Convert equation 8x=8y to polar form

[Elissa89]

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

 

[topsquark]

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: \(\displaystyle cos( \theta ) = sin( \theta )\). There are several possible values for \(\displaystyle \theta\).

-Dan

 

[Elissa89]

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: \(\displaystyle cos( \theta ) = sin( \theta )\). There are several possible values for \(\displaystyle \theta\).

-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)

 

[MarkFL]

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:

\(\displaystyle \tan(\theta)=1\)

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

 

[Elissa89]

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

\(\displaystyle \tan(\theta)=1\)

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.

 

[MarkFL]

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:

\(\displaystyle y=ax\)

will correspond to a polar equation of the form:

\(\displaystyle \tan(\theta)=a\)

or:

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

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

Did you try inputting:

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

 

[Elissa89]

Any Cartesian line of the form:

\(\displaystyle y=ax\)

will correspond to a polar equation of the form:

\(\displaystyle \tan(\theta)=a\)

or:

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

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

Did you try inputting:

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

Yes, it didn't take it

 

[MarkFL]

Yes, it didn't take it

Perhaps it wants:

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

 

[Elissa89]

Perhaps it wants:

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

It still didn't take it

 

[MarkFL]

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.

 

[Elissa89]

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