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legacyofpiracy

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Code: 
Consider the curve defined by  (2y^3)+(6x^2)y-(12x^2)+6y=1

dy/dx= (4x-2xy)/((x^2)+(y^2)+1)

Write and equation of each horizontal tangent line to the curve

I am at a loss as to how to go about the second part of the problem (I first had to find the derivative) because generally I graph it on my calculator just to get an idea of where the tangents would lie...but this includes two variables so I am somewhat puzzled. Suggestions would be greatly appreciated
 
I'm stumped by this now?! I whittled it down to saying that
dy/dx=0 and so;
4x-2xy=0 which I factorised to;
2x(2-y)=0
so y=2 and x=0
But they don't work with the initial equations, so I've gone wrong somewhere.
All I can think of is partial derivatives but isn't that 3D equations?
 
Just tried it using partial derivatives, and you don't get a real solution, namely the stationary point is at (2, sqrt[-10]) :shock:
Which is wrong wrong wrong :lol:
Unless it doesn't have stationary points?
 
As you can probably see from your calculator, the curve approaches y=2 asymptotically, but there is no horizontal tangent to any point on the curve because of that asymptotic behaviour.

Edit: you can't see it on your calculator. I'll post a graph.

Here:

legacyofpiracy1.gif
 
Confusion!
I too came up with y=2 as the zero. I then tried replacing Y with a constant A and graphed it on TI-83.
Using Y=A=1.8 I got a downward opening parobola, vertex (.5,23.7).
Using Y=A=2 I got a horizontal line at y=27???
Using Y=A=2.2 I got an upward opening parobola, vertex (0,33.5).
Not sure what it means except there is a horizontal tangent? at y=2
------------------
Gene
 
legacyofpiracy said:
Code: 
Consider the curve defined by  (2y^3)+(6x^2)y-(12x^2)+6y=1

dy/dx= (4x-2xy)/((x^2)+(y^2)+1)

Write and equation of each horizontal tangent line to the curve

I am at a loss as to how to go about the second part of the problem (I first had to find the derivative) because generally I graph it on my calculator just to get an idea of where the tangents would lie...but this includes two variables so I am somewhat puzzled. Suggestions would be greatly appreciated
Click to expand...



Hi there.


dy/dx= (4x-2xy)/((x^2)+(y^2)+1)


Horizontal tangent means that dy/dx=0.


4x-2xy=0
2x(2-y)=0

x=0 or y=2


When x=0,

2y^3+6y=1; y=0.165


When y=2


8+12x^2-12x^2+12=1


The x's cancel, so there is no point on the curve with y coordinate of 2.


Thus, there is only one horizontal line tangent to the curve, and it's equation is y=0.165
 
Great stuff, Daniel. That is a local minimum on the graph.
 
:shock: wow thank you everyone for the effort! At least I wasn't the only one to get confused. I do get it now though, thank you especially Daniel!
 
As much as I would like to see this thing put to rest, I'm still at sea. If I graph 2a^3+6x^2a-12x^2+6a-1
with a=y=.165... I get a downward opening parabola with the apex at (0,0). Dan says the tangent is y=.165 which doesn't touch the parabola.
As a (or y) increases the parabola moves up and opens till it is a line at f(2) => y=27 then continues upward as an upward opening parabola with the apex at (0,2a^3+6a-1)
I thought the horizontal tangent would be the line at 2y^3+6y-1 which agrees with both x=0 and y=2 of dy/dx=0
I don't understand how it ties in with Maple's plot though so its probably wrong. Speaking of which, Unco, what does Maple do with -.55 < x < .55
----------------------
Gene
 
Thanks Unco, apologies Ted,
I'm persuaded :oops:
I still don't know why my making y=a=.165... a constant didn't work. Gotta think about that, but now I can sleep tonight. :roll:
------------------
Gene
 
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