1. Acceleration vector and speed

Ap Calc exam tomorrow, and I really need to know how to do this...

It is given that an object is moving along a curve in an xy-plane at position (x(t), y(t)), where dx/dt = arcsin(1-2(e^-t)) and dy/dt = (4t)/(1+(t^3)).
At t=2, the object is at (6,3).

The problem asks to find the acceleration vector and speed of object at t=2.

2. Originally Posted by BeiW
Ap Calc exam tomorrow, and I really need to know how to do this...

It is given that an object is moving along a curve in an xy-plane at position (x(t), y(t)), where dx/dt = arcsin(1-2(e^-t)) and dy/dt = (4t)/(1+(t^3)).
At t=2, the object is at (6,3).

The problem asks to find the acceleration vector and speed of object at t=2.

Well acceleration is the first derivative of velocity and therefore it is the second position of position.

So just differentiate $\displaystyle \frac{dx}{dt}$ to get the acceleration in the x direction and differentiate $\displaystyle \frac{dy}{dt}$ to get the acceleration in the y direction.

To find the speed at t=2, use the parametric differentiation formula $\displaystyle \frac{dy}{dx} =\frac {\frac{dy}{dt}}{\frac{dx}{dt}}$ and fill in t = 2