Unit tangent vector

**icelated** · June 15th 2012, 03:34 PM

I am trying to find the unit tangent vector T(t)
Heres the problem

$r(t) = ti + \frac{1}{t}j, t = 1$

$v(t) = i - \frac {1}{t^2} j$

then

$a(t) = \frac {2}{t^3} j$

I think the magnitude is:

$||v(t)|| = \sqrt \frac{2} {t^4} j$

then the solutions manual has this:

$T(t) = \frac {v(t)}{ || v(t) ||} = \frac{t^2} {\sqrt{ t^4 + 1}} (i - \frac{ 1}{t^2}j )$

I am not sure how they are getting this.
Can someone help me with the steps? I would appreciate this!!

**Reckoner** · June 15th 2012, 04:20 PM

Originally Posted by icelated

I think the magnitude is:

$||v(t)|| = \sqrt \frac{2} {t^4} j$

This is way off. You do know that the magnitude of a vector is a scalar, right? What is that $\mathbf j$ doing there? And where did the 2 come from?

**icelated** · June 15th 2012, 04:32 PM

so the magnitude of v(t)

$v(t) = i - \frac {1}{t^2} j$

would =

$v(t) = \sqrt { ( 1- \frac { 1 }{t^2} )^2}$

then,

$v(t) = \sqrt { ( \frac {1 + 1 }{t^4} )}$

I think i see a mistake? should be

$||v(t)|| = \sqrt { ( \frac {t^4 + 1 }{t^4} )}$

How would i take the magnitude of v(t) then?

**Reckoner** · June 15th 2012, 04:37 PM

Originally Posted by icelated

would =

$v(t) = \sqrt { ( 1- \frac { 1 }{t^2} )^2} j$

No, we square the components individually.

Originally Posted by icelated

$v(t) = \sqrt { ( \frac {1 + 1 }{t^4} )} j$

No, that's not how we add fractions.

$\mathbf v(t) = \mathbf i - \frac1{t^2}\mathbf j$

$\left\|\mathbf v(t)\right\| = \sqrt{1^2 + \left(-\frac1{t^2}\right)^2}$

$= \sqrt{1 + \frac1{t^4}}$

$= \sqrt{\frac{t^4+1}{t^4}}$

$= \frac{\sqrt{t^4+1}}{t^2}$

**icelated** · June 15th 2012, 04:50 PM

how are they getting the t squared on top?

**Reckoner** · June 15th 2012, 05:00 PM

Originally Posted by icelated

how are they getting the t squared on top?

Division. What happens when you divide

$\frac1{\sqrt{t^4+1}/t^2}?$

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