Expert: Paul Klarreich - 10/27/2005

Question
Hi Paul,
For what value of x does the function 5x^2+2xy+y^2=16 have a vertical tangent line?
a. x=2 only
b. x=2 and x=0 only
c. x=0, x=-1, and x=-2
d. x=2 and x=-2 only
e. never
The answer is d, but don't have any idea how to get that.(I would think the answer is "e" because a vertical tangent line's slope is undefined).
Please explain.
Thanks again.I really appreciate it.

Answer
Hi, Jeff,
(SORRY THE TEXT GOT MESSED UP BEFORE.)

In your question:

For what value of x does the function (you mean the graph of the equation) 5x^2+2xy+y^2=16 have a vertical tangent line?
a. x=2 only
b. x=2 and x=0 only
c. x=0, x=-1, and x=-2
d. x=2 and x=-2 only
e. never
The answer is d, but don't have any idea how to get that.(I would think the answer is "e" because a vertical tangent line's slope  is undefined).
Please explain.

You are correct in that you want to find points on the graph where the slope is undefined; so you want to find the derivative and find where IT is undefined.

You could proceed by:

A. solving the equation for  y in terms of x.  Since the equation is a quadratic and has no obvious factorization, you would have to use the quadratic formula.  (Not my favorite way of doing it.)

B. Using implicit differentiation.  The answer will involve y, so you will need the y for each of your values of x.  There will be a bit of work involved.

Starting by differentiating  5x^2 + 2xy + y^2 = 16, using
implicit differentiation.  Recall that the process is:

------- TAKEN FROM AN ANSWER SENT TO SOMEONE ELSE ------
1. DON'T solve for y. (Part of your grade will depend on how well you don't do this.)

2. Differentiate each SIDE of the equation, using Whatever rules apply -- product, quotient, etc. AND the chain rule, assuming that y is some function of x, even though we can't see it.

2A. When you see the 'y' appear, you use the appropriate rule like this:

    D(y) = dy/dx
    D(y^2) = 2y dy/dx
    D(sqrt(y)) = D(y^1/2) = 1/2y^-1/2 dy/dx
    D(xy) = x dy/dx + y

etc. In other words, every term that involves y will produce an instance of dy/dx. If it doesn't you made a mistake.

3. Solve algebraically for dy/dx. (You can write y', but a little bit of sloppiness here can cost you.)

------------- YOUR EXAMPLE ---------------
10x + 2x dy/dx + 2y + 2y dy/dx = 0
        2x dy/dx   + 2y dy/dx = -10x - 2y
         x dy/dx   +  y dy/dx = -5x - y

         dy/dx = (-5x - y)/(x + y)

Now this will be undefined when the denominator is zero; that is, for any point where x = -y.  So we can test the points, if we like, or see a clever way to do it at the end.

x = 2:  20 + 4y + y^2 = 16
       y^2 + 4y + 4 = 0
       y = -2
YES, THIS IS SUCH A POINT.


x = 0:  y^2 = 16
       y = +- 4
NO, THEY ARE NOT

x = -2: 20 - 4y + y^2 = 16
       y^2 - 4y + 4 = 0
       y = 2
YES, IT IS.

x = -1: 5 - 2y + y^2 = 16
       y^2 - 2y     = 11
       y^2 - 2y + 1 = 12
        (y - 1)^2   = 12
            y = 1 +- 2sqrt(3)
These would be hard to compute, but the values are certainly not equal to +1, so NO, THEY ARE NOT.

So x= 2, and x = -2 are the points with vertical tangents.

=========== ANOTHER WAY ================
Start with your equation:

5x^2 + 2xy + y^2 = 16

and put  y = -x

5x^2 - 2x^2 + x^2 = 16
4x^2 = 16
x^2 = 4
x = +-2

That's it.