yuki
  • yuki
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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yuki
  • yuki
woops, the last sentence should have been csc^2(x)
anonymous
  • 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
  • anonymous
\[\frac{dy}{dx}=-\frac{\cos 2t}{\sin 2t}=-\cot 2t\]

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yuki
  • yuki
right, so I found dx/dt and dy/dt
yuki
  • yuki
yep
anonymous
  • anonymous
\[\frac{d}{dt}(-\cot 2t)=2\csc ^2 x\]
anonymous
  • anonymous
^^ that's your numerator.
anonymous
  • anonymous
\[\frac{dx}{dt}=-2\sin 2t\]
anonymous
  • anonymous
2x in csc^2
yuki
  • yuki
woops, my derivative of cot was completely wrong. let me try it again
anonymous
  • anonymous
Man I hate this editor.
anonymous
  • anonymous
Guys who programme this site - stop mucking around with frigging medals and shields and spend your time building something user-friendly!
anonymous
  • anonymous
\[\frac{d^2y}{dx^2}=\frac{2 \csc^2 2t}{-2 \sin 2t}=-\csc^3 2t\]
anonymous
  • anonymous
yuki, are you typing a thesis?
yuki
  • 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
  • anonymous
The acceleration vector is the double time derivative of each coordinate. The acceleration is the magnitude of that vector.
anonymous
  • anonymous
\[a=(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2})\]
anonymous
  • anonymous
\[|a|=\sqrt{x''(t)^2+y''(t)^2}\]
anonymous
  • anonymous
\[a=(-4\cos 2t, -4\sin 2t)=-4r\]where r is your position vector.
anonymous
  • anonymous
The acceleration points in a direction opposite to your position vector.
yuki
  • 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
  • anonymous
\[|a|=4\]
anonymous
  • anonymous
This is calculus.
anonymous
  • 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
  • anonymous
Hope it helped, yuki :)

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