# Difficulty Calculating the Principal Unit Normal Vector.

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• Oct 20th 2011, 06:47 PM
SC313
Difficulty Calculating the Principal Unit Normal Vector.
Hi,

I am having difficulties find the principal unit normal vector N(t) for the vector function

$\displaystyle \\r(t)=(t,4-t^2,0)$.

I have worked up to this point:

$\displaystyle \\r(t)=(t,4-t^2,0); -2\leq t\leq 2$
$\displaystyle \\r'(t)=(1,-2t,0)$
$\displaystyle \\r'(t) \right \|=\sqrt(1+4t^2)$
$\displaystyle \\T(t)=<\frac{1}{\sqrt(1+4t^2)},\frac{-2t}{\sqrt(1+4t^2)},0>$

I know that I need to take the derivative of this, and devide it by its length to find the princiipal normal vector; however, I am not sure how to do this!

Thanks!
• Oct 20th 2011, 07:30 PM
SC313
Re: Difficulty Calculating the Principal Unit Normal Vector.
I have either made progress or gone the wrong direction. Either way, here is the work that I have done since I made this question:
I first factored the $\displaystyle (1+4t^2)^(-1/2)$, leaving <1,-2t,0> as the vector. Then, to multiply the two, I set $\displaystyle (1+4t^2)^(-1/2)$ as “u” and <1,-2t,0> as “v” to use the product rule. I found the derivative of “u” to be $\displaystyle frac {-(1+4t^2)^{-3/2}}{2}8t$, which further simplifies to $\displaystyle frac {-4t(1+4t^2)^{-3/2}}$, which I multiplied against the scalar. I did the same with taking the derivative of the vector, and multiplying it by the $\displaystyle (1+4t^2)^(-1/2)$.
At the end, I know have:
$\displaystyle <-4t(1+4t^2)\tfrac{-3}{2},8t^3,0> + <0,-2(1+4t^2)^{^{1/2}},0>\$
• Oct 20th 2011, 08:26 PM
SammyS
Re: Difficulty Calculating the Principal Unit Normal Vector.
Quote:

Originally Posted by SC313
Hi,

I am having difficulties find the principal unit normal vector N(t) for the vector function

$\displaystyle \\r(t)=(t,4-t^2,0)$.

I have worked up to this point:

$\displaystyle \\r(t)=(t,4-t^2,0); -2\leq t\leq 2$
$\displaystyle \\r'(t)=(1,-2t,0)$
$\displaystyle \\r'(t) \right \|=\sqrt(1+4t^2)$
$\displaystyle \\T(t)=<\frac{1}{\sqrt(1+4t^2)},\frac{-2t}{\sqrt(1+4t^2)},0>$

I know that I need to take the derivative of this, and devide it by its length to find the princiipal normal vector; however, I am not sure how to do this!

Thanks!

$\displaystyle \frac{dT}{dt}=\left<\frac{-4 t}{(4 t^2+1)^{3/2}}\,,\ \frac{-2}{(4 t^2+1)^{3/2}}\,,\ 0\right>$

Divide this by it's magnitude.
• Oct 20th 2011, 09:20 PM
SC313
Re: Difficulty Calculating the Principal Unit Normal Vector.
Quote:

Originally Posted by SammyS
$\displaystyle \frac{dT}{dt}=\left<\frac{-4 t}{(4 t^2+1)^{3/2}}\,,\ \frac{-2}{(4 t^2+1)^{3/2}}\,,\ 0\right>$

Divide this by it's magnitude.

Where do you get this from? Thanks!
• Oct 20th 2011, 09:22 PM
SammyS
Re: Difficulty Calculating the Principal Unit Normal Vector.
Quote:

Originally Posted by SC313
Where do you get this from? Thanks!

I took the derivative of T(t) that you had in your OP.
• Oct 20th 2011, 09:40 PM
SC313
Re: Difficulty Calculating the Principal Unit Normal Vector.
Okay, I see now. And how do I find the magnitude of this..., and divide the number by this? :( It's so confusing. thanks by the way
• Oct 20th 2011, 10:16 PM
SammyS
Re: Difficulty Calculating the Principal Unit Normal Vector.
Magnitude: Square root of the sum of the squares of the components.

At least they have a common denominator.