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find the curvature k(t) for y= 1/x
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It seems like you're being asked to find the curvature as a function of some parameter, t. You can parametrize your equation by setting\[x=t \]which will then force y to be\[y=\frac{1}{t}\]For a plane curve given parametrically in Cartesian coordinates, \[r(t)=(x(t),y(t))\]the curvature is given by\[\kappa (t) = \frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}=\frac{1.\frac{2}{t^3}+\frac{1}{t^2}.0}{\left( 1^2+\frac{1}{t^4} \right)^{3/2}}=\frac{2t^3}{(1+t^4)^{3/2}}\]
Where x', x'', y', y'' are the first and second derivatives of x and y with respect to t. Hope this helps. Let me know if you need any clarification.
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