# Equation of Tangent Line

1. May 24, 2013

### Jimbo57

1. The problem statement, all variables and given/known data
Give the equation of the line tangent to the curve at the given point.

ytan^-1x = xy at (sqrt3,0)

2. Relevant equations

3. The attempt at a solution

Would it be right to do an implicit differentiation or to isolate for y here?

I isolated for y and got y=1/(tan^-1x-x)

deriving: y' = x^2/((1+x^2)(tan^-1x-x))

That's where I'm confused, at the initial differentiation.

Any guidance would be greatly appreciated!
Jim

2. May 24, 2013

### LCKurtz

There is something funny about your equation. Are you sure it is copied correctly? If $y\ne 0$ you have $\arctan(x) = x$ and if $y=0$, then $x$ can be anything.

3. May 24, 2013

### Jimbo57

Yep, it's copied correctly.

y*tan^-1(x) = x*y

4. May 24, 2013

### Jimbo57

LCKurtz, when I plug this into Wolfram Alpha, I get a straight vertical line... There is no tangent line for a straight vertical line am I right?

5. May 24, 2013

### LCKurtz

Look at what I observed earlier. If $y=0$, then $x$ can be anything. That is the $x$ axis, not the $y$ axis.

6. May 24, 2013

### Jimbo57

Hmm, I am a little confused. To my uneducated eye, if y= 0, doesn't that just make x=0 as well?

0*arctanx = x*0
0=0

Is this wrong?

7. May 24, 2013

### LCKurtz

Please quote the post to which you are replying, like I just did. No it doesn't "make x = 0". When $y=0$ it doesn't tell you anything about x because it gives 0=0 no matter what x is. Every value of $x$ works if $y=0$.

8. May 24, 2013

### Jimbo57

Oh sorry about the quote thing!

I see what you're saying. Do you think there is a typo? I haven't encountered a problem like this and I doubt it's a "thinker" question if it's throwing a retired math professor a curve ball hah.

9. May 24, 2013

### LCKurtz

Isn't that what I suggested in my first post? But given the question as posted, you can still answer it since the only function that equation represents is $y=0$.

10. May 24, 2013

### Skins

How did you isolate y ? Let's try it

Assume $$y \neq 0$$

We begin with

$$y arctan(x) \, = \, xy$$

Moving all the terms to the left hand side (subtracting xy)

$$yarctan(x) \, - \, xy \, = \, 0$$

Factoring out y on the left hand side..

$$y(arctan(x) \, - \, x) \, = \, 0$$

Dividing both sides by (arctan(x) - x)

$$y \, = \, \frac{0}{arctan(x) \, - \, x}$$

====>

$$y \, = \, 0$$

Of course you could have seen this by inspecting the original equation. If you divide both sides by y you get

$$arctan(x) \, - \, x \, = \, 0$$

11. May 24, 2013

### Jimbo57

Ah yes. In my original factoring process, I left a 1 on the right hand side for some reason. So now, thanks to the help of both you and LCKurtz, I have y = 0.

Can the question even be answered? For y= 0 there is no tangent line. Unless, that is the answer they want...

12. May 24, 2013

### LCKurtz

13. May 24, 2013

### Skins

Well, y=0 is a linear equation.We could write it as y = 0x, it's slope being 0. Geometrically its the x-axis as for any x, y = 0. So it's tangent line is y=0, in essence the "tangent line" of a linear function is the function itself, although it is not as intuitive to visualize geometrically as with many nonlinear functions.

I stumbled across this paper which is kind of interesting. If you like reading about mathematical stuff you may enjoy this.

14. May 24, 2013

### Staff: Mentor

Can you go back and double check, right back to the original source?

15. May 24, 2013

### Jimbo57

Yeah, I quadruple checked the original source. I'll probably speak with my tutor (it's a distance course) about it.

I really appreciate the input folks!