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Given $r(t) = (7\sqrt{2}t, e^{7t}, e^{-7t})$, is there any way to find the unit tangent vector more simply: $$T(t) = \frac{r'(t)}{|r'(t)|}$$

When I calculate $|r'(t)|$, I get something horrendous looking. There's no way I will be able to take the derivative of that to get the normal vector after, $$N(t) = \frac{T'(t)}{|T'(t)|}$$

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John Pocha
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1 Answers1

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I do not believe there is a simpler way to find the unit tangent and normal vectors. However, the expression for $|r'(t)|$ can be written more simply using the hyperbolic cosine function, $$ |r'(t)| = \sqrt{72+2\cosh t}.$$ You can find the derivatives of the hyperbolic functions here.

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