MATH 1121 Calculus

Project #2

Rotated Conic Sections and Implicit Differentiation.

(This project actually sounds a lot more complicated than it is.  Work carefully through the steps, and you'll see that the actual math involved is not too difficult.)

Here's an applet demonstrating the general situation, just for you to play with:

Applet:

How do you find the points where the ellipse "just touches" the box?

On the top and on the bottom, the derivative must be zero, while on the left and right sides, the derivative is infinite (the denominator of the derivative function goes to zero.)

You've already done a few of these in the exercises for section 2.5.

Let's make life a little easier by putting the center of the ellipse at the origin, and looking at a fixed bounding rectangle, the one bounded by x = +/- 2 and y = +/- 1.

In this special case, h = k = 0, and C = 4.

Applet:

What your project will be is to determine the formula for finding "D" in the equation:

x^2 + B*x*y + 4*y^2 = D

as a function of "B."  In the applet, you're actually doing the work that I did to create the applet; that is, you're finding the restriction on "D" that keeps the ellipse just touching the box.

Your first step will be to find the derivative of y with respect to x  in the formula above, remembering that B is a constant, and D is a constant, while y is a function of x.

Solve this for dy/dx.  Your answer will have "B," "x," and "y" in it.  (Since the derivative of "D" is zero, it disappears.)

That is:

Next, you're going to need to look at the values that give you the horizontal and vertical tangents.  Remember that a rational function is zero if the top is zero, and it's undefined if the bottom is zero.  So, for the horizontal tangents, notice that these happen when the top of the fraction you get from dy/dx is zero.

Because you want these to happen when y = +/- 1, substitute 1 in for y in the equation, and solve for x.  (It turns out that you get exactly the same final answer if you use -1 for y, eventually, so we only need the case y = 1).

Now, go back to the original equation, and substitute in  for x, the value you got in part 2.

Once you've done this, solve the equation for D, so that you know how D must change as a function of B, in order for the edges of the ellipse to stay on the box.

Next, if you play around with the applet, sliding the values for "B," you'll see that if you move the slider too far, the shape "escapes" the box, and becomes an hyperbola, although the points of tangency stay on the lines.

Finally, we apparently never used the vertical tangents.  The truth is, that I already used those for you, in determining that if the ratio of the bounding box (horizontal to vertical width) is 2:1, then the constant "C" in front of y^2 is 4.  If you want, you can try the whole exercise again (you DON'T HAVE TO!!!!), using the fact that you want the denominator of the derivative to be zero when x = +/- 2, solve for y, substitute and see that you get exactly the same relationship between B and D.

If you're really interested in this, see if you can figure out how I made the first GeoGebra applet.  That is, you'll need to think about how, by sliding the corners of the rectangle around, you can find the center (h, k) -- that's pretty easy, the coefficient "C," and a formula for D in terms of B and C.  It's not especially difficult, but it is somewhat messy, and you'll need to use both the horizontal and the vertical tangents.  You'll also want simplify your work by looking at the case where (h,k) is the origin; then, simply translate your final results.

Page Last Modified: 13 February, 2008

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