# Homework Help: Find the velocity vector of a curve at point r(t)

1. Sep 7, 2013

### Unart

1. The problem statement, all variables and given/known data
What is the velocity vertor of a particle traveling to the right along the hyperbola y=x-1 wich constant 5 cm/s when the particles location is (2, $\frac{1}{2}$)?

2. Relevant equations
The Length of path forumula.
$$s\,=\int_a^b ||r'(t)||\,dt$$
Please don't make me write you the magnitude formula, LaTeX is a pain, it's going to take me a while to type out everything.

3. Relevant example
This is an example finding the Arc Length which is need it order to find the unit velocity vector of the Arc Length. There were a few steps I didn't understand well. I'll point them out to you. I have a feeling is probably some Calculus 2 or 1 I'm forgetting.
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Find the arc length Parametrization of r(t) = <t2,t3>
assume t>0

r'(t) = <2t,3t2>
||r'(t)|| =(4t2+9t4)1/2 $\Rightarrow$ t (4+9t2)1/2

So now we integrate....
S(t) = $\int_0^t t(4+9t^2)^{1/2}\,dt$ u=t2; du=2t dt $\Rightarrow\frac{du}{2} = t\,dt$

This is where he completely looses me. How did t become t2? Where does the 1/9 come from, is this like as if he did u=9t2 instead?
S(t) = $\frac{1}{2}\int_0^{t^2} (4+9u)^{1/2}\,du$
S(t) = $\frac{1}{2}\cdot\frac{2}{3}(4+9u)\cdot\frac{1}{9}\int_0^{t^2}$
S(t) = $\frac{1}{27}[(4+9(t)^2)^{3/2}-8]$
Invert
27s = (4+9t2)3/2-8 $\Rightarrow$ 9t^2=(27s+8)2/3-4
t2=(s+ $\frac{8}{27}$)-$\frac{4}{9}$
Lastly, how did he get these coordinates?
δ(s)= $\left( (s+\frac{8}{27})^{2/3}-\frac{4}{9},[(s+\frac{8}{27})^{2/3}-\frac{4}{9}]^{3/2}\right)$

4. The attempt at a solution
This problem has little to do with this problem other than I have to do the first step in order to find the velocity vector at said point on the curve when the velocity is a constant 5cm/s.

if X=t then y=t-1 $\Rightarrow\hspace {30mm} R(t)=<t,t^{-1}>$
R'(t) = <1,-t-2>$\Rightarrow$ ||R'(t)||=(1+$\frac{1}{t^4}$)1/2
$\int_0^t (1+\frac{1}{t^4})^{1/2})\Rightarrow u=t^{-4}\, and \,du= -4t^{-5} dt\, or\, \frac{du}{-4t^-5}=dt$

$\frac {-4}{t^{5}}\int_0^t (1+u)du$
5. Summary
So, did I do something wrong? Where do I go from here? And, could someone help explain what the teacher did?

2. Sep 7, 2013

### HallsofIvy

What, exactly, is the question here? You ask about the velocity vector but then give as a "relevant example" "finding the arc length". One has nothing to do with the other.

The curve y= 1/x can be written as parametric equations x= s y= 1/s where I have just used "s" as an "alias" for "x". We can write the "position vector" as <t, 1/t>. Differentiating, the tangent vector is <1, -1/t^2>. In particular, the object will be at <2, 1/2> when t= 2 and there the tangent vector is <1, -1/4>. That has length ("magnitude") sqrt(1+ 1/16)= sqrt(17/16)= sqrt(17)/4. Divide the vector <1, -1/4> by that to get the "unit tangent vector" and multiply by 5 cm/s to get a vector tangent to the curve with "magnitude" 5 cm/s- the velocity vector.

3. Sep 7, 2013

### Unart

Whew! Couldn't see the trees for the forest on that one. I had the right idea before I went to class on Friday, but somehow did some arithmetic or algebra error, and thought about the teacher not going over the chapter. Yesterday he said/ and also read on the book that the derivative of the Parameterization of arc length will give you the unit speed. So I thought you had to do that to get the unit velocity vector along the curve.
I realize what I've done wrong and am moving on.

However I would still like some help understanding the sample problem on the teacher gave us. Please, anyone??

4. Sep 7, 2013

### LCKurtz

Yes, the appropriate substitution would have been $u = 9t^2,~du=18t\, dt$. Instead he first does $u=t^2,~du=2tdt$ and has to make up for it later with $u=9t,~du=9 dt$, which he doesn't write out explicitly.