Finding Unit Normal Vector

**melanie89** · February 4th 2009, 11:40 AM

Alright, so here's a question from my online Calc 3 assignment. I know the formula for N(t), it is T'(t)/|T'(t)|. However I am still very stuck and confused as to how to find N. I can find B easily, it's just getting to N that is giving me problems.

Find the vectors T, N, and B at the given point.

.

(I found T correctly, it was <6/19, 18/19, 1/19>.)

**Danny** · February 4th 2009, 12:40 PM

Originally Posted by melanie89

Alright, so here's a question from my online Calc 3 assignment. I know the formula for N(t), it is T'(t)/|T'(t)|. However I am still very stuck and confused as to how to find N. I can find B easily, it's just getting to N that is giving me problems.

Find the vectors T, N, and B at the given point.

.

(I found T correctly, it was <6/19, 18/19, 1/19>.)

$\bold{r' } = < 2t, 2t^2,1>$ and

$|| \bold{r' } || = \sqrt{4t^2 +4 t^4 + 1} = 2t^2+1$

so

$\bold{N} = \frac{< 2t, 2t^2,1>}{2t^2+1} = \left< \frac{2t}{2t^2+1} ,\frac{2t^2}{2t^2+1},\frac{1}{2t^2+1} \right>$

**Plato** · February 4th 2009, 12:51 PM

Originally Posted by danny arrigo

$\bold{r' } = < 2t, 2t^2,1>$ and
$|| \bold{r' } || = \sqrt{4t^2 +4 t^4 + 1} = 2t^2+1$ so
$\bold{{\color{red}N}} = \frac{< 2t, 2t^2,1>}{2t^2+1} = \left< \frac{2t}{2t^2+1} ,\frac{2t^2}{2t^2+1},\frac{1}{2t^2+1} \right>$

You have found $\bold{T}$ not $\bold{N}$

$\bold{N} = \frac{{T'}}<br /> {{\left\| T' \right\|}} = \frac{{r' \times \left( {r'' \times r'} \right)}}<br /> {{\left\| {r' \times \left( {r'' \times r'} \right)} \right\|}}$

**Danny** · February 4th 2009, 02:53 PM

Originally Posted by Plato

You have found $\bold{T}$ not $\bold{N}$

$\bold{N} = \frac{{T'}}<br /> {{\left\| T' \right\|}} = \frac{{r' \times \left( {r'' \times r'} \right)}}<br /> {{\left\| {r' \times \left( {r'' \times r'} \right)} \right\|}}$

Your absolutely right. I did stop short didn't I (and with a typo no doubt)

**melanie89** · February 6th 2009, 09:33 PM

I found the correct N using the formula r' x (r'' x r') / |r' x (r'' x r')|. Thanks so much for your help!

**acg716** · February 19th 2009, 11:37 AM

My homework is almost identical to this except my point is <9, -18, -3>.
I got the same T as you all did, except I'm not sure what to plug into it to get my point. I plugged in 9 for my x value but I got it wrong. I got 0.11. What am I doing wrong?

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