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# Principle domain of a polar equation?

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# Principle domain of a polar equation?

August 4, 2010 6:50 PM Subscribe

I need help understanding the "principle domain" (a term only my professor seems to use) of a polar function. That is, how do I find the smallest value delta such that [0,delta] plots all of the unique points on the curve; any values greater than delta re-trace points. I suspect that if MeFi could identify the more common name for this, uh, procedure, I'd be set.

Ah, yes, probably is. Still no luck googling though.

This is for a Calculus II class, if it's relevant.

posted by phrontist at 6:58 PM on August 4, 2010

This is for a Calculus II class, if it's relevant.

posted by phrontist at 6:58 PM on August 4, 2010

Well, if you imagine a picture of an arbitrary differentiable function, a portion of the function which is one-to-one (has only one of any particular y-value) is one in which the derivative of the function does not change sign. As soon as the derivative passes through zero, a maximum or minimum occurs, which means the function will start repeating y-values.

So, to find a domain where the function is one-to-one, you could take the derivative, and select a region between two zero-crossings of the derivative which spans the entire range of the function. For instance, for sin(x)

d(sin(x))/dx = cos(x)

cos(x) has zeros at x = -pi/2 and pi/2

Between those values, sin(x) spans the whole range of sin(x) (from -1 to 1)

Therefore the domain [-pi/2, pi/2] is the domain you're looking for - I guess you could call it a "principal domain".

posted by Salvor Hardin at 7:04 PM on August 4, 2010

So, to find a domain where the function is one-to-one, you could take the derivative, and select a region between two zero-crossings of the derivative which spans the entire range of the function. For instance, for sin(x)

d(sin(x))/dx = cos(x)

cos(x) has zeros at x = -pi/2 and pi/2

Between those values, sin(x) spans the whole range of sin(x) (from -1 to 1)

Therefore the domain [-pi/2, pi/2] is the domain you're looking for - I guess you could call it a "principal domain".

posted by Salvor Hardin at 7:04 PM on August 4, 2010

[0, 2*pi] wouldn't fulfill the uniqueness criteria, since sin(x) repeats y-values for that domain.

posted by Salvor Hardin at 7:05 PM on August 4, 2010

posted by Salvor Hardin at 7:05 PM on August 4, 2010

Salvor has a more general approach, though I'll point out that technically one end of the interval should be open, e.g., [-π/2, π/2).

posted by jedicus at 7:13 PM on August 4, 2010

posted by jedicus at 7:13 PM on August 4, 2010

The original poster seems to be talking about

In general I don't know of a non-tedious procedure for doing this. You could potentially 'take the easy way out' and graph it with a graphing device or software and determine a domain for theta in that way. But mainly all I can think of is to graph it yourself with pencil and paper: for different values of theta, you get different values of r to plot -- at some point you start retracing the curve. And certainly determining the periodicity of your function is helpful. And sometimes, if you can convert your polar function to rectangular (e.g., 1/(sin(theta) + cos(theta)) is much easier to work with in cartesian coordinates) that can help with the graphing process.

posted by evinrude at 7:43 PM on August 4, 2010

*polar*functions, e.g. r= sin(theta), for which the 'principal domain' would indeed be [0, 2*pi); or r = cos(2*theta), in which case we cover all the unique points on the polar curve in the domain [0, pi). If I'm wrong, disregard the next paragraph (which is unhelpful anyway).In general I don't know of a non-tedious procedure for doing this. You could potentially 'take the easy way out' and graph it with a graphing device or software and determine a domain for theta in that way. But mainly all I can think of is to graph it yourself with pencil and paper: for different values of theta, you get different values of r to plot -- at some point you start retracing the curve. And certainly determining the periodicity of your function is helpful. And sometimes, if you can convert your polar function to rectangular (e.g., 1/(sin(theta) + cos(theta)) is much easier to work with in cartesian coordinates) that can help with the graphing process.

posted by evinrude at 7:43 PM on August 4, 2010

Uh, wait -- the principal domain of r = sin(theta) would just be [0, pi) after all -- as theta goes from pi to 2*pi, we traverse the circle a second time. And the principal domain of r=cos(2*theta) is [0, pi/2). (And I only knew that by actually sketching the graph in my head, not by following some general procedure -- sorry)

posted by evinrude at 7:46 PM on August 4, 2010

posted by evinrude at 7:46 PM on August 4, 2010

Look in a complex analysis book. What your prof is trying to get at is also known as the principle value of a complex valued function (which, as you may know, can be visualized as a polar function by using the euler formula, r * exp(i*phi) =r*( cos(phi) + i * sin(phi)))

see: Principal Value from Wolfram Mathworld

posted by scalespace at 7:47 PM on August 4, 2010 [1 favorite]

see: Principal Value from Wolfram Mathworld

posted by scalespace at 7:47 PM on August 4, 2010 [1 favorite]

See p. 71 of this text (PDF), also this doc, p. 1 here, here.

Or in fact look at all these.

In short, I think "principal domain" is in fact a very common name for this concept--part of the trouble is it also seems to be a very common phrase so when you google it you find tons of non-math related results unless you also add some other math-related search terms (trig, polar coordinates, etc.).

posted by flug at 11:06 PM on August 4, 2010

Or in fact look at all these.

In short, I think "principal domain" is in fact a very common name for this concept--part of the trouble is it also seems to be a very common phrase so when you google it you find tons of non-math related results unless you also add some other math-related search terms (trig, polar coordinates, etc.).

posted by flug at 11:06 PM on August 4, 2010

This thread is closed to new comments.

posted by box at 6:56 PM on August 4, 2010