Monday, May 14, 2012

Cat Ear Parabolas

So, I promise this isn't going to turn into a cat blog.  I mean, I am obsessed with my cat, I ain't gonna lie, but I have a modicum of control.

However, I was thinking recently about how Copernicus' ears look like 3-D conic sections.



And then I thought about the parabola that is the front outline of his ear, and I decided, since I am done with school and can do random fun things like this now, that it had to be done.

So I took a picture of his ear.
And then I traced it in Adobe Illustrator (badly), rotated the drawing so that the vertex was at the top, and overlaid a graph.  The points that I found were based on the 8-by grid that was visible for editing on Illustrator.  I placed it so that my graph could have 2 x-intercepts.  For the graph I'm going for, it doesn't matter where on the coordinate system it lies.  It's about the shape.

Then, I looked up how to find the equation of a parabola from points on the graph because I couldn't remember it.  Turns out you need the vertex and one other point for the simplest process.

For a vertically symmetrical parabola, you start with y=a(x-h)^2 + k, where (h, k) is the vertex.

My vertex was (0, 25), so I started with that:  y=a(x-0)^2 + 25, i.e., y=ax^2 + 25.  Then input another point and solve for a.  I chose (-8, 16), because I was fairly sure of my sketch at that point on the graph.

16=a[(-8)^2] + 25 . . . 16= 64a + 25 . . . -9 = 64a . . . a = -0.138461538

So my equation for that view of Buddy's ear is


 and the graph of that function is ...



 Which, if you imagine some angling and perspective isn't bad.

I really want to try one with a better head-on picture.  Getting Bud to sit still for that, however, is the hard part.













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