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Consider the following space curve: $$ \gamma(x)=(e^x\cos(x), e^x\sin(x), e^x). $$ My main goal is to find the Frenet Frame T,N,B.

So far I have found the arc-length using the following formula: $$ s(x)=\int_{x_0}^x |\gamma'(x)| dx. $$ i.e. $$ s(x)=\int_{x_0}^x \sqrt{3} \sqrt{e^{2x}} dx $$

I am now unsure of where to go from here, any help would be greatly appreciated.

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1 Answer 1

$\gamma'(x)=(e^{x}\cos(x)-e^{x}\sin(x),e^{x}\sin(x)+e^{x}\cos(x),e^{x})$ and $|\gamma'(x)|=\sqrt{3e^{2x}}$ so you can take $T$ to be $\frac{\gamma'(x)}{|\gamma'(x)|}=\frac{1}{\sqrt{3}}(\cos(x)-\sin(x),\sin(x)+\cos(x),1)$. Next, $\gamma''(x)=(-2e^{x}\sin(x),2e^{x}\cos(x),e^{x})$ which is not orthogonal to $\gamma'(x)$ but I think you can use it and perform Gram-Schmidt to get something that is orthonormal to $T$ and that can be taken to be $N$. After that, $B=T\times N$.

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