Implicit Differentiation: Finding Points on an Ellipse Tangent to a Box

[ACLerok]

My TA did not get the chance to go over this problem. I know that I'm supposed to differentiate both sides of the equation. But I have not the slightest idea what to do after that. I was told that I am supposed to get out 4 points that lie on the ellipse and the sides of the box are tangent to the ellipse. Also, the horizontal lines have a slope of 0 but as for the vertical lines, i don't know. So how should I solve this problem? after differentiating, am i supposed to solve for y' or y? Any hints or tips? Thanks a lot!

http://www.eden.rutgers.edu/~cjjacob/images/workshop6.gif

 

[Hurkyl]

What is the equation of a vertical line? What does its derivative look like?

 

[ACLerok]

Originally posted by Hurkyl
What is the equation of a vertical line? What does its derivative look like?

is the equation of a vert. line x= something? is the derivative undef?

 

[cookiemonster]

Yes. dy/dx = undefined corresponds to a vertical line. Remember that slope is rise over run. A vertical line has infinite rise and zero run, so m = (infinity)/0 = undefined.

cookiemonster

 

[Hurkyl]

s the equation of a vert. line x= something? is the derivative undef?

The equation is right. The assertion is not.

If we have the equation x = a (for some constant a), then if we differentiate we get x' = 0.


(This problem is much easier if your differentiations are with respect to some unnamed third variable, instead of differentiating with respect to x)


edit: fixed my quoting problems.

 

[ACLerok]

so let me get this straight. in order to do this problem, you differentiate both sides of the equation with respect to x and y is a functin of x. then you solve for y'=0 for the vert tangent lines of the ellipse and y'= something else for the horizontal tangent lines? can you help guide me through this please? this thing is due tomorrow!

 

[Hurkyl]

What -I- would do (essentially) is differentiate with respect to some unnamed third variable.

Then, since horizontal lines are y = a (for some a), I'm looking for y' = 0 to get the horizontal ones. Since vertical lines are x = b (for some b), I'm looking for x' = 0 to get vertical ones.

Cookiemonster was suggesting you look for y' = undefined.


(note, though, if x' and y' are both 0, we have a singularity and can't tell what the corresponding tangent line, if any, would be)

 

[HallsofIvy]

I think Hurkyl's response could be a little misleading. The equation x= 1 says that x is a constant and so, thinking of x as a function of some other variable, the derivative of x with respect to that other variable is 0. In this case, we are looking for a point on the graph at which the figure has a vertical tangent. A vertical line has undefined slope and so the derivative of y with respect to x must be undefined.

In this particular case, the equation is x²- xy+ y²= 3.

"Implicit differentiation" gives 2x- y+ xy'+ 2yy'= 0 or
y'(x+2y)= y- 2x. The derivative will be 0 when y- 2x= 0 or y= 2x.
Since x and y satisfy x²- xy+ y²= 3, replacing x with y/2, this becomes y²/4- 2y²/4+ y²= 3²/4= 3 and so y= 2 or -2. When y= 2, x²- xy+ y²= x²- 2x+ 4= 3 or x²-2y+ 1= (x-1)²= 0 so x= 1. When y= -2, x²- xy+ y²= x²+ 2x+ 4= 3 or x²+ 2x+ 1= (x+1)²= 0 so x= -1. The horizontal tangents are at (1, 2) and (-1, -2), and so the top and bottom of the rectangle are y= 2 and y= -2 respectively.

y' is undefined when x+2y= 0. Because of the symmetry, it should be clear that the two sides of the rectangle are at x= 2 and x= -2.

 

[ACLerok]

Originally posted by HallsofIvy
y' is undefined when x+2y= 0. Because of the symmetry, it should be clear that the two sides of the rectangle are at x= 2 and x= -2.

can you just explain and justify this statement. I understand the entire problem except this one line kind of confuses just a wee bit.. Thanks to all of you guys!

 

[HallsofIvy]

We got, by implicit differentiation: y'(x+2y)= y- 2x.
Then y'= (y-2x)/(x+2y) which does not exist if the denominator
x+2y= 0. My remark about "symmetry" was based on the fact that we had just shown that when y'= 0, y- 2x= 0 gives y= 2 and -2 as upper and lower bounds.

The original equation was x²- xy+ y²= 3 which is "symmetric" in x and y. Clearly, using x+ 2y= 0 instead of y- 2x= 0 will give exactly the same results but with -x in place of y.

If that is not clear to you, go ahead an substitute 2y in place of x in the equation x²- xy+ y²= 4y²- 2y²+ y²= 3y²= 3 gives y= +/- 1 just as before we had x= +/- 1. Putting y= +1 into x+2y= 0 gives x= -2, putting y= -1 into x+ 2y= 0 gives x= 2. The bounding rectangle is the square bounded by x= 2, x= -2, y= 2, and y= -2.