# Thread: Help on this question?

1. ## Help on this question?

Hello, I was wondering if anyone could help me with this? If there seems to be some information lacking, please ask me. Maybe to solve (d) I need some of the things from (a), (b), or (c). I have those done already, but I don't think they are that relevant to part (d). Thanks.

An object moving along a curve in the xy-plane has position (x(t),y(t)) at time t ≥ 0 with dx/dt = 3 + cos(t^2). The derivative dy/dt is not explicitly given. At time t=2, the object is at position (1,8).

d) for t ≥ 3, the line tangent to the curve at (x(t),y(t)) has a slope of 2t+1. Find the acceleration vector of the object at time t=4.

2. Originally Posted by BoombaWoomba
Hello, I was wondering if anyone could help me with this? If there seems to be some information lacking, please ask me. Maybe to solve (d) I need some of the things from (a), (b), or (c). I have those done already, but I don't think they are that relevant to part (d). Thanks.

An object moving along a curve in the xy-plane has position (x(t),y(t)) at time t ≥ 0 with dx/dt = 3 + cos(t^2). The derivative dy/dt is not explicitly given. At time t=2, the object is at position (1,8).

d) for t ≥ 3, the line tangent to the curve at (x(t),y(t)) has a slope of 2t+1. Find the acceleration vector of the object at time t=4.
You will have:

$\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$

(the slope of the curve is $\displaystyle dy/dx$)

So now you have enough information to find $\displaystyle (\dot{x},\dot{y})$ (the velocity) as a
function of $\displaystyle t$, now just differentiate again to get the acceleration $\displaystyle (\ddot{x},\ddot{y})$

RonL