Find the total distance traveled by the object

[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.

 

[M98Ranger]

a. The equation for the line tangent to the curve at (4,5) can be found using the point-slope form, where m is the slope at the given point: y - y1 = m(x - x1). Substituting the given values, we get y - 5 = (3sin(4^2))/(cos(2^3))(x - 4). Simplifying, the equation is y = 3.9x - 15.6.

b. The speed of the object at time t=2 can be found using the given formula: speed = sqrt[(dx/dt)^2 + (dy/dt)^2]. Substituting the given values, we get speed = sqrt[(cos(2^3))^2 + (3sin(2^2))^2] = 2.3166.

c. To find the total distance traveled by the object over the time interval 0<=t<=1, we can use the formula for distance: d = integral (0 to 1) sqrt[(dx/dt)^2 + (dy/dt)^2] dt. Substituting the given values, we get d = integral (0 to 1) sqrt[(cos(t^3))^2 + (3sin(t^2))^2] dt = 1.458.

d. To find the position of the object at time t=3, we can use the given position equation (x(t), y(t)) and substitute t=3. This gives us the position (x(3), y(3)) = (cos(3^3), 3sin(3^2)) = (0.998, 3.746).