MHB Convert equation 8x=8y to polar form

AI Thread Summary
The discussion revolves around converting the equation 8x = 8y into polar form. Initial attempts involved using the polar coordinates x = r*cos(θ) and y = r*sin(θ), but these were deemed incorrect. The correct approach involves dividing by r, leading to the equation cos(θ) = sin(θ), which simplifies to tan(θ) = 1. Despite trying values for θ, the user struggled to find an acceptable polar equation format. The final suggestion was to consult the professor for clarification before the assignment deadline.
Elissa89
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Convert the equation to polar form

8x=8y

I thought it would be

8*r*cos(theta)=8*r*sin(theta)

Said it was incorrect

then I thought I needed to divide by 8 to remove it, giving me:

r*cos(theta)=r*sin(theta)

But that was also incorrect and now I am stuck
 
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Elissa89 said:
Convert the equation to polar form

8x=8y

I thought it would be

8*r*cos(theta)=8*r*sin(theta)

Said it was incorrect

then I thought I needed to divide by 8 to remove it, giving me:

r*cos(theta)=r*sin(theta)

But that was also incorrect and now I am stuck
Either there's a typo somewhere or this is too simple a problem.

Your difficulty is that you stopped too soon. You can also divide by r (assuming that it's not zero). Thus you have the equation: [math]cos( \theta ) = sin( \theta )[/math]. There are several possible values for [math]\theta[/math].

-Dan
 
topsquark said:
Either there's a typo somewhere or this is too simple a problem.

Your difficulty is that you stopped too soon. You can also divide by r (assuming that it's not zero). Thus you have the equation: [math]cos( \theta ) = sin( \theta )[/math]. There are several possible values for [math]\theta[/math].

-Dan
Nope, no typo. I also tried cos(theta)=sin(theta). Then I thought I need to get the r alone and I got r=r*tan(theta)
 
Elissa89 said:
Nope, no typo. I also tried cos(theta)=sin(theta). Then I thought I need to get the r alone and I got r=r*tan(theta)

If you divided through by \(r\), you have:

$$\tan(\theta)=1$$

What does this imply for \(\theta\)?
 
MarkFL said:
If you divided through by \(r\), you have:

$$\tan(\theta)=1$$

What does this imply for \(\theta\)?

I tried inputting pi/4 and 5pi/4, but it doesn't want an answer, it wants the question converted to a polar equation.
 
Elissa89 said:
I tried inputting pi/4 and 5pi/4, but it doesn't want an answer, it wants the question converted to a polar equation.

Any Cartesian line of the form:

$$y=ax$$

will correspond to a polar equation of the form:

$$\tan(\theta)=a$$

or:

$$\theta=\arctan(a)+k\pi$$

Only a line not passing through the origin will have a polar equation involving both \(r\) and \(\theta\).

Did you try inputting:

$$\tan(\theta)=1$$ ?
 
MarkFL said:
Any Cartesian line of the form:

$$y=ax$$

will correspond to a polar equation of the form:

$$\tan(\theta)=a$$

or:

$$\theta=\arctan(a)+k\pi$$

Only a line not passing through the origin will have a polar equation involving both \(r\) and \(\theta\).

Did you try inputting:

$$\tan(\theta)=1$$ ?

Yes, it didn't take it
 
Elissa89 said:
Yes, it didn't take it

Perhaps it wants:

$$\theta=\frac{\pi}{4}(4k+1)$$
 
MarkFL said:
Perhaps it wants:

$$\theta=\frac{\pi}{4}(4k+1)$$

It still didn't take it
 
  • #10
Elissa89 said:
It still didn't take it

At this point, I would recommend you speak to the professor, and let him/her know what you've done.
 
  • #11
MarkFL said:
At this point, I would recommend you speak to the professor, and let him/her know what you've done.

Yeah I shot him a message, I just hope he gets back to me in time, assignment is due tonight at midnight
 
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