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### Function problem

```
Date: Wed, 7 Dec 1994 14:48:43 -0500 (EST)
From: Aklilu Habte - ASNG/F94
Subject: Help, Math Problem

Hi...there,

I have been working on the problem written below
and am not able to solve it, need help.

Q- Find the slope of the tangent line for the function
y2 = x2 - x4 at (0, 0), also explain what is going on
at the origin.

NB: The function read as " y SQUARE = x SQUARE - x RAISE 4"

I appreciate your help, thank you.

Aklilu Habte.
Ryerson Polytechnic University.
Aero-Space Engineering.
```

```
Date: Thu, 8 Dec 1994 12:28:46 -0500 (EST)
From: Dr. Ken
Subject: Re: Help, Math Problem

Hello there!

This problem is one that makes use of the concept of implicit
differentiation.  I'll go ahead and use the concept, and please write back
if you don't understand what I've done.

The standard way to write your equation is y^2 = x^2 - x^4.  Note that this
is not really a function in the strictest sense of the word, since you get
two values of y for almost every value of x you plug in.

So we'll take the derivative with respect to x of both sides:  we use the
chain rule on the left side, and it spits out a dy/dx.

2*y*dy/dx = 2*x - 4*x^3.

So solving for dy/dx (the derivative of y with respect to x, i.e. the slope of
the tangent line), we get

dy/dx = (x - 2x^3)/y

This is the form in which most mathematicians would leave the answer.  It's
kind of a strange answer, since it has both x's and y's in it, but it's the
best we can do.

So to find the value of dy/dx at (0,0), we'd like to just plug in (0,0) for
x and y.  But that would be really really bad, since then we'd divide by
zero.  So we'll look at the limit as (x,y) approaches (0,0) instead.

Ordinarily, we'd solve for y in the original equation, and plug that value
into the derivative equation, and then take the limit.  I love to take
limits.  The tricky part here, though, is that we can't really solve for y,
since there are two values for y.  But we can get around that by using both
of the values

y = Sqrt{x^2 - x^4}

and

y = -Sqrt{x^2 - x^4}.

So substituting into the derivative equation, we see that

dy/dx  = +- (x - 2x^3)/Sqrt{x^2 - x^4}

Where the +- means "positive or negative."  Note at this point that our only
hope of having a well-defined tangent line is if its slope is zero, since
that's the only time the limit as x->0 will exist.  This makes sense since
the graph of our original equation is symmetric about the y-axis.

So I'll leave it to you to verify that this limit as x->0 is indeed zero.
Here's a hint: try squaring (if the square of something goes to zero, the
something must go to zero too).

Good luck!  Please write back if you have more questions!

-Ken "Dr." Math
```

```
Date: Thu, 8 Dec 1994 14:06:04 -0500 (EST)
From: Aklilu Habte - ASNG/F94
Subject: Re: Help, Math Problem

Hello Dr.

I have tried in the same way too. However, when I set up, the
"limit" still didn't make any sense to me. Your hint: "Try squaring
(if the square of something goes to zero, the something must go to
zero too)." Can you be a little more specific, please!

I also tried to approach by graphing the equation, and at the origin
(0,0) the slope of the tangent line seems to be "vertical", i.e. the
slope is undefined! How does this sound to you?

Aklilu  Habte. "Student"
```

```
Date: 9 Dec 1994 15:41:46 GMT
From: Dr. Math
Subject: Re: Help, Math Problem

Hello there!

When we last spoke, we had gotten to the formula

dy/dx  = +- (x - 2x^3)/Sqrt{x^2 - x^4}.

Now we want to take the limit as x->0.  I must apologize for misleading
you about what to expect this limit to be.  I think I suggested that it
was going to be zero, and in fact it's not.  But let's find out what it is.

Notice that we can pull an x^2 out of the square root sign.  So we can
cancel an x with the top, and we get

dy/dx  = +- (1 - 2x^2)/Sqrt{1 - x^2}.

If we let x go to zero, then we see that this limit approaches either 1 or
-1, depending on what direction we come from.  So in answer to the
original question, there are _two_ tangent lines at the origin, one that
has slope 1, and one that has slope -1.

If you plot the picture, you should get something that looks like a figure
eight turned on its side, something that looks kind of like an infinity
sign.  The figure eight has two lines that cross at the origin, one that
has slope 1, and one that has slope -1.

I hope this helps you!  Please write back again if you need more help.
```
Associated Topics:
College Calculus

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