Who likes Calculus? :)
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Who likes Calculus? :)

[From: ] [author: ] [Date: 11-06-10] [Hit: ]
First, you could separate the variables so that you have all the y on one side, and all the x terms on the other and then differentiate, but this seems hard, so you could just differentiate it all (Im assuming you know how to differentiate terms with y in them? If not,......
At which point in the first quadrant is the tangent line vertical to the graph of x^3+y^3=6xy Thank you so much...trying to remember this stuff for the finals :O

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I assume by 'first quadrant' you mean in the x>0, y>0 part of the x-y plane.
Well, you want the tangent line, and you want it vertical, so you need to find at which points
dx/dy=infinity.
First, you could separate the variables so that you have all the y on one side, and all the x terms on the other and then differentiate, but this seems hard, so you could just differentiate it all (I'm assuming you know how to differentiate terms with y in them? If not, go by rule of thumb that the differential of a y^n is ny^(n-1)dy/dx I know the notation looks a bit nasty but see that it's exactly the same method as differentiating x terms, but you just bung a dy/dx on it too)
So differentiated:
3x^2+3y^2(dy/dx)=6x(dy/dx)+6y
Rearranged:
dy/dx=(6y-3x^2)/(6x-3y^2)
So where is this infinity? When the bottom equals 0!
So see where 2x=y^2
when y=0.5x^0.5!!
Now substitute this into the equation and solve!

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First of all this would be a very simple question to do but what is the function that you are trying to find a tangent line to? Beside calculus is fun, because it helps us deal with calculations in REAL life, not in the theoretical world.
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