Question

Find the point(s) on the curve of 4x^2+9y^2=36 at which the curvature is largest.

Answer 1

\[k(x)=\frac{ \left| y'' \right| }{ [1+(y')^2]^{3/2} }\]

Answer 2

8x+18y(dy/dx)=0
18y(dy/dx)=-8x
dy/dx=-8x/18y=-4x/9y
@SithsAndGiggles

Answer 3

Are you having trouble with the second derivative or something?

Answer 4

\[4x^2+9y^2=36~~\implies~~y=\pm\frac{1}{3}\sqrt{36-4x^2}\]
You can also consider the ellipse as a function of \(y\):
\[x=\pm\frac{3}{2}\sqrt{4-y^2}\]
and find the curvature
\[k(y)=\frac{|x''|}{(1+(x')^2)^{3/2}}\]
(which you might have to do anyway, judging by the shape of this ellipse).

Answer 5

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