Parametric

[Leighton]

An object moving along a curve in the xy-plane has position (x(t), y(t)) at time t with dx/dt = cos(t^3) and dy/dt 3sin(t^2) for 0<= t<= 3. At time t=2, the object is at position (4,5).
a. Write an equation for the line tangent to the curve at (4,5).
b. Find the speed of the object at time t=2.
c. Find the total distance traveled by the object over the time interval 0<=t<=1.
d. Find the position of the object at time t=3.

a. I found the slope by 3sin(t^2)/cos(t^3) (t->2)
I get 15.6
so y-5=15.6(x-4)?

b. Sqrt[(x'(t))^2 + (y'(t)^2)]
so I just squared the givens (with t=2)
= 2.3166 ?

c. Integral (0 to 1) Sqrt[(cos(t^3))^2 + (3sin(t^2))^2]
= 1.458

d. Not sure what to do about this part...

[Ebolamonk3y]

haha! Irony... I was doing the same exact test! You ready for the BC exam Leighton? I got this test off from AP Central... they also posted the grading sheet...

part d...

Integral (2 to 3) cos(x^3) dx = [f(3) - f(2)]i = -0.046
Integral (2 to 3) 3sin(x^2) dx = [f(3) - f(2)]j = -0.093

[f(3)-f(2)+f(2)]i=f(3)i=-0.046+4=3.95i
[f(3)-f(2)+f(2)]j=f(3)j=-0.094+5=4.91j

(3.95,4.91) is the final answer... :)

ARG, I was so mad at the first part because I forgot about the parametric equations... that the dx/dt=Vi dy/dt=Vj and V=Vi+Vj, argggggggggg!!! :) Anyways.