Question

Another Calculus Trajectory problem involving a helix.

Found below.

Answer 1

The equation r(t) = <3cost,3sint,4t> for 0 ≤ t ≤ 2 π
Which is helix

Again doing the same thing as the last problem using the equation below to solve for speed:\[S(t) = \sqrt{f'(u(t))^{2}+g'(u(t))^{2}+h'(u(t))^{2}}\]:

I found that the speed is 5 for the problem above using that equation.
However the next part of the question asks this:

Suppose the helical path of the problem above is to be traversed at speed of\[S(t) = V_{o}e ^{-\frac{ t }{ 4 }} m/s\]

Find the trajectory if \[V _{o}= 15πm/s\]

On what interval of
\[0 \le t \le T\]is one loop of the helix traversed?

Answer 2

@oldrin.bataku

I hope this one isn't as hard to understand.

Answer 3

hmm

Answer 4

well acknowledge from before that we can use the chain rule when computing \(\frac{d}{dt}\vec{r}(u(t))=u'(t)\vec{r}'(t)\) hence \(\|\frac{d}{dt}\vec{r}(u(t))\|=\|u'(t)\vec{r}'(t)\|=|u'(t)|\|\vec{r}'(t)\|\). Here you found \(\|\vec{r}'(t)\|=5\) hence \(15\pi e^{-t/4}=5|u'(t)|\); let's assume \(u'(t)<0\) hence \(u'(t)=-3\pi e^{-t/4}\) and therefore integrating yields \(u(t)=3\pi\int-e^{-t/4}\,dt=12\pi e^{-t/4}\)

Answer 5

Our trajectory is just then \(\vec{r}(u(t))\) (cbf to write it all out but substitute \(12\pi e^{-t/4}\) for \(t\)).

Answer 6

Anyway, observe our first loop of our helix completes where \(u(t)=2\pi\). $$12\pi e^{-T/4}=2\pi\\e^{-T/4}=\frac16\\-\frac{T}4=\log\frac16=-\log6\\T=4\log 6$$... :-) by the way, earlier you could also have assumed \(u'(t)>0\) but then the helix would have opposite orientation.

Answer 7

hmm i disagree here...the domain is 0<t<2pi
u(0) = 12pi
1st loop completes when u(t) = 10pi
correct?

Answer 8

anyway hes not replying...im pretty sure im right since 4log6 > 2pi
by setting u(T) = 10pi you get
T = -4log(5/6)

Answer 9

hmm alright so back up top for the trajectory, where do i plug \[12πe ^{-t/4} \] into.. sorry if that's a dumb question really tired.

Answer 10

@dumbcow ah it appears you're correct. Since we begin with \(u(0)=12\pi\) we should complete with \(u(t)=10\pi\) (because we let \(u'<0\) i.e. \(u\) is decreasing):$$10\pi=12\pi e^{-T/4}\\\frac56=e^{-T/4}\\\frac{T}4=\log\frac65\\T=4\log\frac65$$

Answer 11

@charlotte you plug that into \(r\) in place of \(t\) to get the parameterization with the desired speed

Answer 12

so back up into this: r(t) = <3cost,3sint,4t>so where r(t) is plug that in and solve?

Answer 13

sorry @oldrin.bataku I am really confusing my self lol

Answer 14

replace "t" with u(t)
cos(12pi*e^-t/4)
and so on ...

Answer 15

thanks @dumbcow