# Equation of tangent to curve

1. Jan 18, 2012

### ArcanaNoir

1. The problem statement, all variables and given/known data
"Find the equation of the plane tangent to the surface $(x^2-y^2)(x^2+y^2)=15$ at the point $(2,1)$"

If only it really were a plane and a surface, I could do that. I have a formula for that. Unfortunately, this is a curve and I'm looking for tangent line.

2. Relevant equations

In three dimensions, the formula for the equation of the tangent plane to the surface z=f(x,y) at the point $P(x_0,y_0,z_0)$ is $z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

where $f_a$ is the partial derivative of f wrt a.

3. The attempt at a solution

Well, pretending it's in three variables, I can do
$$f_x=4x^3$$
$$f_y=-4y^3$$

$$z-z_0=f_x(2,1)(x-2)+f_y(2,1)(y-1)$$
$$z-z_0=32(x-2)-4(y-1)$$

So, how do I repair this situation/make the formula work in two dimensions/try something else?

2. Jan 18, 2012

### I like Serena

I'm afraid you equation is not of the form z=f(x,y).
If you want to compare it with a 3D case, it's of the form g(x,y,z)=0.

Do you have other methods available?

3. Jan 18, 2012

### ArcanaNoir

I can not thank you enough for all your help! I really need to start paying you ;)

I can use anything through vector calculus. Something about the gradient? I think I might be supposed to take the gradient and then evaluate at the point. Does that sound reasonable?

4. Jan 18, 2012

### I like Serena

:)

But what would you get?

5. Jan 18, 2012

### ArcanaNoir

well, the gradient is $$4x^3i-4y^3j$$ isn't it?

6. Jan 18, 2012

### I like Serena

Yes, it is.

7. Jan 18, 2012

### ArcanaNoir

Well how do I evaluate that at a point? Plug in the x and y values?
If so, I get 32i-4j

8. Jan 18, 2012

### I like Serena

Yes.
This is a vector.
Can you say anything about its direction and how it relates to the curve?

9. Jan 18, 2012

### ArcanaNoir

Is the gradient normal to the curve perhaps? I can't find it in my calc book.

10. Jan 18, 2012

### I like Serena

Yes, it is.

11. Jan 18, 2012

### ArcanaNoir

Okay so now I just need a line normal to the normal.

How do I make a vector that is perpendicular to 32i-4j ?

12. Jan 18, 2012

### I like Serena

You don't need to.

Suppose you have a line given by h(x,y)=ax+by-c=0.

EDIT: to answer your question, a perpendicular vector would have a dot product that comes out as zero.

13. Jan 18, 2012

### ArcanaNoir

But, if a gradient is normal to the curve, don't I need to find the line perpendicular to the gradient? oh.. wait, for the equation of a line, do I use a point and a normal vector? I think I do... *looking up*

[EDIT] no no, I see you use a parallel vector. So, what do I do now?

[EDIT] now I'm reading about normal vectors and equations of lines and checking my notes....

14. Jan 18, 2012

### I like Serena

A line is also a curve.

If you have 2 curves that both have a normal vector that point in the same direction, wouldn't they have to be tangential to each other?

15. Jan 18, 2012

### ArcanaNoir

I'm confused. I have my curve and I have my normal. I don't have another curve.

16. Jan 18, 2012

### I like Serena

The parametric representation of a curve (or a surface), has a (partial) derivative that yields a vector tangential to the curve (or surface).

The equation of a curve (or surface) has a gradient that is normal to the curve (or surface).

17. Jan 18, 2012

### I like Serena

The line given by by h(x,y)=ax+by-c=0 is also a curve.
Its gradient is normal to the line.

18. Jan 18, 2012

### ArcanaNoir

So are you saying I need to parametrize instead and take the partial derivative?

19. Jan 18, 2012

### ArcanaNoir

I'm going to call it quits for tonight, I still have to iron clothes, and rest! I'll pick this back up tomorrow. Thank you so much for your time tonight!

20. Jan 18, 2012

### I like Serena

Okay. Thanks for telling me.
See you later.