Direct and Inverse Variation

Understanding how quantities change together

What is Variation?

Variation describes how one quantity changes in relation to another. When two quantities are related, a change in one causes a predictable change in the other.

Types of Variation

Direct Variation

\(y \propto x\)

\(y = kx\)

As \(x\) increases, \(y\) increases

As \(x\) decreases, \(y\) decreases

Graph: Straight line through origin

Inverse Variation

\(y \propto \frac{1}{x}\)

\(y = \frac{k}{x}\)

As \(x\) increases, \(y\) decreases

As \(x\) decreases, \(y\) increases

Graph: Rectangular hyperbola

The Constant of Variation (k)

In both types of variation, \(k\) is called the constant of variation or constant of proportionality.

For direct variation: \(k = \frac{y}{x}\)

For inverse variation: \(k = xy\)

Direct Variation

Example 1: Finding the Constant

If \(y\) varies directly as \(x\), and \(y = 15\) when \(x = 3\), find:

(a) the constant of variation, (b) the value of \(y\) when \(x = 7\)

1 Write the variation equation:

\[y = kx\]

2 Find k using given values:

\[15 = k(3)\]

\[k = \frac{15}{3} = 5\]

3 Write the complete equation:

\[y = 5x\]

4 Find y when x = 7:

\[y = 5(7) = 35\]

Example 2: Direct Variation with Powers

The area \(A\) of a circle varies directly as the square of its radius \(r\). If \(A = 78.5\) cm² when \(r = 5\) cm, find \(A\) when \(r = 8\) cm.

1 Write the variation:

\[A \propto r^2 \quad \Rightarrow \quad A = kr^2\]

2 Find k:

\[78.5 = k(5)^2 = 25k\]

\[k = \frac{78.5}{25} = 3.14\]

3 Find A when r = 8:

\[A = 3.14(8)^2 = 3.14(64) = 200.96 \text{ cm}^2\]

Inverse Variation

Example 3: Inverse Variation

If \(y\) varies inversely as \(x\), and \(y = 6\) when \(x = 4\), find \(y\) when \(x = 8\).

1 Write the inverse variation equation:

\[y = \frac{k}{x}\]

2 Find k:

\[6 = \frac{k}{4}\]

\[k = 6 \times 4 = 24\]

3 Find y when x = 8:

\[y = \frac{24}{8} = 3\]

Notice: When x doubled (4 to 8), y halved (6 to 3). This is characteristic of inverse variation.

Example 4: Real-World Inverse Variation

The time \(t\) to complete a journey varies inversely as the speed \(s\). If it takes 4 hours at 60 km/h, how long would it take at 80 km/h?

1 Set up equation:

\[t = \frac{k}{s}\]

2 Find k:

\[4 = \frac{k}{60}\]

\[k = 240\]

(This represents the distance: 240 km)

3 Find t when s = 80:

\[t = \frac{240}{80} = 3 \text{ hours}\]

Interactive Variation Explorer

Visualize Variation

See how direct and inverse variation differ graphically

Equation & Properties

y = 10x

x 1 2 3 4 5
y 10 20 30 40 50

Joint and Combined Variation

Joint Variation

When a quantity varies directly as the product of two or more other quantities:

\[z \propto xy \quad \Rightarrow \quad z = kxy\]

Example 5: Combined Variation

The force \(F\) between two magnets varies directly as the product of their strengths \(m_1\) and \(m_2\), and inversely as the square of the distance \(d\) between them.

Write the equation for this relationship.

1 Express the variation:

\[F \propto \frac{m_1 m_2}{d^2}\]

2 Write as an equation:

\[F = \frac{k m_1 m_2}{d^2}\]

Practice Problems

Question 1: If \(y\) varies directly as \(x\), and \(y = 12\) when \(x = 4\), find \(y\) when \(x = 10\).

Show Solution

\(y = kx\)

\(12 = k(4) \Rightarrow k = 3\)

\(y = 3x\)

When \(x = 10\): \(y = 3(10) = 30\)

Question 2: The number of workers \(w\) needed to complete a job varies inversely as the time \(t\) taken. If 6 workers can complete the job in 10 days, how many workers are needed to complete it in 4 days?

Show Solution

\(w = \frac{k}{t}\)

\(6 = \frac{k}{10} \Rightarrow k = 60\)

When \(t = 4\): \(w = \frac{60}{4} = 15\) workers

Question 3: The volume \(V\) of a cone varies jointly as its height \(h\) and the square of its radius \(r\). If \(V = 48\pi\) when \(h = 9\) and \(r = 4\), find \(V\) when \(h = 12\) and \(r = 5\).

Show Solution

\(V = khr^2\)

\(48\pi = k(9)(16) = 144k\)

\(k = \frac{48\pi}{144} = \frac{\pi}{3}\)

When \(h = 12, r = 5\):

\(V = \frac{\pi}{3}(12)(25) = 100\pi\)

Question 4: The gravitational force \(F\) between two objects varies directly as the product of their masses and inversely as the square of the distance between them. If \(F = 100\) N when \(m_1 = 10\) kg, \(m_2 = 20\) kg, and \(d = 2\) m, find \(F\) when \(d\) is doubled.

Show Solution

\(F = \frac{km_1m_2}{d^2}\)

\(100 = \frac{k(10)(20)}{4} = 50k\)

\(k = 2\)

When \(d = 4\): \(F = \frac{2(10)(20)}{16} = \frac{400}{16} = 25\) N

(Force becomes 1/4 when distance doubles)

CSEC Exam Tips

1. "varies directly" or "is proportional to" means \(y = kx\)
2. "varies inversely" means \(y = \frac{k}{x}\)
3. Always find \(k\) first using given values
4. Check your answer makes sense (inverse: bigger x = smaller y)
5. Watch for "varies as the square" which means \(x^2\)

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