Variation Equations (page 2 of 3)

• Suppose that y is inversely proportional to x, and that y = 0.4 when x = 2.5.  Find y when x = 4.

Translating, I get the formula:

y = k/x

Plugging in the data point they gave me, I can solve for the value of k:

y = k/x
0.4 = k/(2.5)
(0.4)(2.5) = k = 1

Now that I have found the value of the variation constant, I can plug in the x-value they gave me, and find the value of y:

y = 1/x
y = 1/4

Then the answer is:  y = 1/4

Most word problems, of course, do not come all neatly arranged like the above examples. Instead, you have to figure out what values go where.

• According to Hooke's Law, the force needed to stretch a spring is proportional to the amount the spring is stretched. If fifty pounds of force stretches a spring five inches, how much will the spring be stretched by a force of 120 pounds?

"Is proportional to" means "varies directly with", so the formula for Hooke's Law is "F = kd", where "F" is the force and "d" is the distance. (Note that, in physics, "weight" is a force. These Hooke's Law problems are often stated in terms of weight, and the weight is the force.)

First I have to solve for the value of k. They've given me the data point (d, F) = (5, 50), so I'll plug this in to the formula:

F = kd
50 = k×5
10 = k

Now I know that the formula for this particular spring is "F = 10d". (Hooke's Law doesn't change, but each spring is different, so each spring will have its own "k".) Once I know the formula, I can answer their question: "How much will the spring be stretched by a force of 120 pounds?"

F = 10d
120 = 10d
12 = d

Note that they did not ask "What is the value of 'd'?". Be sure to answer the question they actually asked. The final answer is:

The spring will stretch twelve inches.

• Kepler's third law of planetary motion states that the square of the time required for a planet to make one revolution about the sun varies directly as the cube of the average distance of the planet from the sun.  If you assume that Mars is 1.5 times as far from the sun as is the earth, find the approximate length of a Martian year.

This one is a bit different. The variation relationship is between the square of the time and the cube of the distance. The formula is: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

t2 = kd3

If I take "d = 1" to mean "the distance is one AU", an AU being an "astronomical unit" (the distance of earth from the sun), then the distance for Mars is 1.5 AU. Also, I will take "t = 1" to stand for "one earth year". Then, in terms of the planet Earth, I get:

(1)2 = k(1)3
1 = k

Then the formula, in terms of Earth, is:

t2 = d3

Now I'll plug in the information for Mars (in comparison to earth):  d = 1.5:

t2 = (1.5)3
t = sqrt(3.375)

In other words, the Martian year is approximately the length of 1.837 earth years (or just over one year and ten months).

<< Previous  Top  |  1 | 2 | 3  |  Return to Index  Next >>

 Cite this article as: Stapel, Elizabeth. "Variation Equations & Word Problems." Purplemath. Available from     http://www.purplemath.com/modules/variatn2.htm. Accessed [Date] [Month] 2016

MathHelp.com Courses
This lesson may be printed out for your personal use.