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Slope of a tangent line
- Thread starter xsever
- Start date
D
Deleted member 4993
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Consider this;
h = 4xy(x^2 - y^2) + 1
x^2 + y^2 = 1
\(\displaystyle h \, = \, 4x\sqrt{1 \, - \, x^2}\cdot (2x^2 \, - \, 1) \, + \, 1\)
Now what is it that you really meant by slope of a tangent line?
h = 4xy(x^2 - y^2) + 1
x^2 + y^2 = 1
\(\displaystyle h \, = \, 4x\sqrt{1 \, - \, x^2}\cdot (2x^2 \, - \, 1) \, + \, 1\)
Now what is it that you really meant by slope of a tangent line?
This is confusing to me also because at any point (x,y,h) there could be more than one tangent line (many tangent lines in different planes) so I'm trying to figure this one out. I am starting to think that I should get the equation of a plane that is tangent to a point (x,y,h) and all the lines in that plane which pass by the point of tangency would be tangent to that the curve at that point.
Or I might start with the assumption that the tangent line lies in a plane that is parallel to the x-y plane and try to get that first.
Or I might start with the assumption that the tangent line lies in a plane that is parallel to the x-y plane and try to get that first.