Friday, December 30, 2022

Normal Form Example

It is necessary to practice how to give appropriate signs for sine and cosine terms in the normal form. We have seen an example in section 10.8. A similar example is given below:

Equation of a line is 3x - 5y - 15 = 0. Reduce this equation to normal form. Find the values of p and 𝜔.
Solution:
1. From the given equation, we get:
A = 3, B = -5 and C = -15
2. Calculation of p

$\begin{array}{ll}
{}&{p}
&{}={}& {\pm \frac{C}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{-15}{\sqrt{3^2~+~(-5)^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{15}{\sqrt{34}}}
&{} \\

\end{array}$

2. Calculation of cos 𝜔

$\begin{array}{ll}
{}&{\cos \omega }
&{}={}& {\pm \frac{A}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{3}{\sqrt{3^2~+~(-5)^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{3}{\sqrt{34}}}
&{} \\

\end{array}$

3. Calculation of sin 𝜔

$\begin{array}{ll}
{}&{\sin \omega }
&{}={}& {\pm \frac{B}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{-5}{\sqrt{3^2~+~(-5)^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{5}{\sqrt{34}}}
&{} \\

\end{array}$

4. Now we determine the signs:
(i) Slope of the line = $m=-\frac{A}{B}~=~-\frac{3}{-5}~=~\frac{3}{5}$
• This is +ve. So we move along the left arrow of the chart in fig.13.36.
• The chart is shown again below:


(ii) x-intercept of the line = $a=-\frac{C}{A}~=~-\frac{-15}{3}~=~5$
• This is +ve. So sin 𝜔 is -ve and cos 𝜔 is +ve.

5. So we can write:

$\begin{array}{ll}
{}&{\sin \omega }
&{}={}& {- \frac{5}{\sqrt{35}}}
&{} \\

{}&{\cos \omega}
&{}={}& {\frac{3}{\sqrt{35}}}
&{} \\

\end{array}$

6.  The sign of p will be always +ve because, it is a distance.

7. Now we can write the normal form:

$\begin{array}{ll}
{}&{x \cos \omega~+~y \sin \omega}
&{}={}& {p}
&{} \\

{\Rightarrow}&{x × \frac{3}{\sqrt{35}}~+~y × - \frac{5}{\sqrt{35}}}
&{}={}& {\frac{15}{\sqrt{35}}}
&{} \\

{\Rightarrow}&{ \frac{3x}{\sqrt{35}}~+~ - \frac{5y}{\sqrt{35}}}
&{}={}& {\frac{15}{\sqrt{35}}}
&{} \\

{\Rightarrow}&{\frac{3x}{\sqrt{35}}~-~ \frac{5y}{\sqrt{35}}}
&{}={}& {\frac{15}{\sqrt{35}}}
&{} \\

\end{array}$

8. Calculation of 𝜔:
(i) From (5), we have: tan 𝜔 = -5/3
• Using a scientific calculator, we get: 𝜔 = -59.04o
(ii) In our present case, 𝜔 is in the fourth quadrant (sin is -ve and cos is +ve).
• -59.04o indeed lies in the fourth quadrant. But we want the +ve angle. The +ve angle corresponding to -59.04o is (360 - 59.04) = 300.96o
• We get: tan 𝜔 = tan -59.04 = tan 300.96
(iii) So the value of 𝜔 is 300.96o

9. So the normal form can be written as:
x cos 300.96o + y sin 300.96o = $\frac{15}{\sqrt{35}}$

• The actual plot is shown below:


 

 

No comments:

Post a Comment