x = cos(2t) and y = sin(2t)
For this parametric eqn. I would like to find the second derivative, but I don't seem to be doing it
right.
This is what I tried so far.
dx/dt = -2sin(2t)
dy/dt = 2cos(2t)
thus dy/dx = -cot(2t).
for the second derivative,
d/dx(dy/dx) = -csc(2t)cot(2t)
so
d^2/dx^2 [y] = sec^2(x)

- yuki

- chestercat

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- yuki

woops, the last sentence should have been csc^2(x)

- anonymous

You should understand that\[\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]

- anonymous

\[\frac{dy}{dx}=-\frac{\cos 2t}{\sin 2t}=-\cot 2t\]

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## More answers

- yuki

right, so I found dx/dt and dy/dt

- yuki

yep

- anonymous

\[\frac{d}{dt}(-\cot 2t)=2\csc ^2 x\]

- anonymous

^^ that's your numerator.

- anonymous

\[\frac{dx}{dt}=-2\sin 2t\]

- anonymous

2x in csc^2

- yuki

woops, my derivative of cot was completely wrong.
let me try it again

- anonymous

Man I hate this editor.

- anonymous

Guys who programme this site - stop mucking around with frigging medals and shields and spend your time building something user-friendly!

- anonymous

\[\frac{d^2y}{dx^2}=\frac{2 \csc^2 2t}{-2 \sin 2t}=-\csc^3 2t\]

- anonymous

yuki, are you typing a thesis?

- yuki

right? I thought that was the answer too.
Okay maybe my problem was not set up right.
The actual problem is.
The motion of a particle with x =cos(2t) and y=sin(2t).
What is the magnitude of the acceleration at any time t ?
I thought the acceleration had to do with f".

- anonymous

The acceleration vector is the double time derivative of each coordinate. The acceleration is the magnitude of that vector.

- anonymous

\[a=(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2})\]

- anonymous

\[|a|=\sqrt{x''(t)^2+y''(t)^2}\]

- anonymous

\[a=(-4\cos 2t, -4\sin 2t)=-4r\]where r is your position vector.

- anonymous

The acceleration points in a direction opposite to your position vector.

- yuki

Do we learn this in Calculus? It sounds more like multivariable.
Anyways, I did get the answer and it was huge help.
Thanks ! :)

- anonymous

\[|a|=4\]

- anonymous

This is calculus.

- anonymous

This is calculus of parametric equations. You need to parametrize a path since in the x-y plane, a path can form loops and cut into itself, which isn't a function. If you parametrize each of the coordinates, you set it up so that each coordinate is a function of some parameter, with that parameter allowing you to plot your x-y points, but at the same time, the x and y are *functions* of t, so you may perform calculus.

- anonymous

Hope it helped, yuki :)

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