# Thread: More Unit Tangent and Unit Normal Vectors

1. ## More Unit Tangent and Unit Normal Vectors

Find the unit tangent vector.

r (t) = 4t^2i

r'(t) = 8ti

||r'(t)|| = root {(8t)^2} = root {64t^2} = 8t

So, T (t) = 8ti/8t = i.

Yes? No?

I now need the unit normal vector.

I am stuck here again.

I know that T (t) = i (see above).

I need to find T'(t) and ||T'(t)||.

2. ## Re: More Unit Tangent and Unit Normal Vectors

you cannot learn math like this.

Math within a given subject domain has to be learned in order.

You cannot start playing the tangent and normal vectors until you understand how to differentiate a vector valued function.

If you can't immediately see that $\dfrac{d}{dt} (1i + 0j + 0k )= 0i + 0j + 0k$

you're just not ready to proceed further.

3. ## Re: More Unit Tangent and Unit Normal Vectors

According to the textbook, my answer for T (t) is correct.
So, T (t) = i is correct.

I need N (t) = T'(t)/||T'(t)||

I found T'(t) = 0. Then the magnitude is root {0^2} = 0.

I concluded that N (t) = 0/0 = 0 or is it really zero?

According to the textbook, "N (t) is undefined. The path is a line and the speed is a variable."

Why is N (t) undefined? Can we say that 0/0 is undefined?

4. ## Re: More Unit Tangent and Unit Normal Vectors

Originally Posted by USNAVY
According to the textbook, my answer for T (t) is correct.
So, T (t) = i is correct.

I need N (t) = T'(t)/||T'(t)||

I found T'(t) = 0. Then the magnitude is root {0^2} = 0.

I concluded that N (t) = 0/0 = 0 or is it really zero?

According to the textbook, "N (t) is undefined. The path is a line and the speed is a variable."

Why is N (t) undefined? Can we say that 0/0 is undefined?
it's undefined because there is no way you can express the 0 vector as a unit vector.

or as you said.. $\dfrac{T^\prime}{\|T^\prime\|} = \dfrac {(0,0,0)} 0 \text{ which is undefined.}$

5. ## Re: More Unit Tangent and Unit Normal Vectors

Originally Posted by romsek
it's undefined because there is no way you can express the 0 vector as a unit vector.

or as you said.. $\dfrac{T^\prime}{\|T^\prime\|} = \dfrac {(0,0,0)} 0 \text{ which is undefined.}$
Cool.

6. ## Re: More Unit Tangent and Unit Normal Vectors

Originally Posted by romsek
you cannot learn math like this.

Math within a given subject domain has to be learned in order.

You cannot start playing the tangent and normal vectors until you understand how to differentiate a vector valued function.

If you can't immediately see that $\dfrac{d}{dt} (1i + 0j + 0k )= 0i + 0j + 0k$

you're just not ready to proceed further.
You appear to have misread the OP's question Romsek. The question is to find the unit tangent vector to \displaystyle \begin{align*} 4\,t^2\,\mathbf{i} \end{align*}, which is \displaystyle \begin{align*} \mathbf{i} \end{align*}, not to find the unit tangent vector to \displaystyle \begin{align*} \mathbf{i} \end{align*}.

What the OP posted is correct.

7. ## Re: More Unit Tangent and Unit Normal Vectors

Surely you realized that this "curve" is a straight line, the x- axis? Its unit tangent vector is the unit vector in the direction of the x- axis, i. And there is no single unit normal vector- every vector in the yz-plane is normal to it.