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