportional to the square of either side [latex](x).[/latex]

[latex]\begin{array}{c} \text{Area }(A) =\text{ constant }(k)\text{ times side}^2\text{ } (x^2) \\ \\ \text{or} \\ \\ A=kx^2 \end{array}[/latex]

Example 2.7.2

When looking at two buildings at the same time, the length of the buildings’ shadows [latex](s)[/latex] varies directly as their height [latex](h).[/latex] If a 5-story building has a 20 m long shadow, how many stories high would a building that has a 32 m long shadow be?

The equation that describes this variation is:

[latex]h=kx[/latex]

Breaking the data up into the first and second parts gives:

[latex]\begin{array}{ll} \begin{array}{rrl} \\ &&\textbf{1st Data} \\ s&=&20\text{ m} \\ h&=&5\text{ stories} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ h&=&kx \\ 5\text{ stories}&=&k\text{ (20 m)} \\ k&=&5\text{ stories/20 m}\\ k&=&0.25\text{ story/m} \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ s&=&\text{32 m} \\ h&=&\text{find 2nd} \\ k&=&0.25\text{ story/m} \\ \\ &&\text{Find }h\text{:} \\ h&=&kx \\ h&=&(0.25\text{ story/m})(32\text{ m}) \\ h&=&8\text{ stories} \end{array} \end{array}[/latex]

Inverse Variation Problems

Inverse variation problems are reciprocal relationships. In these types of problems, the product of two or more variables is equal to a constant. An example of this comes from the relationship of the pressure [latex](P)[/latex] and the volume [latex](V)[/latex] of a gas, called Boyle’s Law (1662). This law is written as:

[latex]\begin{array}{c} \text{Pressure }(P)\text{ times Volume }(V)=\text{ constant} \\ \\ \text{ or } \\ \\ PV=k \end{array}[/latex]

Written as an inverse variation problem, it can be said that the pressure of an ideal gas varies as the inverse of the volume or varies inversely as the volume. Expressed this way, the equation can be written as:

[latex]P=\dfrac{k}{V}[/latex]

Another example is the historically famous inverse square laws. Examples of this are the force of gravity [latex](F_{\text{g}}),[/latex] electrostatic force [latex](F_{\text{el}}),[/latex] and the intensity of light [latex](I).[/latex] In all of these measures of force and light intensity, as you move away from the source, the intensity or strength decreases as the square of the distance.

In equation form, these look like:

[latex]F_{\text{g}}=\dfrac{k}{d^2}\hspace{0.25in} F_{\text{el}}=\dfrac{k}{d^2}\hspace{0.25in} I=\dfrac{k}{d^2}[/latex]

These equations would be verbalized as:

  • The force of gravity [latex](F_{\text{g}})[/latex] varies inversely as the square of the distance.
  • Electrostatic force [latex](F_{\text{el}})[/latex] varies inversely as the square of the distance.
  • The intensity of a light source [latex](I)[/latex] varies inversely as the square of the distance.

All inverse variation relationship are verbalized in written problems as inverse variations or as inversely proportional. Examples of inverse variation or inversely proportional equations are:

  • [latex]x[/latex] varies inversely as [latex]y[/latex]
  • [latex]x[/latex] varies as the inverse of [latex]y[/latex]
  • [latex]x[/latex] varies inversely proportional to [latex]y[/latex]
  • [latex]x[/latex] is inversely proportional to [latex]y[/latex]
  • [latex]x[/latex] varies inversely as the square of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as [latex]y[/latex] squared
  • [latex]x[/latex] is inversely proportional to the square of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as the cube of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as [latex]y[/latex] cubed
  • [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex]
  • [latex]x[/latex] varies inversely as the square root of [latex]y[/latex]
  • [latex]x[/latex] varies as the inverse root of [latex]y[/latex]
  • [latex]x[/latex] is inversely proportional to the square root of [latex]y[/latex]

Example 2.7.3

The force experienced by a magnetic field [latex](F_{\text{b}})[/latex] is inversely proportional to the square of the distance from the source [latex](d_{\text{s}}).[/latex]

[latex]F_{\text{b}} = \dfrac{k}{{d_{\text{s}}}^2}[/latex]

Example 2.7.4

The time [latex](t)[/latex] it takes to travel from North Vancouver to Hope varies inversely as the speed [latex](v)[/latex] at which one travels. If it takes 1.5 hours to travel this distance at an average speed of 120 km/h, find the constant [latex]k[/latex] and the amount of time it would take to drive back if you were only able to travel at 60 km/h due to an engine problem.

[latex]t=\dfrac{k}{v}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} &&\textbf{1st Data} \\ v&=&120\text{ km/h} \\ t&=&1.5\text{ h} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ k&=&tv \\ k&=&(1.5\text{ h})(120\text{ km/h}) \\ k&=&180\text{ km} \end{array} & \hspace{0.5in} \begin{array}{rrl} \\ \\ \\ &&\textbf{2nd Data} \\ v&=&60\text{ km/h} \\ t&=&\text{find 2nd} \\ k&=&180\text{ km} \\ \\ &&\text{Find }t\text{:} \\ t&=&\dfrac{k}{v} \\ \\ t&=&\dfrac{180\text{ km}}{60\text{ km/h}} \\ \\ t&=&3\text{ h} \end{array} \end{array}[/latex]

Joint or Combined Variation Problems

In real life, variation problems are not restricted to single variables. Instead, functions are generally a combination of multiple factors. For instance, the physics equation quantifying the gravitational force of attraction between two bodies is:

[latex]F_{\text{g}}=\dfrac{Gm_1m_2}{d^2}[/latex]

  • [latex]F_{\text{g}}[/latex] stands for the gravitational force of attraction
  • [latex]G[/latex] is Newton’s constant, which would be represented by [latex]k[/latex] in a standard variation problem
  • [latex]m_1[/latex] and [latex]m_2[/latex] are the masses of the two bodies
  • [latex]d^2[/latex] is the distance between the centres of both bodies

To write this out as a variation problem, first state that the force of gravitational attraction [latex](F_{\text{g}})[/latex] between two bodies is directly proportional to the product of the two masses [latex](m_1, m_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating the two masses. From this information, the necessary equation can be derived. All joint variation relationships are verbalized in written problems as a combination of direct and inverse variation relationships, and care must be taken to correctly identify which variables are related in what relationship.

Example 2.7.5

The force of electrical attraction [latex](F_{\text{el}})[/latex] between two statically charged bodies is directly proportional to the product of the charges on each of the two objects [latex](q_1, q_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating these two charged bodies.

[latex]F_{\text{el}}=\dfrac{kq_1q_2}{d^2}[/latex]

Solving these combined or joint variation problems is the same as solving simpler variation problems.

First, decide what equation the variation represents. Second, break up the data into the first data given—which is used to find [latex]k[/latex]—and then the second data, which is used to solve the problem given. Consider the following joint variation problem.

Example 2.7.6

[latex]y[/latex] varies jointly with [latex]m[/latex] and [latex]n[/latex] and inversely with the square of [latex]d[/latex]. If [latex]y = 12[/latex] when [latex]m = 3[/latex], [latex]n = 8[/latex], and [latex]d = 2,[/latex] find the constant [latex]k[/latex], then use [latex]k[/latex] to find [latex]y[/latex] when [latex]m=-3[/latex], [latex]n = 18[/latex], and [latex]d = 3[/latex].

[latex]y=\dfrac{kmn}{d^2}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ && \textbf{1st Data} \\ y&=&12 \\ m&=&3 \\ n&=&8 \\ d&=&2 \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ 12&=&\dfrac{k(3)(8)}{(2)^2} \\ \\ k&=&\dfrac{12(2)^2}{(3)(8)} \\ \\ k&=& 2 \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ y&=&\text{find 2nd} \\ m&=&-3 \\ n&=&18 \\ d&=&3 \\ k&=&2 \\ \\ &&\text{Find }y\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ y&=&\dfrac{(2)(-3)(18)}{(3)^2} \\ \\ y&=&12 \end{array} \end{array}[/latex]

For questions 1 to 12, write the formula defining the variation, including the constant of variation [latex](k).[/latex]

  • [latex]x[/latex] is jointly proportional to [latex]y[/latex] and [latex]z[/latex]
  • [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex]
  • [latex]x[/latex] is jointly proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
  • [latex]x[/latex] is inversely proportional to [latex]y[/latex] to the sixth power
  • [latex]x[/latex] is jointly proportional with the cube of [latex]y[/latex] and inversely to the square root of [latex]z[/latex]
  • [latex]x[/latex] is inversely proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
  • [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex] and is inversely proportional to the cube of [latex]p[/latex]
  • [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex] and square of [latex]z[/latex]

For questions 13 to 22, find the formula defining the variation and the constant of variation [latex](k).[/latex]

  • If [latex]A[/latex] varies directly as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=15[/latex] and [latex]B=5.[/latex]
  • If [latex]P[/latex] is jointly proportional to [latex]Q[/latex] and [latex]R,[/latex] find [latex]k[/latex] when [latex]P=12, Q=8[/latex] and [latex]R=3.[/latex]
  • If [latex]A[/latex] varies inversely as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=7[/latex] and [latex]B=4.[/latex]
  • If [latex]A[/latex] varies directly as the square of [latex]B,[/latex] find [latex]k[/latex] when [latex]A=6[/latex] and [latex]B=3.[/latex]
  • If [latex]C[/latex] varies jointly as [latex]A[/latex] and [latex]B,[/latex] find [latex]k[/latex] when [latex]C=24, A=3,[/latex] and [latex]B=2.[/latex]
  • If [latex]Y[/latex] is inversely proportional to the cube of [latex]X,[/latex] find [latex]k[/latex] when [latex]Y=54[/latex] and [latex]X=3.[/latex]
  • If [latex]X[/latex] is directly proportional to [latex]Y,[/latex] find [latex]k[/latex] when [latex]X=12[/latex] and [latex]Y=8.[/latex]
  • If [latex]A[/latex] is jointly proportional with the square of [latex]B[/latex] and the square root of [latex]C,[/latex] find [latex]k[/latex] when [latex]A=25, B=5[/latex] and [latex]C=9.[/latex]
  • If [latex]y[/latex] varies jointly with [latex]m[/latex] and the square of [latex]n[/latex] and inversely with [latex]d,[/latex] find [latex]k[/latex] when [latex]y=10, m=4, n=5,[/latex] and [latex]d=6.[/latex]
  • If [latex]P[/latex] varies directly as [latex]T[/latex] and inversely as [latex]V,[/latex] find [latex]k[/latex] when [latex]P=10, T=250,[/latex] and [latex]V=400.[/latex]

For questions 23 to 37, solve each variation word problem.

  • The electrical current [latex]I[/latex] (in amperes, A) varies directly as the voltage [latex](V)[/latex] in a simple circuit. If the current is 5 A when the source voltage is 15 V, what is the current when the source voltage is 25 V?
  • The current [latex]I[/latex] in an electrical conductor varies inversely as the resistance [latex]R[/latex] (in ohms, Ω) of the conductor. If the current is 12 A when the resistance is 240 Ω, what is the current when the resistance is 540 Ω?
  • Hooke’s law states that the distance [latex](d_s)[/latex] that a spring is stretched supporting a suspended object varies directly as the mass of the object [latex](m).[/latex] If the distance stretched is 18 cm when the suspended mass is 3 kg, what is the distance when the suspended mass is 5 kg?
  • The volume [latex](V)[/latex] of an ideal gas at a constant temperature varies inversely as the pressure [latex](P)[/latex] exerted on it. If the volume of a gas is 200 cm 3 under a pressure of 32 kg/cm 2 , what will be its volume under a pressure of 40 kg/cm 2 ?
  • The number of aluminum cans [latex](c)[/latex] used each year varies directly as the number of people [latex](p)[/latex] using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in a city that has a population of 1,000,000?
  • The time [latex](t)[/latex] required to do a masonry job varies inversely as the number of bricklayers [latex](b).[/latex] If it takes 5 hours for 7 bricklayers to build a park wall, how much time should it take 10 bricklayers to complete the same job?
  • The wavelength of a radio signal (λ) varies inversely as its frequency [latex](f).[/latex] A wave with a frequency of 1200 kilohertz has a length of 250 metres. What is the wavelength of a radio signal having a frequency of 60 kilohertz?
  • The number of kilograms of water [latex](w)[/latex] in a human body is proportional to the mass of the body [latex](m).[/latex] If a 96 kg person contains 64 kg of water, how many kilograms of water are in a 60 kg person?
  • The time [latex](t)[/latex] required to drive a fixed distance [latex](d)[/latex] varies inversely as the speed [latex](v).[/latex] If it takes 5 hours at a speed of 80 km/h to drive a fixed distance, what speed is required to do the same trip in 4.2 hours?
  • The volume [latex](V)[/latex] of a cone varies jointly as its height [latex](h)[/latex] and the square of its radius [latex](r).[/latex] If a cone with a height of 8 centimetres and a radius of 2 centimetres has a volume of 33.5 cm 3 , what is the volume of a cone with a height of 6 centimetres and a radius of 4 centimetres?
  • The centripetal force [latex](F_{\text{c}})[/latex] acting on an object varies as the square of the speed [latex](v)[/latex] and inversely to the radius [latex](r)[/latex] of its path. If the centripetal force is 100 N when the object is travelling at 10 m/s in a path or radius of 0.5 m, what is the centripetal force when the object’s speed increases to 25 m/s and the path is now 1.0 m?
  • The maximum load [latex](L_{\text{max}})[/latex] that a cylindrical column with a circular cross section can hold varies directly as the fourth power of the diameter [latex](d)[/latex] and inversely as the square of the height [latex](h).[/latex] If an 8.0 m column that is 2.0 m in diameter will support 64 tonnes, how many tonnes can be supported by a column 12.0 m high and 3.0 m in diameter?
  • The volume [latex](V)[/latex] of gas varies directly as the temperature [latex](T)[/latex] and inversely as the pressure [latex](P).[/latex] If the volume is 225 cc when the temperature is 300 K and the pressure is 100 N/cm 2 , what is the volume when the temperature drops to 270 K and the pressure is 150 N/cm 2 ?
  • The electrical resistance [latex](R)[/latex] of a wire varies directly as its length [latex](l)[/latex] and inversely as the square of its diameter [latex](d).[/latex] A wire with a length of 5.0 m and a diameter of 0.25 cm has a resistance of 20 Ω. Find the electrical resistance in a 10.0 m long wire having twice the diameter.
  • The volume of wood in a tree [latex](V)[/latex] varies directly as the height [latex](h)[/latex] and the diameter [latex](d).[/latex] If the volume of a tree is 377 m 3 when the height is 30 m and the diameter is 2.0 m, what is the height of a tree having a volume of 225 m 3 and a diameter of 1.75 m?

Answer Key 2.7

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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examples of joint variation problem solving

JOINT VARIATION WORD PROBLEMS

Problem 1 :

z varies directly with the sum of squares of x and y. z = 5 when x = 3 and y = 4. Find the value of z when x = 2 and y = 4.

Since z varies directly with the sum of squares of x and y,

z ∝ x 2 +  x 2

z = k(x 2  + y 2 ) ----(1)

Substitute z = 5, x = 3 and y = 4 to find the value k.

5 = k(3 2  + 4 2 )

5 = k(9 + 16)

Divide both sides by 25.

Substitute k = 1/5 in (1).

z = (1/5)(x 2  + y 2 )

Substitute  x = 2, y = 4 and evaluate z.

z = (1/5)( (2 2  + 4 2 )

z = (1/5)( (4   + 16)

z = (1/5)( (20)

Problem 2 :

M varies directly with the square of d and inversely with the square root of x. M = 24 when d = 4 and x = 9. Find the value of M when d = 5 and x = 4.

Since m varies directly with the square of d and inversely with the square root of x

M ∝ d 2 √ x

M = kd 2 √ x ----(1)

Substitute M = 24, d = 4 and x = 9 to find the value k.

24 = k4 2 √9

24 = k(16)(3)

Divide both sides by 48.

Substitute k = 1/2 in (1).

M = (1/2)(d 2 √ x )

Substitute d  = 5, x = 4 and evaluate M.

M = (1/2) (5 2 √4 )

M = (1/2)( (25)(2)

Problem 3 :

Square of T varies directly with the cube of a and inversely with the square of d. T = 2 when a = 2 and d = 4. Find the value of s quare of T when a = 4 and d = 2

Since square of T varies directly with the cube of a and inversely with the square of d

T 2 ∝ a 3 d 2

T 2  = ka 3 d 2  ----(1)

Substitute T = 2, a = 2 and d = 4 to find the value k.

2 2  = k2 3 4 2

4 = k(4)(16)

Divide both sides by 64.

Substitute k = 1/16 in (1).

T 2  = (1/16)a 3 d 2

Substitute a = 4, d = 2 and evaluate T 2 . 

T 2  = (1/16)(4 3 )(2 2 )

T 2  = (1/16)(64)(4)

T 2  = 16

Problem 4 :

The area of a rectangle varies directly with its length and square of its width. When the length is 5 cm and width is 4 cm, the area is 160 cm 2 . Find the area of the rectangle when the length is 7 cm and the width is 3 cm.

Let A represent the area of the rectangle, l represent the length and w represent width.

Since the area of the rectangle varies directly with its length and square of its width,

A ∝ lw 2

A = klw 2  ----(1)

Substitute A = 160, l = 5 and d = 4 to find the value k.

160 = k(5)(4 2 )

160 = k(5)(16 )

160 = 80k

Divide both sides by 80.

Substitute k = 2 in (1).

Substitute l  = 7, w = 3 and evaluate A. 

A = 2(7)(3 2 )

A = 2(7)(9)

Area of the rectangle = 126 cm 2

Problem 5 :

The volume of a cylinder varies jointly as the square of radius and two times of its height. A cylinder with radius 4 cm and height 8 cm has a volume 128 π cm 3 . Find the volume of a cylinder with radius 3 cm and height 10 cm.

Let V represent volume of the cylinder, r represent radius and h represent height.

Since t he volume of a cylinder varies jointly as the radius and the sum of the radius and the height.

V ∝ r 2 (2h)

V = kr 2 (2h)   ----(1)

Substitute V = 128 π , r = 4 and h = 8 to find the value of k.

128π   = k(4 2 )(2 ⋅  8)

128π   = k(16)(16)

128π  = 256k

Divide both sides by 256.

π/2  = k

Substitute k = π/2 in (1).

V = ( π/2) r 2 (2h)

V = π r 2 h

Substitute r = 3, h = 10 and evaluate V.

V = π(3 2 )(10)

V = π(9) (10)

Volume of the cylinder = 90 π cm 3

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Big Ideas Math Answers

Joint Variation – Formula, Examples | How to Solve Problems Involving Joint Variation?

Joint Variation definition, rules, methods and formulae are here. Check the joint variation problems and solutions to prepare for the exam. Refer to problems of direct and inverse variations and the relationship between the variables. Know the different type of variations like inverse, direct, combined and joint variation. Go through the below sections to check definition, various properties, example problems, value tables, concepts etc.

Joint Variation – Introduction

Joint Variation refers to the scenario where the value of 1 variable depends on 2 or more and other variables that are held constant. For example, if C varies jointly as A and B, then C = ABX for which constant “X”. The joint variation will be useful to represent interactions of multiple variables at one time.

Most of the situations are complicated than the basic inverse or direct variation model. One or the other variables depends on the multiple other variables. Joint Variation is nothing but the variable depending on 2 or more variables quotient or product. To understand clearly with an example, The amount of busing candidates for each of the school trip varies with the no of candidates attending the distance from the school. The variable c (cost) varies jointly with n (number of students) and d (distance).

Joint Variation problems are very easy once you get the perfection of the lingo. These problems involve simple formulae or relationships which involves one variable which is equal to the “one” term which may be linear (with just an “x” axis), a quadratic equation (like “x²) where more than one variable (like “hr²”), and square root (like “\sqrt{4 – r^2\,}4−r2​”) etc.

Functions of 2 or More Variables

It is very uncommon for the output variable to depend on 2 or more inputs. Most of the familiar formulas describe the several variables functions. For suppose, if the rectangle perimeter depends on the length and width. The cylinder volume depends on its height and radius. The travelled distance depends on the time and speed while travelling. The function notation of the formulas can be written as

P = f(l,w) = 2l + 2w where P is the perimeter and is a function of width and length

V = f(r,h) = Πr²h where V is the volume and is a function of radius and height

d = f(r,t) = rt where d is the distance and is a function of time and rate.

Tables of Values

Just for the single variable functions, we use the tables to describe two-variable functions. The heading of the table shows row and column and it shows the value if two input variables and the complete table shows the values of the output variable.

You can easily make graphs in three dimensions for two-variable functions. Instead of representing graphs, we represent functions by holding two or one variable constants.

Also, Read:

  • What is Variation
  • Practice Test on Ratio and Proportion

How to Solve Joint Variation Problems?

Follow the step by step procedure provided below to solve problems involving Joint Variation and arrive at the solution easily. They are along the lines

Step 1: Write the exact equation. The problems of joint variation can be solved using the equation y =kxz. While dealing with the word problems. you should also consider using variables other than x,y and z. Use the variables which are relevant to the problem being solved. Read the problem carefully and determine the changes in the equation of joint variation such as cubes, squares or square roots.

Step 2: With the help of the information in the problem, you have to find the value of k which is called the constant of proportionality and variation.

Step 3: Rewrite the equation starting with 1 substituting the value of k and found in step 2.

Step 4: Use the equation in step 3 and the information in the problem to answer the question. While solving the word problems, remember including the units in the final answer.

Joint Variation Problems with Solutions

The area of a triangle varies jointly as the base and the height. Area = 12m² when base = 6m and height = 4m. Find base when Area = 36m² and height = 8m?

The area of the triangle is represented with A

The base is represented with b

Height is represented with h

As given in the question,

A = 12m² when B = 6m and H = 4m

We know the equation,

A = kbh where k is the constant value

12 = k(6)(4)

Divide by 24 on both sides, we get

12/24 = k(24)/24

The value of k = 1/2

As the equation is

To find the base of the triangle of A = 36m² and H = 8m

36 = 1/2(b)(8)

Dividing both sides by 4, we get

36/4 = 4b/4

The value of base = 9m

Hence, the base of the triangle when A = 36m² and H = 8m is 9m

Wind resistance varies jointly as an object’s surface velocity and area. If the object travels at 80 miles per hour and has a surface area of 30 square feet which experiences 540 newtons wind resistance. How much fast will the car move with 40 square feet of the surface area in order to experience a wind resistance of 495 newtons?

Let w be the wind resistance

Let s be the object’s surface area

Let v be the object velocity

The object’s surface area = 80 newtons

The wind resistance = 540 newtons

The object velocity = 30

w = ksv where k is the constant

(540) = k (80) (30)

540 = k (2400)

540/2400 = k

The value of k is 9/40

To find the velocity of the car with s = 40, w = 495 newtons and k = 9/40

Substitute the values in the equation

495 = (9/40) (40) v

The velocity of a car is 55mph for which the object’s surface area is 40 and wind resistance is 495 newtons

Hence, the final solution is 55mph

For the given interest, SI (simple interest) varies jointly as principal and time. If 2,500 Rs left in an account for 5 years, then the interest of 625 Rs. How much interest would be earned, if you deposit 7,000 Rs for 9 years?

Let i be the interest

Let p be the principal

Let t be the time

The interest is 625 Rs

The principal is 2500

The time is 5 hours

i = kpt where k is the constant

Substituting the values in the equation,

(625) = k(2500)(5)

625 = k(12,500)

Dividing 12,500 on both the sides

625/12,500 = k (12,500)/12,500

The value of k = 1/20

To find the interest where the deposit is 7000Rs for 9 years, use the equation

i = (1/20) (7000) (9)

i = (350) (9)

Therefore, the interest is 3,150 Rs, if you deposit 7,000 Rs for 9 years

Thus, the final solution is Rs. 3,150

The volume of a pyramid varies jointly as its height and the area of the base. A pyramid with a height of 21 feet and a base with an area of 24 square feet has a volume of 168 cubic feet. Find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet?

Let v be the volume of a pyramid

Let h be the height of a pyramid

Let a be the area of a pyramid

The volume v = 168 cubic feet

The height h = 21 feet

The area a = 24 square feet

V = Kha where K is the constant,

168 = k(21)(24)

168 = k(504)

Divide 504 on both sides

168/504 = k(504)/504

The value of k = 1/3

To find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet

h = 18 feet

a = 42 square feet

V = (1/3) (18) (42)

V = (6) (42)

V = 252 ft³

The volume of the pyramid = 252 ft³ which has a height of 18 feet and a base with an area of 42 square feet

Therefore, the final solution is 252 ft³

The amount of oil used by a ship travelling at a uniform speed varies jointly with the distance and the square of the speed. If the ship uses 200 barrels of oil in travelling 200 miles at 36 miles per hour, determine how many barrels of oil are used when the ship travels 360 miles at 18 miles per hour?

No of barrels of oil = 200

The distance at which the oil is travelling = 200 miles

The distance at which the ship is travelling = 36 miles per hour

A = kds² where k is constant

200 = k.200.(36)²

Dividing both sides by 200

200/200 = k.200.(36)²/200

1 = k.(36)²

The value of k is 1/1296

To find the no of barrels when the ship travels 360 miles at 18 miles per hour

A = 1/1296 * 360 * 18²

Therefore, 90 barrels of oil is used when the ship travels 360 miles at 18 miles per hour

Thus, the final solution is 90 barrels

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6.3 Proportion and Variation

A used-car company has just offered their best candidate, Uki, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships like this one between earnings, sales and commission rate.

Solving Direct Variation Problems

e=.16s

Direct Variation

x

Many times we are given a description of a direct variation problem and need to find an unknown.  They key to these problems is to first find the constant of variation using what you know and then you are able to use your new formula to find the unknown quantity.

Example Solving a Direct Variation Problem

y=25

and we can plug in x:

y=\dfrac{25}{8}(6)^3=\dfrac{(25)(216)}{8}=25(27)=675

Note that our direct variation graphs look very different as the first one is linear and the previous example is cubic. These functions can be linear, quadratic, cubic, quartic or even radical, but they all pass through (0,0).

Try it Now 1

y=24

Solving Inverse Variation Problems

t=\dfrac{0.7}{r}

We notice in the relationship between these variables that, as one quantity was doubled, the other was cut in half. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportion relationships are called inverse variations . See the graph below:

Graph showing Inverse proportion

Inverse Variation

\[y=\dfrac{k}{x^n}\]

Example Writing a Formula for an Inversely Proportional Relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.

d=rt

Time varies inversely with rate of speed.

Example Solving an Inverse Variation Problem

First we want to find k:

k=x^ny = 2^3(25)=8(25)=200

Now we have the formula:

y=\dfrac{200}{x^n}

Try it Now 2

y=8

Solving Problems Involving Joint Variation

c

Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

z

Example Solving a Problem Involving Joint Variation

x=6

We start with our general equation based on the description of the variation:

x=\dfrac{ky^2}{\sqrt[3]{z}}

Now write our formula:

x=\dfrac{3y^2}{\sqrt[3]{z}}

Try it Now 3

x=40

Try it  Now Answers

k=\dfrac{24}{9}

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  1. Solve problems involving joint variation

    Example 4: Solving Problems Involving Joint Variation A quantity x varies directly with the square of y and inversely with the cube root of z. If x = 6 when y = 2 and z = 8, find x when y = 1 and z = 27. Solution Begin by writing an equation to show the relationship between the variables.

  2. Joint Variation: Solving Joint Variation Problems in Algebra

    Example 1: Finding an Equation of Joint Variation Find an equation of variation where a varies jointly as b and c, and a = 30 when b = 2 and c =3. Solution Write the joint variation equation that resembles the general joint variation formula y = kxz. Let a = y, x = b, z = c. y = kxz a = kbc Recommended The Madness of Doomsday Cults

  3. Joint Variation

    For example, if C varies jointly as A and B, then C = ABX for which constant "X". The joint variation will be useful to represent interactions of multiple variables at one time. Most of the situations are complicated than the basic inverse or direct variation model. One or the other variables depends on the multiple other variables.

  4. Joint And Combined Variation Word Problems (video lessons, examples and

    Example 1: A quantity varies inversely as two or more other quantities. The figure below shows a rectangular solid with a fixed volume. Express its width, w, as a joint variation in terms of its length, l, and height, h. Solution: w ∝ 1/ (lh) In other words, the longer the length l or the height h, the narrower is the width w.

  5. Joint Variation

    1. The variable x is in joint variation with y and z. When the values of y and z are 4 and 6, x is 16. What is the value of x when y = 8 and z =12? Solution: The equation for the given problem of joint variation is x = Kyz where K is the constant. For the given data 16 = K × 4 × 6 or, K = 46 4 6. So substituting the value of K the equation becomes

  6. JOINT VARIATION (Definition, Examples, Solving Problems) Made Easy

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  7. 1.8: Variation

    Solving Problems involving Direct, Inverse, and Joint variation ... Joint variation is a relationship in which one quantity is proportional to the product of two or more quantities. Combined variation exists when combinations of direct and/or inverse variation occurs . Example \(\PageIndex{3}\): Joint Variation. The area of an ellipse varies ...

  8. Joint or Combined Variation (video lessons, examples and solutions)

    Show Video Lesson Joint Variation Application Example: The energy that an item possesses due to its motion is called kinetic energy. The kinetic energy of an object (which is measured in joules) varies jointly with the mass of the object and the square of its velocity.

  9. 5.8 Modeling Using Variation

    Solving Problems Involving Joint Variation. Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing ...

  10. Joint Variation: Definition & Examples

    Let's look at some examples of joint variation problems. If y varies jointly with x, z, and w, and the value of y is 60 when x = 2, z = 3, and w = 5, what is the value of y when x = -3, z =...

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    Beginning Algebra Direct, Inverse, Joint and Combined Variation Direct, Inverse, Joint and Combined Variation When you start studying algebra, you will also study how two (or more) variables can relate to each other specifically. The cases you'll study are: Direct Variation, where one variable is a constant multiple of another.

  13. Joint and Combined Variation

    Combined variation is a mix of direct and indirect variation. The joint variation equation is z = k x m y n where k ≠ 0 and m > 0, n > 0. Review. For questions 1-5, write an equation that represents relationship between the variables. w varies inversely with respect to x and y. r varies inversely with the square of q. z varies jointly with x ...

  14. Lesson 9.5

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  16. Joint Variation Word Problems

    JOINT VARIATION WORD PROBLEMS. z varies directly with the sum of squares of x and y. z = 5 when x = 3 and y = 4. Find the value of z when x = 2 and y = 4. Substitute z = 5, x = 3 and y = 4 to find the value k. Divide both sides by 25. Substitute k = 1/5 in (1). Substitute x = 2, y = 4 and evaluate z. M varies directly with the square of d and ...

  17. Joint Variation

    Joint Variation refers to the scenario where the value of 1 variable depends on 2 or more and other variables that are held constant. For example, if C varies jointly as A and B, then C = ABX for which constant "X". The joint variation will be useful to represent interactions of multiple variables at one time.

  18. Joint Variation Examples and Word Problems

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  20. Math 9 Module 4

    Joint variation - the statement "a varies jointly as b and c" means a = kbc, or 𝑘 = 𝑎 𝑏𝑐 , where k is the constant of variations. The volume (V) of a parallelogram varies jointly as the length (l), width (w), and its height (h). The volume (V) of a cylinder varies jointly as height (h) and the square of the radius (r).