Equation of Tangent Line for y = arctan(x) at (sqrt3,0)

[Jimbo57]

Homework Statement

Give the equation of the line tangent to the curve at the given point.

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

Homework Equations

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

 

[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.

 

[Jimbo57]

Yep, it's copied correctly.

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

 

[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?

 

[LCKurtz]

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

 

[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?

 

[LCKurtz]

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?

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 $y=0$.

 

[Jimbo57]

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 $y=0$.

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.

 

[LCKurtz]

I see what you're saying. Do you think there is a typo?

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 $y=0$.

 

[Skins]

Homework Statement

Give the equation of the line tangent to the curve at the given point.

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

Homework Equations

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

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

 

[Jimbo57]

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

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...

 

[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 $y=0$.

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

Didn't I already answer that? Do straight lines have tangent lines?

 

[Skins]

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...

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.

http://www.google.com/url?sa=t&rct=j&q=tangent+line+to+a+linear+function&source=web&cd=4&ved=0CEkQFjAD&url=http%3A%2F%2Fmathdl.maa.org%2Fimages%2Fupload_library%2F22%2FPolya%2F07468342.di020721.02p01112.pdf&ei=4wOgUfftN8ry0gH-koCIBw&usg=AFQjCNG0Zz-29qqQn3j4KTXx1SbusB3jPA&bvm=bv.47008514,d.dmQ

 

[NascentOxygen]

Yep, it's copied correctly.

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

 

[Jimbo57]

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

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!