Unit Tangent Vector

**adkinsjr** · July 30th 2009, 10:02 PM

Given the vector function $r(t)=<\sqrt{2}t,e^t,e^{-t}>$ I need to find the tangent unit vector $T(t)=\frac{r'(t)}{\sqrt{x'^2+y'^2+z'^2}}$ . The denominator is the absolute value of the vector derivative $r'(t)=<\sqrt{2},e^t,-e^{-t}>$ . I write an expression for $T(t)$ :

$T(t)=\frac{1}{\sqrt{2+e^{2t}+e^{-2t}}}<\sqrt{2},e^t,-e^{-t}>$

$=\frac{1}{\sqrt{2(1+\frac{e^{2t}+e^{-2t}}{2}})}<\sqrt{2},e^t,-e^{-t}>$

$=\frac{1}{\sqrt{(1+cosh2t)}}(i+\frac{e^t}{\sqrt{2} }j-\frac{e^{-t}}{\sqrt{2}}k)$

I don't think this could be wrong since it's pretty straight forward. It requires zero intelligence, you just apply the definition directly. However, I can't seem to obtain the same expression given in the book:

$T(t)=\frac{1}{e^{2t}+1}<\sqrt{2}e^t,e^{2t},-1>$

I can see that $e^tr'(t)=<\sqrt{2}e^t,e^{2t},-1>$ , which suggests that they just multiplied the numerator and denominator of $T(t)$ by $e^t$ . However I can't get the demonator in the form $e^{2t}+1$ . I'm confused by the algebra, I can't show that $e^t\sqrt{2+e^{2t}+e^{-2t}}=e^{2t}+1$ .

????

**Opalg** · July 30th 2009, 11:46 PM

One of the addition formulae for hyperbolic functions (very similar to the addition formulae for trig functions) is $\cosh2x = 2\cosh^2x-1$ . As a consequence, $\sqrt{1+\cosh2x} = \sqrt2\,\cosh x = \frac{e^{2x}+1}{\sqrt2\,e^x}$ .

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