Solving real world examples of Inverse variation

I introduced variation as a topic not too long ago. I explained what it is and listed the different types of variation. I also solved some examples of direct and inverse variation. It was well understand I guess, by those that saw it. If you missed the publications, why not take a look below:

• Variation - the mathematical relationship between two quantities
• Inverse variation - two quantities whose value go up and down

In this article, we will see real world examples of inverse variation and how to tackle them. You remember that in inverse variation, two quantities are involved. The rise in value of one of the commodities, lead to the decrease in value of the second commodity and vice versa. Now we will see and solve real world examples of this.

Example 1

The number of pipes required required to fill a pool with water varies inversely as the cube root of the time involved. When 3 pipes were used, it took 64 minutes to fill the pool. Find how many pipes will be used to fill the pool in 27 minutes. Also find how long it will take 2 pipes to fill the pool.

Solution: We will use the first data to solve or the constant, before solving for the other variables.

Let n = number of pipe = 3
t = time = 64
k = constant

(First Step: Expressing it mathematically, we have:)
n ∝ 1/∛t
(Next Step: Introducing the constant k, we have)
n =k/∛t
(Next Step: Having derived the formular, we can solve for the constant k)

n =k/∛t
3 = k/(∛64)
(Next Step: Solve the bracket or finding the cube root of 64 we have)
3 = k/4
(Next Step: Cross-multiply to make k the subject of the formula)
k = 4x3 = 12.

Having solved for the constant, we can easily substitute and find the unknown variables as follows:

To find the number of pipes needed to fill the pull in 27 minutes

n =k/∛t
n = 12/∛27
(Next Step: evaluating the right side, we have)
n = 12/3
n = 4

Also , finding how long it will take 2 pipes to fill the pool

n =k/∛t
2 =12/∛t
(Next Step: Cross multiply to make t the subject of the formula)
2 x ∛t = 12
(Next Step: Collect like terms)
∛t = 12/2
∛t = 6
(Next Step: Introduce cube on the right side to cancel the cube root on the left side)
t = 6³
t = 216

Example 2

A truck travelling on a highway has a resistance R (in Newton) which varies inversely as the square of its speed (in km/h). At a speed of 4km/h, its resistance is 4N. Find the resistance at a speed of 3 km/h

Solution
R = resistance = 4N
Speed = S = 4
Let k = constant.

R ∝ 1/S²
(Next Step: Introducing the constant k and finding it)
R = k/S²
4 = k/4²
4 = k/16
(Next Step: Cross-multiply to make k the subject of the formula)
k = 4 x 16
k = 64

Finding the truck resistance at a speed of 3 km/h
R = k/S²
R = 64/3²
R = 64/9
R = 7.1N

Conclusion

We have seen two real-world examples of inverse variation. We have also seen how to solve the arithmetic. Its not really difficult. Just understand that for the two quantity involved, as one increases in value, the other decreases and vice versa. One you find the constant with the initial values given, its easier then to find the values of the other quantities given in the example.