Chapter 9: Q. 63 (page 732)

. Prove that the graph of the equation $r = k c o e s c θ$ , $0 < θ < π$ , is a horizontal line for any value of $k \neq 0$

Short Answer

Expert verified

Equation in rectangular form is $y = k$ , hence it represent a horizontal line which is at a distance of $k$ units from $x -$ axis.

Step by step solution

Step 1. Given information

The equation in polar form : $r = k \cos e c θ$

$0 < θ < π$

Step 2. Convert the polar form of equation into rectangular form:

Since in $y = r \sin θ$

$\sin θ = \frac{y}{r} \cos e c θ = \frac{r}{y}$

$r = k \cos e c θ r = k (\frac{r}{y}) y = k$

This is an equation of horizontal line .

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. Prove that the graph of the equation $r = k c o e s c θ$ , $0 < θ < π$ , is a horizontal line for any value of $k \neq 0$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Convert the polar form of equation into rectangular form:

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. Prove that the graph of the equation r=kcoescθ, 0<θ<π, is a horizontal line for any value of k≠0

Short Answer

Step by step solution

Step 1. Given information

Step 2. Convert the polar form of equation into rectangular form:

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. Prove that the graph of the equation $r = k c o e s c θ$ , $0 < θ < π$ , is a horizontal line for any value of $k \neq 0$