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anonymous
 5 years ago
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)
anonymous
 5 years ago
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)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0woops, the last sentence should have been csc^2(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}=\frac{\cos 2t}{\sin 2t}=\cot 2t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, so I found dx/dt and dy/dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dt}(\cot 2t)=2\csc ^2 x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^^ that's your numerator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dx}{dt}=2\sin 2t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0woops, my derivative of cot was completely wrong. let me try it again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Man I hate this editor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Guys who programme this site  stop mucking around with frigging medals and shields and spend your time building something userfriendly!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d^2y}{dx^2}=\frac{2 \csc^2 2t}{2 \sin 2t}=\csc^3 2t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yuki, are you typing a thesis?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right? 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
 5 years ago
Best ResponseYou've already chosen the best response.0The acceleration vector is the double time derivative of each coordinate. The acceleration is the magnitude of that vector.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a=(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a=\sqrt{x''(t)^2+y''(t)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a=(4\cos 2t, 4\sin 2t)=4r\]where r is your position vector.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The acceleration points in a direction opposite to your position vector.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do we learn this in Calculus? It sounds more like multivariable. Anyways, I did get the answer and it was huge help. Thanks ! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is calculus of parametric equations. You need to parametrize a path since in the xy 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 xy points, but at the same time, the x and y are *functions* of t, so you may perform calculus.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hope it helped, yuki :)
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