Direct and Inverse Variation

y varies directly with x (or: "x and y are directly proportional") if there is a constant k such that

$$y = kx.$$

(You can put the "k" on either side --- that is, you can write $x = ky$ instead --- as long as you pick one or the other for a given problem.)

y varies inversely with x (or "x and y are inversely proportional") if there is a constant k such that

$$y = \dfrac{k}{x}.$$


Example. Here is an example where x varies directly with y:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & 1 & & 2 & & 3 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & y & & 6 & & 12 & & 18 & & 24 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

In this case, $y = kx$ , and $k =
   6$ --- that is, $y = 6x$ .

Here's another

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & 1 & & 2 & & 3 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & y & & -4 & & -8 & & -12 & & -16 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

In this case, $y = kx$ , and $k =
   -4$ --- that is, $y = -4x$ . Notice that the equation of a direct variation between two variables is the equation of a line.

Here is a case of inverse variation:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & 1 & & 2 & & 3 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & y & & 2 & & 1 & & $\dfrac{2}{3}$ & & $\dfrac{1}{2}$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

In this case, $y = \dfrac{k}{x}$ , and $k = 2$ .


Example. Values of x and y are given in the table below.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & x & & 1 & & 5 & & 9 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & y & & 5 & & 17 & & 29 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Are x and y are directly proportional?

When $x = 1$ and $y = 5$ , I get $\dfrac{y}{x} = \dfrac{5}{1} = 5$ . But when $x = 9$ and $y = 29$ , I get $\dfrac{y}{x} = \dfrac{29}{9}$ , and $5
   \ne \dfrac{29}{9}$ . Therefore, x and y are not directly proportional.


Example. y varies directly with x. When $x = 10$ , $y = 25$ . Find x when $y = 40$ .

y varies directly with x: $y = kx$ .

When $x = 10$ , $y = 25$ : $25 =
   k\cdot 10$ , or $25 = 10k$ . Solving for k, I get $k = \dfrac{25}{10} = \dfrac{5}{2}$ .

Thus, $y = \dfrac{5}{2}x$ .

When $y = 40$ , I have $40 =
   \dfrac{5}{2}x$ . Solving for x, I get $x = 16$ .


Example. y varies inversely with x. When $x = 20$ , $y = -3$ . Find y when $x = 120$ .

y varies inversely with x: $y =
   \dfrac{k}{x}$ .

When $x = 20$ , $y = -3$ : $-3 =
   \dfrac{k}{20}$ . Solving for k, I get $k = -60$ .

Thus, $y = -\dfrac{60}{x}$ .

When $x = 120$ , I have $y =
   -\dfrac{60}{120} = -\dfrac{1}{2}$ .


Other kinds of variation are possible.

z varies jointly with x and y if there is a constant k such that

$$z = kxy.$$

y varies directly with the square of x if there is a constant k such that

$$y = kx^2.$$

y varies inversely with the square of x if there is a constant k such that

$$y = \dfrac{k}{x^2}.$$


Example. The area of a cylinder varies jointly with the radius and the height. When the radius is 3 and the height is 6, the area is $36\pi$ . Find the area when the radius is 4 and the height is 5.

Let A be the area, let r be the radius, and let h be the height.

The area of a cylinder varies jointly with the radius: $A = krh$ .

When the radius is 3 and the height is 6, the area is $36\pi$ : $36\pi = k\cdot
   3\cdot 6$ , or $36\pi = 18k$ . Solving for k, I get $k = 2\pi$ .

Thus, $A = 2\pi r h$ .

When $r = 4$ and $h = 5$ , I get $A
   = 2\pi\cdot 4\cdot 5 = 40\pi$ .


Example. A force varies inversely with the square of the distance from an object. When the distance from the object is 6, the force is 144. Find the force when the distance is 9.

Let F be the force and let d be the distance from the object.

A force varies inversely with the square of the distance from an object: $F = \dfrac{k}{d^2}$ .

When the distance from the object is 6, the force is 144: $144 = \dfrac{k}{6^2}$ , or $144 =
   \dfrac{k}{36}$ . Solving for k, I get $k = 5184$ .

Thus, $F = \dfrac{5184}{d^2}$ .

When $d = 9$ , $F =
   \dfrac{5184}{9^2} = \dfrac{5184}{81} = 64$ .


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