## yuki 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)

1. Yuki

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

2. anonymous

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

3. anonymous

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

4. Yuki

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

5. Yuki

yep

6. anonymous

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

7. anonymous

8. anonymous

$\frac{dx}{dt}=-2\sin 2t$

9. anonymous

2x in csc^2

10. Yuki

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

11. anonymous

Man I hate this editor.

12. anonymous

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

13. anonymous

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

14. anonymous

yuki, are you typing a thesis?

15. 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".

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. Yuki

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

22. anonymous

$|a|=4$

23. anonymous

This is calculus.

24. 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.

25. anonymous

Hope it helped, yuki :)