How many times will the particle stop in an hour?

[devanlevin]

a particle move according to the vector
\vec{r}=3cos^{2}(6t)\hat{x}-5sin(\frac{PI}{6}t)\hat{y}

what is the formula for the route it takes, y(x)? when will the particle stop and how many times will it stop in an hour??

i said, the vectors in the direction x and y
x(t)=3cos^{2}6t=3(1-sin^{2}6t)
y(t)=-2sin6t
-------------
sin6t=\frac{-y}{2}

x=3(1-(\frac{-y}{2})^{2})
x(y)=3-\frac{3}{4}y^{2}

this course is a parabula lying on its side, (minus infinity) on its x-axis with its maximum at Max(3,0)

to find out when the particle stops i say
the particle will stop when the parabula peaks-- \frac{dx}{dy}=0
x(y)=3-0.75y^{2}
\frac{dx}{dy}=-1.5y
y=0
the particle will stop every time y=0
y(t)=-2sin(6t)=0
sin(6t)=0
6t=PI*K
t=\frac{PI}{6}*K (K being a positive whole number)

does this mean that the particle will stop every \frac{PI}{6} seconds??

how can this be, since the route the particle moves on is a parabula and doesn't peak more than once, i realize that where i have misunderstood something is in the transition from the trigonometric equations x(t) and x(y) which repeat themselves to the parabula x(y).

 

[Hootenanny]

a particle move according to the vector
\vec{r}=3cos^{2}(6t)\hat{x}-5sin(\frac{PI}{6}t)\hat{y}

what is the formula for the route it takes, y(x)? when will the particle stop and how many times will it stop in an hour??

i said, the vectors in the direction x and y
x(t)=3cos^{2}6t=3(1-sin^{2}6t)
y(t)=-2sin6t

Careful

 

[devanlevin]

sorry, the -5sin...was a typo, , meant to be -2sin(6t)y like i wrote in all the calculations

 

[Hootenanny]

sorry, the -5sin...was a typo, , meant to be -2sin(6t)y like i wrote in all the calculations

What about the \pi/6 in the question? Was that a typo as well?

 

[devanlevin]

correct equation is r=3cos^2(6t)x-2sin(6t)y
but that doesn't really make a diffeerence to what I am asking, its more of a technical question, the end result as a number isn't of that much importance,,,
after reading over what i wrote i now think that i was wrong, and that the particle will stop when |v|=0 meaning when cos(6t)=0
6t=PI/2 +PI*K
t=PI/12 + (PI/6)*K

from this how do i know how many times it will stop in an hour??

 

[Hootenanny]

The particle doesn't stop when

\frac{dy}{dx}=0

Rather, it stops when

\frac{d\bold{r}}{dt}=0

 

[devanlevin]

thats essentially what i did by saying Vx=0 and Vy=0

but then i get to t=PI/12 + (PI/6)*K how do i translate that into how many times it will stop in an hour