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!

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>\$

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.

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!

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.

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

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.