Inverse Proportion Worksheet 3 featuring exercises on direct and inverse proportion, including tables, equations, and graph sketches.

Inverse Proportion Worksheet 3 Solution

#### 1) Explain in words the difference between direct and inverse proportion.

- Direct Proportion: Two quantities \( x \) and \( y \) are directly proportional if an increase in one quantity causes a proportional increase in the other, and vice versa. Mathematically, this is expressed as:
\[
y \propto x \quad \text{or} \quad y = kx
\]
where \( k \) is a constant.

- Inverse Proportion: Two quantities \( x \) and \( y \) are inversely proportional if an increase in one quantity causes a proportional decrease in the other, and vice versa. Mathematically, this is expressed as:
\[
y \propto \frac{1}{x} \quad \text{or} \quad y = \frac{k}{x}
\]
where \( k \) is a constant.

---

#### 2) Fill in the tables below using the statements written above.

##### Table 1: \( y \propto \frac{1}{x} \)

Given:
\[
y = \frac{k}{x}
\]
We know that when \( x = 2 \), \( y = 8 \). So,
\[
8 = \frac{k}{2} \implies k = 8 \times 2 = 16
\]
Thus, the equation is:
\[
y = \frac{16}{x}
\]

Now, fill in the table:
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -6 & -4 & -2 & 2 & 6 \\
\hline
y & \frac{16}{-6} = -\frac{8}{3} & \frac{16}{-4} = -4 & \frac{16}{-2} = -8 & \frac{16}{2} = 8 & \frac{16}{6} = \frac{8}{3} \\
\hline
\end{array}
\]

##### Table 2: \( y \propto \frac{1}{x^2} \)

Given:
\[
y = \frac{k}{x^2}
\]
We know that when \( x = 2 \), \( y = 8 \). So,
\[
8 = \frac{k}{2^2} \implies 8 = \frac{k}{4} \implies k = 8 \times 4 = 32
\]
Thus, the equation is:
\[
y = \frac{32}{x^2}
\]

Now, fill in the table:
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -6 & -4 & -2 & 2 & 6 \\
\hline
y & \frac{32}{(-6)^2} = \frac{32}{36} = \frac{8}{9} & \frac{32}{(-4)^2} = \frac{32}{16} = 2 & \frac{32}{(-2)^2} = \frac{32}{4} = 8 & \frac{32}{2^2} = \frac{32}{4} = 8 & \frac{32}{6^2} = \frac{32}{36} = \frac{8}{9} \\
\hline
\end{array}
\]

##### Table 3: \( y \propto \frac{1}{x^3} \)

Given:
\[
y = \frac{k}{x^3}
\]
We know that when \( x = 2 \), \( y = 8 \). So,
\[
8 = \frac{k}{2^3} \implies 8 = \frac{k}{8} \implies k = 8 \times 8 = 64
\]
Thus, the equation is:
\[
y = \frac{64}{x^3}
\]

Now, fill in the table:
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -6 & -4 & -2 & 2 & 6 \\
\hline
y & \frac{64}{(-6)^3} = \frac{64}{-216} = -\frac{8}{27} & \frac{64}{(-4)^3} = \frac{64}{-64} = -1 & \frac{64}{(-2)^3} = \frac{64}{-8} = -8 & \frac{64}{2^3} = \frac{64}{8} = 8 & \frac{64}{6^3} = \frac{64}{216} = \frac{8}{27} \\
\hline
\end{array}
\]

##### Table 4: \( y \propto \frac{1}{\sqrt{x}} \)

Given:
\[
y = \frac{k}{\sqrt{x}}
\]
We know that when \( x = 2 \), \( y = 8 \). So,
\[
8 = \frac{k}{\sqrt{2}} \implies k = 8 \sqrt{2}
\]
Thus, the equation is:
\[
y = \frac{8 \sqrt{2}}{\sqrt{x}}
\]

Now, fill in the table:
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 0.5 & 1 & 2 & 4 & 6 \\
\hline
y & \frac{8 \sqrt{2}}{\sqrt{0.5}} = \frac{8 \sqrt{2}}{\frac{1}{\sqrt{2}}} = 8 \cdot 2 = 16 & \frac{8 \sqrt{2}}{\sqrt{1}} = 8 \sqrt{2} & \frac{8 \sqrt{2}}{\sqrt{2}} = 8 & \frac{8 \sqrt{2}}{\sqrt{4}} = \frac{8 \sqrt{2}}{2} = 4 \sqrt{2} & \frac{8 \sqrt{2}}{\sqrt{6}} = \frac{8 \sqrt{2}}{\sqrt{6}} = \frac{8 \sqrt{12}}{6} = \frac{8 \cdot 2 \sqrt{3}}{6} = \frac{8 \sqrt{3}}{3} \\
\hline
\end{array}
\]

---

#### 3) The following statements describe the relationship between \( y \) and \( x \). Sketch a graph showing the relationship.

##### i) \( y \) is inversely proportional to the square of \( x \).
\[
y \propto \frac{1}{x^2} \quad \text{or} \quad y = \frac{k}{x^2}
\]
The graph is a hyperbola symmetric about the y-axis, approaching the x-axis as \( x \to \pm \infty \).

##### ii) \( y \) is inversely proportional to the square root of \( x \).
\[
y \propto \frac{1}{\sqrt{x}} \quad \text{or} \quad y = \frac{k}{\sqrt{x}}
\]
The graph is defined only for \( x > 0 \) and approaches the x-axis as \( x \to \infty \).

##### iii) \( y \) varies as the cube of \( x \).
\[
y \propto x^3 \quad \text{or} \quad y = kx^3
\]
The graph is a cubic function, passing through the origin and increasing rapidly as \( |x| \) increases.

---

#### 4) \( y \propto \frac{1}{\sqrt[3]{x}} \)

##### i) If \( y = 13 \) when \( x = 8 \), write an equation for \( y \) in terms of \( x \).

Given:
\[
y = \frac{k}{\sqrt[3]{x}}
\]
When \( x = 8 \) and \( y = 13 \):
\[
13 = \frac{k}{\sqrt[3]{8}} \implies 13 = \frac{k}{2} \implies k = 13 \times 2 = 26
\]
Thus, the equation is:
\[
y = \frac{26}{\sqrt[3]{x}}
\]

##### ii) Find \( y \) when \( x = -3.375 \).

Using the equation \( y = \frac{26}{\sqrt[3]{x}} \):
\[
y = \frac{26}{\sqrt[3]{-3.375}} = \frac{26}{-1.5} = -\frac{26}{1.5} = -\frac{52}{3} \approx -17.33
\]

##### iii) When \( y \) equals 200, what is the value of \( x \) to 3 sf.

Given \( y = 200 \):
\[
200 = \frac{26}{\sqrt[3]{x}} \implies \sqrt[3]{x} = \frac{26}{200} = 0.13 \implies x = (0.13)^3 = 0.002197 \approx 0.00220
\]

##### iv) When \( y \) is proportional to \( x \), if you double \( x \), you divide \( y \) by 2, true or false?

If \( y \) is proportional to \( x \), then:
\[
y \propto x \quad \text{or} \quad y = kx
\]
If \( x \) is doubled, then:
\[
y' = k(2x) = 2kx = 2y
\]
So, \( y \) is doubled, not divided by 2. Therefore, the statement is false.

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Final Answers:

1. Explanation: Direct proportion means \( y = kx \), and inverse proportion means \( y = \frac{k}{x} \).
2. Tables:
- Table 1: \(\boxed{-\frac{8}{3}, -4, -8, 8, \frac{8}{3}}\)
- Table 2: \(\boxed{\frac{8}{9}, 2, 8, 8, \frac{8}{9}}\)
- Table 3: \(\boxed{-\frac{8}{27}, -1, -8, 8, \frac{8}{27}}\)
- Table 4: \(\boxed{16, 8\sqrt{2}, 8, 4\sqrt{2}, \frac{8\sqrt{3}}{3}}\)
3. Graphs: Sketch the graphs as described.
4. Equation and Calculations:
- i) \( y = \frac{26}{\sqrt[3]{x}} \)
- ii) \( y \approx -17.3 \)
- iii) \( x \approx 0.00220 \)
- iv) False

\(\boxed{-\frac{8}{3}, -4, -8, 8, \frac{8}{3}, \frac{8}{9}, 2, 8, 8, \frac{8}{9}, -\frac{8}{27}, -1, -8, 8, \frac{8}{27}, 16, 8\sqrt{2}, 8, 4\sqrt{2}, \frac{8\sqrt{3}}{3}, -17.3, 0.00220, \text{False}}\)