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Q. Consider the following vector function.

$$ r(t)= \langle 6\sqrt{2}t,e^{6t},e^{-6t} \rangle $$

Find the unit tangent and unit normal vectors T(t) and N(t).

I found $$T(t)= \frac{1}{\sqrt{2+e^{12t}+e^{-12t}}}\langle \sqrt{2},e^{6t},-e^{-6t}\rangle $$

but when I try finding $N(t)=T'(t)/|T'(t)|$ the calculations just go out of hand and I cannot reach an answer.

Can someone please help me in finding the answer.

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  • $\begingroup$ Hint: The terms under the square root in the denominator can be factored into a perfect square. $\endgroup$ – nukeguy Jan 30 '15 at 1:47
  • $\begingroup$ $2+e^{12t}+e^{-12t}=e^{-12t}((e^{12t})^2+2e^{12t}+1)=e^{-12t}(e^{12t}+1)^2$ $\endgroup$ – oldrinb Jan 30 '15 at 1:50
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let $$u = e^{6t}.$$ then we have $$(u^2 + 1)T = (\sqrt 2 u, u^2, 1) \tag 1$$

differentiating $(1)$ with respect to $u,$ gives $2uT + (u^2+1)\frac{dT}{du} = (\sqrt 2, 2u, 0)$ which can be simplifies to give $$(u^2 + 1)^2\frac{dT}{du} = (u^2 + 1)(\sqrt 2, 2u, 0) - 2u(\sqrt 2 u, u^2, 1) =(\sqrt 2(1-u^2),2u,-2u) $$ therefore $$N = \dfrac{1}{(1+u^2)}((1-u^2),\sqrt 2u,-\sqrt 2u)$$

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