Convert equation 8x=8y to polar form

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SUMMARY

The equation 8x=8y can be converted to polar form by first simplifying to r*cos(θ) = r*sin(θ). Dividing both sides by r (assuming r is not zero) leads to the equation cos(θ) = sin(θ), which simplifies to tan(θ) = 1. The polar form of the line represented by this equation is θ = π/4 + kπ, where k is any integer. This conversion is essential for understanding the relationship between Cartesian and polar coordinates.

PREREQUISITES
  • Understanding of polar coordinates and their relationship to Cartesian coordinates
  • Familiarity with trigonometric identities and equations
  • Basic algebraic manipulation skills
  • Knowledge of the tangent function and its properties
NEXT STEPS
  • Study the derivation of polar equations from Cartesian lines
  • Learn about the implications of dividing by r in polar equations
  • Explore the concept of periodicity in trigonometric functions
  • Investigate the graphical representation of polar equations
USEFUL FOR

Students studying mathematics, particularly those focusing on polar coordinates, trigonometry, and algebraic equations. This discussion is beneficial for anyone needing to convert Cartesian equations to polar form.

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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