5.1 Direct Variation (Part 3)

Posted on May 28, 2020 by Myhometuition

(D) Solving problems involving direct variations

1. If

y \propto x^{n}, where n = \frac{1}{2}, 2, 3,

then the equation is

y = k x^{n}

where k is a constant.

2. The graph of y against xⁿ is a straight line passing through the origin.

3. If

y \propto x^{n}

and sufficient information is given, the values of variable x or variable y can be determined.

Example

y varies directly to x³and if y = 54 when x = 3, find

(a) The value of x when y = 16

(b) The value of y when x = 4

Solution:

Given y α x³, y = kx³

When y = 54, x = 3,

54 = k(3)³

54 = 27k

k = 2

Therefore y = 2x³

(a) When y = 16

16 = 2x³

x³ = 8

x = 2

(b) When x = 4

y = 2(4)³ = 128

where k is a constant.

5.4 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

Question 5:
It is given that R varies directly as the square root of S and inversely as the square of T. Find the relation between R, S and T.

Solution:

R α \frac{\sqrt{S}}{T^{2}}

Question 6:

It is given that P varies directly as the square of Q and inversely as the square root of R. Given that the constant is k, find the relation between P, Q and R.

Solution:

\begin{array}{l} P α \frac{Q^{2}}{\sqrt{R}} \\ P = \frac{k Q^{2}}{\sqrt{R}} \end{array}

Question 7:
Given that P varies inversely as the cube root of Q. The relationship between P and Q is

Solution:

\begin{array}{l} P α \frac{1}{\sqrt[3]{Q}} \\ P α \frac{1}{Q^{\frac{1}{3}}} \end{array}

Question 8:
Given that y varies inversely as the cube of x and y = 16 when x = ½. Express y in terms of x.

Solution:

\begin{array}{l} y α \frac{1}{x^{3}} \\ y = \frac{k}{x^{3}} \\ When y = 16, x = \frac{1}{2} \\ 16 = \frac{k}{{(\frac{1}{2})}^{3}} \\ 16 = \frac{k}{\frac{1}{8}} \\ k = 2 \\ y = \frac{2}{x^{3}} \end{array}

Question 9:
W varies directly with X and inversely with the square root of Y. Given that k is a constant, find the relation between W, X and Y.

Solution:

\begin{array}{l} W α \frac{X}{\sqrt{Y}} \\ W = \frac{k X}{\sqrt{Y}} \\ W = \frac{k X}{Y^{^{\frac{1}{2}}}} \end{array}

5.1 Direct Variation (Sample Questions)

Posted on April 24, 2020 by user

Example:

Given that p varies directly as square root of q and p = 12 when q = 36, find

(a) The value of p when q = 16

(b) The value of q when p = 18

Solution:

\begin{array}{l} p \propto \sqrt{q}, p = k \sqrt{q} \\ 12 = k \sqrt{36} \end{array}

12 = k(6)

k = 2

p = 2 \sqrt{q}

(a)

\begin{array}{l} p = 2 \sqrt{q} \\ p = 2 \sqrt{16} \end{array}

p = 8

(b)

\begin{array}{l} p = 2 \sqrt{q} \\ 18 = 2 \sqrt{q} \to 9 = \sqrt{q} \end{array}

9² = q
q = 81

5.1 Direct Variation (Part 1)

Posted on April 24, 2020 by user

(A) Determining whether a quantity varies directly as another quantity

1. If a quantity y varies directly as a quantity x, the

(a) y increases when x increases

(b) y decreases when x decreases

2. A quantity y varies directly as a quantity x if and only if

\frac{y}{x} = k

where k is called the constant of variation.

3. y varies directly as x is written as

y \propto x

.

4. When

y \propto x

, the graph of y against x is a straight line passing through the origin.

(B) Expressing a direct variation in the form of an equation involving two variables

Example 1

Given that y varies directly as x and y = 20 when x = 36 . Write the direct variation in the form of equation.

\begin{array}{l} y \propto x \\ y = k x \\ 20 = k (36) \\ k = \frac{20}{36} = \frac{4}{9} \to F i n d k f i r s t \\ ∴ y = \frac{4}{9} x \end{array}

5.3 Joint Variation

Posted on April 24, 2020 by user

5.3 Joint Variation

5.3a Representing a Joint Variation using the symbol ‘α’.

1. If one quantity is proportional to two or more other quantities, this relationship is known as joint variation.

2. ‘y varies directly as x and z’ is written as y α xz.

3. ‘y varies directly as x and inversely z’ is written as

y α \frac{x}{z} .

4. ‘y varies inversely as x and z’ is written as

y α \frac{1}{x z} .

Example 1:

State the relationship of each of the following variations using the symbol 'α'.

(a) x varies jointly as y and z.

(b) x varies inversely as y and

\sqrt{z} .

(c) x varies directly as r³ and inversely as y.

Solution:

\begin{array}{l} (a) x α y z \\ (b) x α \frac{1}{y \sqrt{z}} \\ (c) x α \frac{r^{3}}{y} \end{array}

5.3b Solving Problems involving Joint Variation

1. If

y α x^{n} z^{n}, then y = k x^{n} z^{n}

, where k is a constant and n = 2, 3 and ½.

2. If

y α \frac{1}{x^{n} z^{n}}, then y = \frac{1}{k x^{n} z^{n}}

, where k is a constant and n = 2, 3 and ½.

3. If

y α \frac{x^{n}}{z^{n}}, then y = \frac{k x^{n}}{z^{n}}

, where k is a constant and n = 2, 3 and ½.

Example 2:

Given that

p α \frac{1}{q^{2} \sqrt{r}}

when p = 4, q = 2 and r = 16, calculate the value of r when p = 9 and q = 4.

Solution:

\begin{array}{l} Given that p α \frac{1}{q^{2} \sqrt{r}}, p = \frac{k}{q^{2} \sqrt{r}} \\ When p = 4, q = 2 and r = 16, \\ 4 = \frac{k}{2^{2} \sqrt{16}} \\ 4 = \frac{k}{16} \\ k = 64 \\ ∴ p = \frac{64}{q^{2} \sqrt{r}} \\ When p = 9 and q = 4, \\ 9 = \frac{64}{4^{2} \sqrt{r}} \\ 9 = \frac{4}{\sqrt{r}} \\ \sqrt{r} = \frac{4}{9} \\ r = {(\frac{4}{9})}^{2} = \frac{16}{81} \end{array}

5.1 Direct Variation (Part 2)

Posted on April 24, 2020 by user

(C) Finding the value of a variable in a direct variation

1. When y varies directly as x and sufficient information is given, the value of y or x can be determined by using:

\begin{array}{l} (a) y = k x, or \\ (b) \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \end{array}

Example 2

Given that y varies directly as x and y = 24 when x = 8, find

(a) The equation relating y to x

(b) The value of y when x = 6

(c) The value of x when y = 36

Solution:

Method 1: Using y = kx

(a) y \propto x

y = kx

when y = 24, x = 8

24 = k (8)

k = 3

y = 3x

(b)
when x = 6,

y = 3 (6)

y = 18

(c)
when y = 36

36 = 3x

x =12

Method 2:

Using \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}}

(a)
Let x₁ = 8 and y₁= 24

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{y_{2}}{x_{2}} \\ \frac{3}{1} = \frac{y_{2}}{x_{2}} \to y_{2} = 3 x_{2} \\ ∴ y = 3 x \end{array}

(b)
Let x₁ = 8 and y₁= 24 and x₂= 6; find y₂.

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{y_{2}}{6} \\ y_{2} = \frac{24}{8} (6) \\ y_{2} = 18 \end{array}

(c)
Let x₁ = 8 and y₁= 24 and y₂= 36; find x₂.

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{36}{x_{2}} \\ 24 x_{2} = 36 \times 8 \\ x_{2} = 12 \end{array}

5.2 Inverse Variation

Posted on April 24, 2020 by user

5.2 Inverse Variation

If a quantity y varies inversely as another quantity x, then

(a) y increases when x decreases,

(b) y decreases when x increases

5.2b Expressing an inverse variation in the form of an equation

An inverse variation can be written in the form of an equation,

y = \frac{k}{x}

where k is a constant which can be determined.

Example 1:

Given y varies inversely as x and y = 4 when x =10. Write an equation which relates x and y.

Solution:

\begin{array}{l} y \propto \frac{1}{x} \to y = \frac{k}{x} \\ 4 = \frac{k}{10} \to k = 40 \\ ∴ y = \frac{40}{x} \end{array}

Example 2:

Given that y = 3 when x = 6, find the equation relates x and y if:

(a) y \propto \frac{1}{x^{2}} (b) y \propto \frac{1}{x^{3}} (c) y \propto \frac{1}{\sqrt{x}}

Solution:

\begin{array}{l} (a) y \propto \frac{1}{x^{2}} \to y = \frac{k}{x^{2}} \\ 3 = \frac{k}{6^{2}} \\ k = 3 \times 36 = 108 \\ ∴ y = \frac{108}{x^{2}} \\ (b) y \propto \frac{1}{x^{3}} \to y = \frac{k}{x^{3}} \\ 3 = \frac{k}{6^{3}} \\ k = 3 \times 216 = 648 \\ ∴ y = \frac{648}{x^{3}} \\ (c) y \propto \frac{1}{\sqrt{x}} \to y = \frac{k}{\sqrt{x}} \\ 3 = \frac{k}{\sqrt{6}} \\ k = 3 \times \sqrt{6} = 3 \sqrt{6} \\ ∴ y = \frac{3 \sqrt{6}}{\sqrt{x}} \end{array}

5.4 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

Question 1:

It is given that y varies directly as the cube of x and y = 192 when x = 4. Calculate the value of x when y = – 24.

Solution:

y α x³
y = kx³

192 = k (4)³

192 = 64 k

k = 3

y = 3 x³

when y = – 24

– 24 = 3 x³

x³ = – 8

x = – 2

Question 2:
It is given that y varies directly as the square of x and y = 9 when x = 2. Calculate the value of x when y = 16.

Solution:

y α x²
y = kx²
9 = k (2)²

\begin{array}{l} k = \frac{9}{4} \\ y = \frac{9}{4} x^{2} \\ When y = 16 \\ 16 = \frac{9}{4} x^{2} \\ x^{2} = \frac{64}{9} \\ x = \frac{8}{3} \end{array}

Question 3:
Given that y varies inversely as w and x and y = 45 when w = 2 and x =

\frac{1}{6}

. Find the value of x when y = 15 and w =

\frac{1}{3}

Solution:

\begin{array}{l} y α \frac{1}{w x} \\ y = \frac{k}{w x} \\ 45 = \frac{k}{(2) (\frac{1}{6})} \\ k = 45 \times \frac{1}{3} = 15 \\ y = \frac{15}{w x} \end{array}

\begin{array}{l} when y = 15, w = \frac{1}{3} \\ 15 = \frac{15}{(\frac{1}{3}) x} \\ \frac{x}{3} = 1 \\ x = 3 \end{array}

Question 4:
Given that

p α \frac{1}{q \sqrt{r}}

and p = 3 when q = 2 and r = 16, find the value of p when q = 3 and r = 4.

Solution:

\begin{array}{l} p α \frac{1}{q \sqrt{r}} \\ p = \frac{k}{q \sqrt{r}} \\ 3 = \frac{k}{(2) \sqrt{16}} \\ k = 24 \\ p = \frac{24}{q \sqrt{r}} \end{array}

\begin{array}{l} when q = 3, r = 4 \\ p = \frac{24}{3 \sqrt{4}} \\ p = \frac{24}{6} = 4 \end{array}