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Find the slope of the normal to y= 2x²-2 at the point x=-4.
Uh... what's a normal?
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the normal to a curve/surface is a line perpendicular to it and passing through the point chosen (as opposed to a tangent which is parellel)
so, y = 2x²-2, dy/dx = 4x, (x = -4) dy/dx = -16
now, you should know from geometry of straight lines, that two lines are perpendicular, when the product of their gradients is -1
so -1 / -16 = 1/16, and that is the gradient of the normal.
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Or just imagine y=2x[sup]2[/sup]-2 is a special case of g(x,y)=y-2x[sup]2[/sup]+2=0
find
(g'[sub]x[/sub],g'[sub]y[/sub]) at the given point (x,y)=(-4,30)
Last edited by George,Y (2007-01-26 16:54:33)
X'(y-Xβ)=0
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