Equation of Tangent Line

  • Thread starter Jimbo57
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  • #1
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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
 

Answers and Replies

  • #2
LCKurtz
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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
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Yep, it's copied correctly.

y*tan^-1(x) = x*y
 
  • #4
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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
LCKurtz
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Look at what I observed earlier. If ##y=0##, then ##x## can be anything. That is the ##x## axis, not the ##y## axis.
 
  • #6
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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
LCKurtz
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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##.
 
  • #8
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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.
 
  • #9
LCKurtz
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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##.
 
  • #10
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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 [tex]y \neq 0[/tex]

We begin with

[tex] y arctan(x) \, = \, xy [/tex]

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

[tex] yarctan(x) \, - \, xy \, = \, 0 [/tex]

Factoring out y on the left hand side..

[tex] y(arctan(x) \, - \, x) \, = \, 0 [/tex]

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

[tex] y \, = \, \frac{0}{arctan(x) \, - \, x} [/tex]

====>

[tex] y \, = \, 0 [/tex]

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

[tex]arctan(x) \, - \, x \, = \, 0 [/tex]
 
  • #11
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How did you isolate y ? Let's try it

Assume [tex]y \neq 0[/tex]

We begin with

[tex] y arctan(x) \, = \, xy [/tex]

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

[tex] yarctan(x) \, - \, xy \, = \, 0 [/tex]

Factoring out y on the left hand side..

[tex] y(arctan(x) \, - \, x) \, = \, 0 [/tex]

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

[tex] y \, = \, \frac{0}{arctan(x) \, - \, x} [/tex]

====>

[tex] y \, = \, 0 [/tex]

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

[tex]arctan(x) \, - \, x \, = \, 0 [/tex]

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
LCKurtz
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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?
 
  • #13
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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
 
  • #14
NascentOxygen
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Yep, it's copied correctly.
Can you go back and double check, right back to the original source?
 
  • #15
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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!
 

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