Object on a curve...

[Mooch22]

An object moving along a curve in the xy-plane has position (x(t), y(t)) at time t (greater than or equal to) 0 with x-coordinate 3.

(dx/xt) = 12t-3t^2 and (dy/dt) = ln(1+(t-4)^4)

At time t=0, the object is at position (-13,5). At time t=2, the object is at point P with x-coordinate 3.

a.) Find the acceleration vector at time t=2 and the speed at time t=2.
**Is the function already a(t)...since it is the derivate? Do I have to combine them to make (dy/dx)?

b.) Find the y-coordinate of P.

c.) Write an equation for the line tangent to the curve at P.

d.) For what value of t, if any, is the object at rest? Explain the reasoning.


**Please help, this is one of the last calculus questions I have to do this year, and I'm just lost! I DESPERATELY need helllllllp!!! :roll:

 

[happy]

What have you done so far?

 

[skeeter]

dx/dt and dy/dt are the x and y components of velocity

d²x/dt² is the x-component of acceleration
d²y/dt² is the y-component of acceleration

speed = \(\displaystyle \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}\)

\(\displaystyle y(2) = y(0) + \int_0^2 ln[1+(t-4)^4] dt\)


slope at P = \(\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) evaluated at t = 2

use point P's x and y coordinates and the slope to find the tangent line equation.


for the object to be at rest, dx/dt and dy/dt have to both be 0 at the same time.