How Do You Determine the Polar Equation of a Line Perpendicular to a Given Ray?

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Homework Help Overview

The discussion revolves around determining the polar equation of a line that is perpendicular to a given ray, specifically the ray defined by the angle ##\theta= \frac {5*\pi}{7}##. The line's shortest distance to the origin is specified as 5, and the task involves finding the polar equation ##r=r(\theta)## along with the angles ##\theta_1## and ##\theta_2## that define the interval for which the polar equation is valid.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the y-intercept in the context of the line's equation and suggest alternative methods to derive the polar equation. There is an emphasis on visualizing the problem by drawing the ray and the line, and some participants propose a simpler expression for ##r(\theta)##.

Discussion Status

Several participants have confirmed similar results regarding the polar equation and have expressed gratitude for the insights shared. There is an ongoing inquiry about the specific values of ##\theta_1## and ##\theta_2##, with one participant indicating that they found these values easily.

Contextual Notes

Participants are operating under the constraints of the problem as stated, focusing on deriving the polar equation without providing complete solutions. The discussion includes checking assumptions related to the calculations and the geometric interpretation of the problem.

Poetria
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Homework Statement


[/B]
Consider the line whose shortest distance to the origin is 5 and that is perpendicular to the ray ##\theta= \frac {5*\pi}{7}## for r>0
Find its polar equation ##r=r(\theta)## and ##\theta_1<\theta_2## in the interval ##[0, 2\pi]## such that ##r(\theta)\geq 0## for all ##\theta_1\leq\theta\leq\theta_2## as ##\theta## increases from ##\theta_1## to ##\theta_2##, the point ##(r(\theta), \theta)## traces the entire line once.

2. The attempt at a solution

m*x+b=y
##m=\tan(3/14*\pi)##
b=5/cos(2/7*pi)

$$-\frac{\frac 5 {\cos(\frac 2 7*\pi)}}{(\tan( \frac 3 {14}*\pi)*\cos(\theta)-\sin(\theta))}$$
 

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You have calculated b incorrectly. If you correct this, then your final answer will be right - but it's a long way round and a complicated expression. Try doing it without going through y = mx + b. Draw the ray, the line, and a point (r,θ) on the line. Can you derive a simpler expression for r(θ)?
 
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mjc123 said:
You have calculated b incorrectly. If you correct this, then your final answer will be right - but it's a long way round and a complicated expression. Try doing it without going through y = mx + b. Draw the ray, the line, and a point (r,θ) on the line. Can you derive a simpler expression for r(θ)?

Something like:

##r=5*sec(\theta-\frac 5 7*\pi)##
 
That's what I got.
 
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mjc123 said:
That's what I got.

Great. :) Many thanks. :)
 
Poetria said:
Great. :) Many thanks. :)
Did you find ##\ \theta _1\ ## and ##\ \theta _2\ ## ?
 
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SammyS said:
Did you find ##\ \theta _1\ ## and ##\ \theta _2\ ## ?

Yes, this was easy:
$$\theta_1=\frac 3 {14}\pi$$
$$\theta_2=\frac {17}{14}\pi$$
 

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