Polar to Rectangular: convert cos(θ)sin(θ) – 2 = 0 to rectangular equation

[BigNate]

Polar to Rectangular: convert cos(θ)sin(θ) – 2 = 0 to rectangular equation

Can someone please help me determine the following question:

The equation cos(θ)sin(θ) – 2 = 0 in polar coordinates is equivalent to which equation in rectangular coordinates?

I know x = rcos(θ) and y = rsin(θ)
Likewise I know tan(θ) = y/x and r = +/- sqrt(x^2+y^2)

But this question is throwing me off. Any help is much appreciated! The - 2 in particular is what has me confused.

Thanks!

 

[Dr.Peterson]

Can someone please help me determine the following question:

The equation cos(θ)sin(θ) – 2 = 0 in polar coordinates is equivalent to which equation in rectangular coordinates?

I know x = rcos(θ) and y = rsin(θ)
Likewise I know tan(θ) = y/x and r = +/- sqrt(x^2+y^2)

But this question is throwing me off. Any help is much appreciated! The - 2 in particular is what has me confused.

Thanks!

The 2 is no problem. What is odd to me is that there is no r in the equation! You can just solve for theta, and the graph of the equation will just be a line or two through the origin (that is, certain values of theta, with r irrelevant). Hint: use a double-angle identity.

If you do want to treat it as a polar equation, just replace cos(θ) with x/r and sin(θ) with y/r, and eventually replace r^2 with x^2 + y^2. You'll eventually find what theta has to be. This is probably the long way; but it does give you practice with the polar-to-rectangular conversion process.

On the other hand, when I actually carried out the first method, I realized that the answer is very simple, and obvious if you think about the equation, and a little confusing. Where did this problem come from?

 

[Deleted member 4993]

Can someone please help me determine the following question:

The equation cos(θ)sin(θ) – 2 = 0 in polar coordinates is equivalent to which equation in rectangular coordinates?

I know x = rcos(θ) and y = rsin(θ)
Likewise I know tan(θ) = y/x and r = +/- sqrt(x^2+y^2)

But this question is throwing me off. Any help is much appreciated! The - 2 in particular is what has me confused.

Thanks!

I think you have made a typo in your post. That equation is NOT possible because |sin(Θ)|≤ 1 and cos(Θ)|≤1.

cos(θ)sin(θ) – 2 = 0 → cos(θ)sin(θ) =2 → sin(2Θ) = 4 → and that is not possible.....

 

[Dr.Peterson]

I think you have made a typo in your post. That equation is NOT possible because |sin(Θ)|≤ 1 and cos(Θ)|≤0.

cos(θ)sin(θ) – 2 = 0 → cos(θ)sin(θ) =2 → sin(2Θ) = 4 → and that is not possible.....

Yes, that is what I was hinting at when I said,

Hint: use a double-angle identity. ... I realized that the answer is very simple, and obvious if you think about the equation, and a little confusing. Where did this problem come from?

It may be a typo, or it may not even be intended as an equation to be converted to rectangular form. I suspect there is something in the context that would show a misinterpretation of some sort.

 

[BigNate]

Yes, that is what I was hinting at when I said,

It may be a typo, or it may not even be intended as an equation to be converted to rectangular form. I suspect there is something in the context that would show a misinterpretation of some sort.

I appreciate the help from both of you. I went back and looked at the problem and it is exactly as I typed it, so a poorly worded question perhaps.

The correct answer is 2x^2 + 2y^2 = xy.

I'm still not sure how this is the right answer. I see what you mean by suggesting I can replace cos(θ) with x/r and sin(θ) with y/r, but doing that does not lead me to the answer of 2x^2 + 2y^2 = xy. Any ideas?

Thanks again so much for your help!

 

[Dr.Peterson]

I appreciate the help from both of you. I went back and looked at the problem and it is exactly as I typed it, so a poorly worded question perhaps.

The correct answer is 2x^2 + 2y^2 = xy.

I'm still not sure how this is the right answer. I see what you mean by suggesting I can replace cos(θ) with x/r and sin(θ) with y/r, but doing that does not lead me to the answer of 2x^2 + 2y^2 = xy. Any ideas?

Thanks again so much for your help!

Here's what I did:

cos(θ) sin(θ) = 2
x/r * y/r = 2
xy = 2r^2
xy = 2(x^2 + y^2)
2x^2 + 2y^2 = xy

How did you not get the same result, following my suggestion?

Note that since x^2 - 2xy + y^2 = (x - y)^2, we have x^2 + y^2 = (x - y)^2 + 2xy, and can rewrite the equation as
x^2 + y^2 = xy/2
(x - y)^2 + 2xy = xy/2
(x - y)^2 = -3xy/2

so x and y must have opposite signs so that both sides are positive; but similarly

(x + y)^2 - 2xy = xy/2
(x + y)^2 = 5xy/2

so that x and y must have the same sign! This is another indication that this is actually an empty relation, as Subhotosh pointed out. (There's probably a more direct way to see this from the rectangular equation, that I am missing at the moment.) The author of the problem probably didn't go farther than the conversion, and missed this fact.