Question

"the position of an object is described by the parametric equations x=ln t and y=5t^2. What is the accerleration of the object in m/sec^2 when t=2?"

Answer 1

Now, what I did was calculuate d^2y/dt^2 and d^2x/dt^2. then I calculated \(\sqrt{a_x^2+a_y^2}\), for t=2...

Answer 2

But, my sister tells me I'm supposed to find d^2y/dx^2...

what am I supposed to do?

Answer 3

@amistre64

Answer 4

http://chat.stackexchange.com/transcript/message/6955366#6955366

Answer 5

Question answered by people at math.stackexchange :)

http://math.stackexchange.com/questions/241312/acceleration-of-a-particle-described-by-parametric-equations

Answer 6

accerleration of the object is the change in velocity of the object with time
velocity is the change in position of the object with time

The position of the object is the vector (x y)
its velocity is the vector (dx/dt dy/dt)
its acceleration is the vector (d^2 x/dt^2 d^2 y/dt^2)
You found the magnitude of the acceleration, which may or may not be what the question is asking for.

Answer 7

@phi , the problem is that the units they want are in m/sec^2 - so direction wouldn't really do anything in here :S