# Thread: [SOLVED] Horizontal tangents to a leminiscate

1. ## [SOLVED] Horizontal tangents to a leminiscate

Find all the points on the leminiscate where the tangent line is horizontal.

Working with the LaTex as image files is proving a little limiting, so hopefully I have the most salient steps included.
Ok, so the derivative of the equation is:

And since I want the horizontals, I want the values (dy/dx)=0. I figured that since 0/anything = 0, I just need the values of x and y that make the numerator 0.
So I set the numerator of the derivative equal to zero and solved for y:

But when I plug that into the original equation, it doesn't work -- I get 4 = 8. Also, just looking at the equations, it seems like (±1,±1) should work, but it makes the orignial eqation untrue as well. Any help is greatly appreciated!

2. Originally Posted by sinewave85
Find all the points on the leminiscate where the tangent line is horizontal.

Working with the LaTex as image files is proving a little limiting, so hopefully I have the most salient steps included.
Ok, so the derivative of the equation is:

And since I want the horizontals, I want the values (dy/dx)=0. I figured that since 0/anything = 0, I just need the values of x and y that make the numerator 0.
So I set the numerator of the derivative equal to zero and solved for y:

But when I plug that into the original equation, it doesn't work -- I get 4 = 8. Also, just looking at the equations, it seems like (±1,±1) should work, but it makes the orignial eqation untrue as well. Any help is greatly appreciated!
Did you try doing what the probable point of the assignment was and convert it to polar coordinates?

3. Originally Posted by Mathstud28
Did you try doing what the probable point of the assignment was and convert it to polar coordinates?
The assignment is on implicit differentiation -- I checked in the index to be sure, and polar coordinated are way ahead of where I am. I can't do it that way.

4. Originally Posted by sinewave85
The assignment is on implicit differentiation -- I checked in the index to be sure, and polar coordinated are way ahead of where I am. I can't do it that way.
you want the change in the x and y direction to be zero...so...

5. Originally Posted by Mathstud28
you want the change in the x and y direction to be zero...so...
Right, I need the (x,y) that make dy/dx = 0, but I am not quite sure how to solve

I tried solving the numerator of the differential set to 0 for y

and plugging that back into the original equation, but that didn't work -- the equation of the leminiscate comes out to 4 = 8, so obviously I went wrong somewhere.

A similar strategy worked for a different problem, so I don't understand what has gone wrong here. (In that problem, I was looking for the vertical tangents, so I set the denominator of the differential to 0 and went from there.)

6. Originally Posted by sinewave85
Right, I need the (x,y) that make dy/dx = 0, but I am not quite sure how to solve

I tried solving the numerator of the differential set to 0 for y

and plugging that back into the original equation, but that didn't work -- the equation of the leminiscate comes out to 4 = 8, so obviously I went wrong somewhere.

A similar strategy worked for a different problem, so I don't understand what has gone wrong here. (In that problem, I was looking for the vertical tangents, so I set the denominator of the differential to 0 and went from there.)
Think about what I said and think about what dy and dx stand for.

7. Originally Posted by Mathstud28
Think about what I said and think about what dy and dx stand for.
Ok, dy is Δy, or f(x + Δx) - f(x); dx is Δx. There may very well by something very obvious I am not seeing here. My brain is really fried -- long day.

EDIT: I think I am going to have to sleep on it and come back to it tomorrow. Maybe the answer will come to me in my dreams! Thanks for helping.

8. Originally Posted by sinewave85
Find all the points on the leminiscate where the tangent line is horizontal.

Working with the LaTex as image files is proving a little limiting, so hopefully I have the most salient steps included.
Ok, so the derivative of the equation is:

And since I want the horizontals, I want the values (dy/dx)=0. I figured that since 0/anything = 0, I just need the values of x and y that make the numerator 0.
So I set the numerator of the derivative equal to zero and solved for y:

But when I plug that into the original equation, it doesn't work -- I get 4 = 8. Also, just looking at the equations, it seems like (±1,±1) should work, but it makes the orignial eqation untrue as well. Any help is greatly appreciated!
Your equation for the lemniscate is wrong. The right hand side should be 4(x^2 - y^2).

By the way, it should be obvious that the equation is wrong: It has common factor of x^2 + y^2. Divide that common factor out and you're left with x^2 + y^2 = 4 which is a circle O not a lemniscate 00.

9. Originally Posted by mr fantastic
Your equation for the lemniscate is wrong. The right hand side should be 4(x^2 - y^2).

By the way, it should be obvious that the equation is wrong: It has common factor of x^2 + y^2. Divide that common factor out and you're left with x^2 + y^2 = 4 which is a circle O not a lemniscate 00.

Ok, sorry, that was just a typo. I have it correctly on paper -- the derivative is correct (I guess). I still don't understand what I am not doing right, though, as far as solving the rest of the problem.

10. Hello, sinewave85!

Find all the points on the leminiscate where the tangent line is horizontal.

Working with the LaTex as image files is proving a little limiting,
so hopefully I have the most salient steps included.
Ok, so the derivative of the equation is:

And since I want the horizontals, I want the values (dy/dx)=0.
I figured that since 0/anything = 0, I just need the values of x and y
that make the numerator 0.

So I set the numerator of the derivative equal to zero and solved for y:

But when I plug that into the original equation, it doesn't work
. . Yes, it does

We have: . .= .2 - x² . . x² + y² .= .2 .[1]

Substitute into the original equation: .(2)² .= .4[x² - (2 - x²)]

. . which simplifies to: .2x² = 3 . . x² = 3/2 .[2]
- . . . . . . . . . . ._
Hence: .x .= .±√6/2

Substitute [2] into [1]: .3/2 + y² .= .2 . . y² = 1/2
- . . . . . . . . . . ._
Hence: .y .= .±√2/2

. . . . . . . . . . _ . . . . _

11. Originally Posted by Soroban
Hello, sinewave85!

We have: . .= .2 - x² . . x² + y² .= .2 .[1]

Substitute into the original equation: .(2)² .= .4[x² - (2 - x²)]

. . which simplifies to: .2x² = 3 . . x² = 3/2 .[2]
- . . . . . . . . . . ._
Hence: .x .= .±√6/2

Substitute [2] into [1]: .3/2 + y² .= .2 . . y² = 1/2
- . . . . . . . . . . ._
Hence: .y .= .±√2/2

. . . . . . . . . . _ . . . . _