Question

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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

yep

Answer 6

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

Answer 7

^^ that's your numerator.

Answer 8

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

Answer 9

2x in csc^2

Answer 10

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

Answer 11

Man I hate this editor.

Answer 12

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

Answer 13

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

Answer 14

yuki, are you typing a thesis?

Answer 15

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".

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

\[|a|=4\]

Answer 23

This is calculus.

Answer 24

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.

Answer 25

Hope it helped, yuki :)