Let C be the curve parametrized by $$x=3sint,\;y=4t,\;z=3cost,\;t\geq 0$$ a)if you start at the point (0,0,3) and move $5\pi$ units along C, you end up at a point $(z,y,z)=(a,b,c)$. Find the coordinates of a,b and c.

b)Find the curvatuce of C at the point (a,b,c) found in part a

Question

Let C be the curve parametrized by $$x=3sint,\;y=4t,\;z=3cost,\;t\geq 0$$ a)if you start at the point (0,0,3) and move $5\pi$  units along C, you end up at a point $(z,y,z)=(a,b,c)$. Find the coordinates of a,b and c.

b)Find the curvatuce of C at the point (a,b,c) found in  part a

richyw · Answer

alright before anyone goes crazy, I have the correct answer for a) I'll quickly explain it as I need it for b (the part I don't get)

richyw · Answer

so we have $$\vec{r}(t)=\langle 3\sin t,4t,3\cos t\rangle$$ then $$\vec{v}(t)=\frac{d\vec{r}}{dt}=\langle3\cos t,\,4\,,-3\sin t\rangle$$and $$u=||\vec{v}(t)||=\sqrt{9\cos^2t+16+9\sin^2t}=5$$

richyw · Answer

so then with arc-length parametrization I also know $$u=\frac{ds}{dt}=5$$ so $ds=5dt$  and  $s=5t$ so $$\vec{r}(s)=\left\langle 3\sin\left({\frac{s}{5}}\right), 4\left(\frac{s}{5}\right),3\cos{\left(\frac {s}{5}\right)} \right\rangle$$

richyw · Answer

and therefore $$\vec{r}(5\pi)=\left\langle\sin\pi,\,4\pi,\,3\cos\pi\right\rangle=\left\langle 0,\,4\pi,\,-3\right\rangle$$

anonymous · Answer

Good.

richyw · Answer

OK so that's where I am. now I know that $$\kappa=\left|\left|\frac{d\vec{T}}{ds}\right|\right| $$ so I'm assuming I need to find $\vec{T}$ which I though was just $$\frac{v(t)}{u(t)}$$

anonymous · Answer

Yes, that is correct.$$\mathbf{T}=\frac{\mathbf{r'}}{\|\mathbf{r'}\|}$$But... once you calculate that, you must differentiate with respect to s, not t. I presume you have a method. Show me.

richyw · Answer

well I got $$v(s)=\frac{1}{5}\left\langle 3\sin(s/5),\,4,\, 3\cos(s/5)\right\rangle$$

richyw · Answer

which the length is also 5 I think

richyw · Answer

so I get $$\vec{T}=\frac{1}{25} \left\langle 3sin\left(\frac{s}{5}\right),\,4,\,3\cos \left(\frac{s}{5}\right)\right\rangle$$

richyw · Answer

but this is incorrect.

richyw · Answer

oops negative for the z component. it's the fraction put front that's wrong though

anonymous · Answer

I keep getting myself confused.

I used the parameter t. Your method is confusing to me, sorry.

anonymous · Answer

The curve is a helix. The curvature at any point is 3/25

richyw · Answer

OK I just figured it out I think. thanks a lot!

richyw · Answer

yup I get the answer you gave now! case closed haha

anonymous · Answer

Now calculate the torsion!