Chapter 9.4, Problem 13E

To determine

To write each equation in polar form and nearest angle in degree

Answer to Problem 13E

The polar form of $21x+20y=87$ is $3=r\cos \left[\theta -43.6°\right]$ and nearest of angle $\phi \approx 43.6°$ in degree

Explanation of Solution

Given information

The given equation is $21x+20y=87$

Formula used: $p=r\cos \left[\theta -\phi \right]$ .

Calculation:

The standard form of the equation is $21x+20y-87=0$ .

Firstly. write the equation in normal form to find the values of $p$ and $\phi$ .To convert to normal form, we need to find the value of $\pm \sqrt{A^2+B^2}$ .

  $\begin{array}{l}\pm \sqrt{A^2+B^2}= \pm \sqrt{{\left(21\right)}^2+{\left(20\right)}^2}\left[A=21,B=20\right] \\ =\pm \sqrt{441+400}=\pm \sqrt{841}=\pm 29\end{array}$

Since $C$ is negative, use $+29$ .The normal form of the equation is $\frac{21}{\left(29\right)}x+\frac{20}{\left(29\right)}y-3=0$ . $x\cos \phi +y\sin \phi -p=0$ We can see from the normal form that $p=3,\cos \phi =\frac{21}{29}\text{ and}\sin \phi =\frac{20}{29}$ .Since $\cos \phi \text{and}\sin \phi$ are both positive, the normal lies in the first quadrant.

  $\begin{array}{l}\tan \phi =\frac{\sin \phi }{\cos \phi }=\frac{\frac{20}{29}}{\frac{21}{29}}=\frac{20}{21}\\ \phi ={\tan }^{\left(-1\right)}\frac{20}{21}\\ \phi \approx 43.6°\end{array}$ [using the arctangent function].

Substitute the values for $p$ and $\phi$ into the polar form.

  $p=r\cos \left[\theta -\phi \right]\to 3=r\cos \left[\theta -43.6°\right]$

The polar form of $21x+20y=87$ is $3=r\cos \left[\theta -43.6°\right]$