Expert: Paul Klarreich - 11/21/2005

Question
1) The graph of the eight curve, x^4=a^2(x^2-y^2), a doesn't = 0.

Determine the points on the curve where the tangent line is horizontal.

2) The graph of the pear-shaped quartic, b^2y^2=x^3(a-x), a,b>0.

Determine the points on the curve where the tangent line is horizontal.

Answer
Hi, Diana,

I know it is a bit late, but:

A reader has discovered an error in my computations, and the second part should be:

The equation was:

b^2y^2=x^3(a-x)
OR:
b^2 y^2 = ax^3 - x^4

Implicit differentiation gives:

2b^2 y dy/dx = 3ax^2 - 4x^3
       3ax^2 - 4x^3
dy/dx = -------------
        2b^2 y

Set the top = 0:

3ax^2 - 4x^3 = 0
x^2(3a - 4x) = 0

x = 0  and  x = 3a/4

If x = 0, y = 0, and dy/dx is undefined.

At 3a/4,

b^2 y^2 = 27a^3/64 (a - 3a/4)

b^2 y^2 = 27a^3/64 (a/4)

b^2 y^2 = 27a^4/256

y^2 = 27a^4/256b^2

y = sqrt(3) 3a^2/16b

Sorry about that, but it is a good example of why I don't like to answer private questions.  It is embarrassing to have errors discovered, but that's life, and I can take a few hits.


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Hi, Diana,
Welcome to the calculus club.

Question:  1) The graph of the eight curve, x^4=a^2(x^2-y^2), a doesn't = 0.
Determine the points on the curve where the tangent line is horizontal.

Do you mean a FIGURE-EIGHT curve?  

You need implicit differentiation.  If you haven't done it before, then: (what I mean is, I haven't seen you before)

Differentiating implicit functions is actually fairly easy, although sometimes we don't like the form of the answer. You will have some EQUATION involving x and y, but the equation is NOT written like y = f(x), i.e. explicitly solved for y. The process is this:

1. DON'T solve for y. (Part of your grade depends 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 chain 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

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.)

OK, here we go with

x^4=a^2(x^2-y^2)

4x^3 = 2a^2 x - 2a^2 y dy/dx

2a^2 y dy/dx = 2a^2 x - 4x^3

dy   2a^2 x - 4x^3
-- = -------------
dx     2a^2 y

For a horizontal tangent line, we want dy/dx = 0, so the numerator must be zero.

2a^2 x - 4x^3 = 0
x(a^2 - 2x^2) = 0

x = 0,  x = +- a/sqrt(2)

If you want POINTS, then you will want to substitute into the original equation:
x^4=a^2(x^2-y^2)

x = 0  -->  0 = -2a^2y^2  --> y = 0
but at this point, the derivative is undefined.

x = +- a/sqrt(2) --> a^4/4 = a^2(a^2/2 - y^2)

a^2/4 = a^2/2 - y^2
y^2 = a^2/4  -->  y = +-a/2

So these are your points.

2) The graph of the pear-shaped quartic, b^2y^2=x^3(a-x), a,b>0.

2b^2 y dy/dx = 3ax^2 - 4x^3
         3ax^2 - 4x^3
dy/dx = --------------
           2b^2 y

Set the numerator = 0:

x^2(3a - 4x) = 0

As before, x = 0 implies y = 0, and dy/dx is undefined.

x = 3a/4

b^2 y^2 = 27a^3/4(a - 3a/4) = 27a^4/16
      27a^4
y^2 = --------
       16b^2

      3sqrt(3)a^2
y = +- -----------
         4b