Reduce Non-Linear Function To Linear Function – Examples (G) To (L)

Posted on April 22, 2020 by user

Examples:
Reduce each of the following equations to the linear form. Hence, state the gradient and the Y-intercept of the linear equations in terms of a and b.

(g)

$k x^{2} + t y^{2} = x$

(h)

$y = \frac{x}{p + q x}$

(i)

$h y = x + \frac{k}{x}$

(j)

$y = a b^{x}$

(k)

$y = a x^{b}$

(l)

$y = a b^{x + 1}$

[Note :
X and Y cannot have constant, but can have the variables (for example x and y)
m and c can only have the constant (for example a and b), cannot have the variables x and y]

Solution:

Gradient Of A Straight Line

Posted on April 22, 2020 by user

(B) Gradient of a Straight Line
If the points

$A (x_{1}, y_{1})$ and

$B (x_{2}, y_{2})$ lie on the straight line

$y = m x + c$ , then gradient of ,the straight line

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} o r \frac{y_{1} - y_{2}}{x_{1} - x_{2}}$

SPM Practice 3 (Linear Law) – Question 3

Posted on April 22, 2020 by user

Question 3
The table below shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation

$\frac{a}{y} = \frac{b}{x} + 1$ , where k and p are constants.

(a) Based on the table above, construct a table for the values of

$\frac{1}{x}$ and

$\frac{1}{y}$ . Plot

$\frac{1}{y}$ against

$\frac{1}{x}$ , using a scale of 2 cm to 0.1 unit on the

$\frac{1}{x}$ - axis and 2 cm to 0.2 unit on the

$\frac{1}{y}$ - axis. Hence, draw the line of best fit.
(b) Use the graph from (b) to find the value of
(i) a,
(ii) b.

Solution
Step 1 : Construct a table consisting X and Y.

Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Steps to draw line of best fit - Click here

Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph

Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required

Steps To Draw The Line Of Best Fit

Posted on April 22, 2020 by user

Steps to draw a line of Best Fit
(i) Select suitable scales for the x-axis and the y-axis, make sure the points plotted accurately and the graph produced is large enough on the graph paper,
(ii) Mark the points correctly,
(iii) Use a long and transparent ruler to draw the line of best fit.

Step 1 : Select the suitable scale on x and y axis
(the graph produced must be more than 50% of the graph paper)

Step 2 : Mark the points correctly

Step 3 : Draw the Line of Best Fit

* Note

-the line passes through four points

-one point is above the line

-one point is below the line

SPM Practice 2 (Question 1 – 3)

Posted on April 22, 2020 by user

Question 1:
Reduce non-linear relation, $y = p x^{n - 1}$ , where k and n are constants, to linear equation. State the gradient and vertical intercept for the linear equation obtained.
[Note : Reduce No-linear function to linear function]

Solution:

Question 2:
The diagram shows a line of best fit by plotting a graph of $y^{2}$ against $\sqrt{x}$ .

Find the equation of the line of best fit.
Determine the value of
1. x when y = 4,
2. y when x = 25.

Solution:

Question 3:
The diagram shows part of the straight line graph obtained by plotting $\sqrt{y}$ against $x^{2}$ .

Express y in terms of x.

Solution:

SPM Practice 3 (Linear Law) – Question 1

Posted on April 22, 2020 by user

Question 1 (10 marks):
Use a graph to answer this question.
Table 1 shows the values of two variables, x and y, obtained from an experiment.
The variables x and y are related by the equation

$y - \sqrt{h} = \frac{h k}{x}$ , where h and k are constants.

(a) Plot xy against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the xy-axis.
Hence, draw the line of best fit.

(b) Using the graph in 9(a), find
(i) the value of h and of k,
(ii) the correct value of y if one of the values of y has been wrongly recorded during the experiment.

Solution:
(a)

(b)

$\begin{array}{l} y - \sqrt{h} = \frac{h k}{x} \\ x y - \sqrt{h} x = h k \\ x y = \sqrt{h} x + h k \\ Y = m X + C \\ Y = x y, m = \sqrt{h}, C = h k \end{array}$

(b)(i)

$\begin{array}{l} m = \frac{36.5}{5.1} \\ \sqrt{h} = \frac{36.5}{5.1} \\ \sqrt{h} = 7.157 \\ h = 51.22 \\ C = - 4 \\ h k = - 4 \\ k = \frac{- 4}{h} \\ k = - \frac{4}{51.22} \\ k = - 0.0781 \end{array}$

(b)(ii)

$\begin{array}{l} x y = 21 \\ 3.5 y = 21 \\ y = \frac{21}{3.5} = 6.0 \\ Correct value of y is 6 .0 . \end{array}$

Tips To Reduce Non-Linear Function To Linear Function

Posted on April 22, 2020 by user

Tips:
(1) The equation must have one constant (without x and y).
(2) X and Y cannot have constant, but can have the variables (for example x and y).
(3) m and c can only have the constant (for example a and b), cannot have the variables x and y.

Examples
(1)
X and Y cannot have constant, but can have the variables (for example x and y)

(2)
m and c can only have the constant (for example a and b), cannot have the variables x and y

SPM Practice 2 (Linear Law) – Question 4

Posted on April 22, 2020 by user

Question 4
The table below shows the corresponding values of two variables, x and y, that are related by the equation

$y = q x + \frac{p}{q x}$ , where p and q are constants.

One of the values of y is incorrectly recorded.
(a) Using scale of 2 cm to 5 units on the both axis, plot the graph of xy against

$x^{2}$ . Hence, draw the line of best fit

(b) Use your graph in (a) to answer the following questions:
(i) State the values of y which is incorrectly recorded and determine its actual value.
(ii) Find the value of p and of q.

Solution
Step 1 : Construct a table consisting X and Y.

Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Steps to draw line of best fit - Click here

(b) (i) State the values of y which is incorrectly recorded and determine its actual value.

Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph

Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required

Distance Between Two Points

Posted on April 22, 2020 by user

Distance between point $A (x_{1}, y_{1})$ and point is $B (x_{2}, y_{2})$ given by

$\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

SPM Practice 3 (Linear Law) – Question 2

Posted on April 22, 2020 by user

Question 2 (10 marks):
Use a graph to answer this question.
Table shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of

$\frac{y^{2}}{x} against \frac{1}{x}$ is plotted.

(a) Based on Table, construct a table for the values of

$\frac{1}{x} and \frac{y^{2}}{x} .$

$\begin{array}{l} (b) Plot \frac{y^{2}}{x} against \frac{1}{x}, using a scale of 2 cm to 0 .1 unit on the \frac{1}{x} -axis \\ and 2cm to 2 units on the \frac{y^{2}}{x} -axis . \\ Hence, draw the line of best fit . \end{array}$

(c) Using the graph in 11(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.

Solution:
(a)

(b)

(c)(i)

$\begin{array}{l} When x = 2.7, \frac{1}{x} = 0.37 \\ From graph, \\ \frac{y^{2}}{x} = 5.2 \\ \frac{y^{2}}{2.7} = 5.2 \\ y = 3.75 \end{array}$

(c)(ii)

$\begin{array}{l} Form graph, y -intercept, c = –4 \\ gradient, m = \frac{16 - (- 4)}{0.8 - 0} = 25 \\ Y = m X + c \\ \frac{y^{2}}{x} = 25 (\frac{1}{x}) - 4 \\ y = \sqrt{25 - 4 x} \end{array}$