# Thread: Unit normal vector function

1. ## Unit normal vector function

I know that the unit vector function is equal to T'(t) / |T'(t)|

If T(t) is < $\frac{2t}{2t^2}$ , $\frac{2t^2}{2t^2}$, $\frac{1}{2t^2}$ >

To find T'(t), would I simply take the derivative of each element using the quotient rule? I did this on paper, and was getting results that were different than the answers in the back of my book. Any ideas? Thanks

Edit: what on earth is wrong with my latex

2. Hello, zellster87

Close LaTex with [/math] . . . not [\math]

I know that the unit vector function is equal to: $\frac{T\,'(t)}{|T\,'(t)|}$

If $T(t) \;=\;\left\langle \frac{2t}{2t^2},\:\frac{2t^2}{2t^2},\:\frac{1}{2t^ 2}\right\rangle$ . . . . Why aren't these reduced?

We have: . $T(t) \;=\;\left\langle\frac{1}{t},\:1,\:\frac{1}{2t^2}\ right\rangle \;=\;\left\langle t^{-1},\:1,\:\frac{1}{2}t^{-2}\right\rangle$

Then: . $T\,'(t) \;=\;\left\langle -t^{-2},\:0,\:-t^{-3}\right\rangle \;=\;\left\langle -\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle$

And: . $|T\,'(t)| \;=\;\sqrt{\left(-\frac{1}{t^2}\right)^2 + 0^2 + \left(-\frac{1}{t^3}\right)^2} \;=\;\sqrt{\frac{1}{t^4} + \frac{1}{t^6}} \;=\;\sqrt{\frac{t^2+1}{t^6}} \;=\;\frac{\sqrt{t^2+1}}{t^3}$

Therefore: . $\frac{T\,'(t)}{|T\,'(t)|} \;=\;\frac{\left\langle-\dfrac{1}{t^2},\:0,\:-\dfrac{1}{t^3}\right\rangle}{\dfrac{\sqrt{t^2+1}}{ t^3}} \;=\;\frac{t^3}{\sqrt{t^2+1}}\left\langle-\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle$

. . . . . . $= \;\left\langle\frac{-t}{\sqrt{t^2+1}},\:0,\:\frac{-1}{\sqrt{t^2+1}}\right\rangle$

3. Originally Posted by Soroban
Hello, zellster87

Close LaTex with [/math] . . . not [\math]

We have: . $T(t) \;=\;\left\langle\frac{1}{t},\:1,\:\frac{1}{2t^2}\ right\rangle \;=\;\left\langle t^{-1},\:1,\:\frac{1}{2}t^{-2}\right\rangle$

Then: . $T\,'(t) \;=\;\left\langle -t^{-2},\:0,\:-t^{-3}\right\rangle \;=\;\left\langle -\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle$

And: . $|T\,'(t)| \;=\;\sqrt{\left(-\frac{1}{t^2}\right)^2 + 0^2 + \left(-\frac{1}{t^3}\right)^2} \;=\;\sqrt{\frac{1}{t^4} + \frac{1}{t^6}} \;=\;\sqrt{\frac{t^2+1}{t^6}} \;=\;\frac{\sqrt{t^2+1}}{t^3}$

Therefore: . $\frac{T\,'(t)}{|T\,'(t)|} \;=\;\frac{\left\langle-\dfrac{1}{t^2},\:0,\:-\dfrac{1}{t^3}\right\rangle}{\dfrac{\sqrt{t^2+1}}{ t^3}} \;=\;\frac{t^3}{\sqrt{t^2+1}}\left\langle-\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle$

. . . . . . $= \;\left\langle\frac{-t}{\sqrt{t^2+1}},\:0,\:\frac{-1}{\sqrt{t^2+1}}\right\rangle$

Thanks a lot Soroban, I appreciate the help.