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Find a formula (provide your answer in terms of $f$ and its derivatives) for the curvature of a curve in $\mathbb{R}^3$ given by $\{(x,y,z)\ | \ x=y, f(x)=z\}$.

How will I be able to do this problem?

I know that a regular parametrization of a curve then the curvature at $\mu(t)$ is given by: $$\kappa(t) = \left|\left|\left(\frac{\mu'(t)}{||\mu'(t)||}\right)'\frac{1}{||\mu'(t)||}\right|\right|.$$

But since $f$ is not a parametrization I don't know how to continue this problem.

Since $x=y$ and there exists a function $f$ such that $f(x)=z$ and $f(y)=z$ is it safe to assume that it is a one to one function?

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Your curve lies in the plane $x=y$; therefore it is a plane curve. Denoting the length along the first main diagonal in the $(x,y)$-plane by $s$ the curve can be written as a graph in the form $$z(s)=f\left({s\over\sqrt{2}}\right)\qquad (s_0\leq s\leq s_1)\ .$$ For such graphs the formula for the curvature reads $$\kappa(s)={|z''(s)|\over\bigl(1+z'^2(s)\bigr)^{3/2}}={\sqrt{2}|f''(s/\sqrt{2})| \over\bigl(2+f'^2(s/\sqrt{2})\bigr)^{3/2}}\ .$$ This means that at the point $P=\bigl(x,x,f(x)\bigr)$ the curvature is given by $$\kappa={\sqrt{2}|f''(x)|\over\bigl(2+f'^2(x)\bigr)^{3/2}}\ .$$

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  • $\begingroup$ I got three different answers and I am confused. I am not sure how you got your results? $\endgroup$ – Lays Sep 1 '13 at 20:58
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You were given a parametric curve, but it's slightly disguised. Let $\mu(x) = (x,x,f(x))$ and then apply the formula you gave in your question, but with $t=x$. To get you started, note that $\mu'(x) = (1,1,f'(x))$. Can you take it from there?

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  • $\begingroup$ Thanks for the useful hint. I am going to apply your suggestion now and get back to you. Thanks for that! $\endgroup$ – Lays Sep 1 '13 at 7:31
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Wikipedia gives the following for a curve $(x(t),y(t),z(t))$: $$\kappa=\frac{\sqrt{(z''y'-y''z')^2+(x''z'-z''x')^2+(y''x'-x''y')^2}}{(x'^2+y'^2+z'^2)^{3/2}}$$

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For a unit speed curve, the curvature can be simply $$ \kappa = || \ddot\gamma || $$ Or if it is regular, then $$ \kappa = \dfrac{||\ddot\gamma \times \dot\gamma ||}{||\dot\gamma||^3} $$

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  • $\begingroup$ How were you able to get that? $\endgroup$ – Lays Sep 1 '13 at 20:58
  • $\begingroup$ The first is simply the definition of curvature for a unit speed curve. The second follows from the fact any regular curve has a unit speed reparameterisation, and it's essentially chain rule and vector identites all the way if i recall correctly. $\endgroup$ – FireGarden Sep 2 '13 at 7:11

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