An object moves in the xy-plane is at position (x(t), y(t)) for time t, , where dx/dt=tan(e^-t) and dy/dt=sec(e^-t)
At time t=1 the position of particle is at (2, -3).

Question

An object moves in the xy-plane is at position (x(t), y(t))  for time t,  , where dx/dt=tan(e^-t)  and  dy/dt=sec(e^-t)
 At time  t=1  the position of particle is at (2, -3).

anonymous · Answer

Is there a time  at which the object is on the y-axis?  Explain why or why not.

anonymous · Answer

helppp
pleasee

anonymous · Answer

Its okay! I'll help ya out!~

anonymous · Answer

First...for the object to ever be on the y axis...all this statement means is that is it's x(t) EVER equal to zero? Because that's what it means for an object to lie on the y-axis. So let's try to find the position function for x position, first you will want to integrate dx/dt, and use the initial condition!

anonymous · Answer

so $$\int\limits_{1}^{t} \tan(e^-t)$$

anonymous · Answer

=x(t)-x(1)

anonymous · Answer

right?

anonymous · Answer

Yes correct! So you want to use that formula to find x(t), then from there you have to figure out if at any time that function will equal zero

anonymous · Answer

alright thank you!

anonymous · Answer

What is this for? Has to be some kind of calculus class, Love it.

anonymous · Answer

its something were doing in calc ab after the ap testing.

anonymous · Answer

Ahhh, I see! What a fun year that was!