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## Homework Statement

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}##)?

## Homework 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.

-------------------------------------------------------------------------------------------------------------

Find the arc length Parametrization of r(t) = <t

^{2},t

^{3}>

assume t>0

r'(t) = <2t,3t

^{2}>

||r'(t)|| =(4t

^{2}+9t

^{4})

^{1/2}##\Rightarrow## t (4+9t

^{2})

^{1/2}

*So now we integrate....*

S(t) = ##\int_0^t t(4+9t^2)^{1/2}\,dt## u=t

^{2}; du=2t dt ##\Rightarrow\frac{du}{2} = t\,dt##

*This is where he completely looses me. How did t become t*

^{2}? Where does the 1/9 come from, is this like as if he did u=9t^{2}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+9t

^{2})

^{3/2}-8 ##\Rightarrow## 9t^2=(27s+8)

^{2/3}-4

t

^{2}=(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##

which leads to

##\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?

Please and thankyou! :-)