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!
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.)
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.
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.
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: .y² .= .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
. . . . . . . . . . _ . . . . _
Answers: .(±√6/2, ±√2/2)