Polar Coordinate Investigation Using Wolfram Alpha (help needed see bottom)

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In my summer school PreCalculus class today we continued our discussion of Polar Coordinates.  I began the class by asking the students to draw the graph of:

r = 2sin(θ)+2 by hand.  Their gaph should have looked like this:

      

We discussed the process of translating between rectangular coordinates and polar coordinates when a student asked “can we just transform the entire function?”  This led to a geeky, yet exciting, hour discussion and investigation.

First, we simplified the problem by dropping the “+2” from r = 2sin(θ)+2 (don’t worry we’ll go back to it…).  So we algebraically converted the equation r = 2sin(θ) into rectangular coordinates (x and y).  After substituting the conversions in (y=rsin(θ) and r2 =x2+y2), the algebra cleans up nicely by completing the square and a circle with center (0,1) and radius of 1 emerges.  After half the class had already plugged the polar coordinate equation into their calculator, we all agreed that this process worked for r=2sinθ, but what about r = 2sin(θ)+2?

The algebra was messy at best (plus I erased the final x,y equation before I was able to plug it into wolfram alpha), but here wass our equation:

Here’s the input into Wolfram Alpha: graph y=2sin^2(aCos(x/(sqrt(x^2+y^2))))+2sin(aCos((x/(sqrt(x^2+y^2))))

Anyway, here was the graph Wolfram Alpha spit out:

Where’s the bottom!!!!

So, in our equation, to find the angle of the polar coordinate in terms of x and y, we had to use inverse cosine, whose range is only zero to pi.  In order to get the bottom, we needed angles from pi to 2pi.  To find that, we had to shift our cosine by pi; that is, we had to add pi to each of the angles:

Wolfram Alpha input: graph y=2sin^2(aCos(x/(sqrt(x^2+y^2)))+pi)+2sin(aCos(x/(sqrt(x^2+y^2)))+pi)

Wolfram Alpha output:

Yay, the bottom.

But how do we get both?  A simple google search (remember this is the middle of class and my computer’s monitor is projected to the entire class, so a little bit of a gamble… good thing Wolfram Alpha isn’t a subsidary of GoDaddy!) landed me on this blog: Gower’s.  Somewhere in the comments, I found that a simple comma would allow my to graph both equations together, so i input the following into Wolfram Alpha:

graph y=2sin^2(aCos(x/(sqrt(x^2+y^2))))+2sin(aCos(x/(sqrt(x^2+y^2)))), y=2sin^2(aCos(x/(sqrt(x^2+y^2)))+pi)+2sin(aCos(x/(sqrt(x^2+y^2)))+pi)

Wolfram Alpha thought for awhile, and returned nothing.

PLEASE HELP US if you know how to get wolfram alpha to graph two intense equations. (note: I can graph two simple graphs:  input:   graph y=2x,y=x^2)

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2 thoughts on “Polar Coordinate Investigation Using Wolfram Alpha (help needed see bottom)

    kid said:
    August 17, 2011 at 2:28 pm

    to get a graph you must initialize the a parameter.

    Adam Lavallee responded:
    November 20, 2013 at 2:08 pm

    Reblogged this on With Respect to x.

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