Parametric equation

Question

A particle is moving around the plane with velocity vector v(t) = (t + sin t+ cos t, 2t−6) for 

−∞< t <∞.

a) If at t = 0, the particle is located at (−1,−1), what is the position function of the particle?

b) At what time(s) will the particle be located on the x -axis?

c) In which general direction is the particle moving at t = 0 (left/right and/or down/up)?

d) Find the equation of the line tangent to the particle’s path at t = 0.

Answer 1

As a matter of notation, let v_x and v_y be the x- and y- components of v.

That is

v_x(t) = t + sin(t) + cos(t)

v_y(t) = 2t - 6

a)

The position function s(t) is the integral of velocity.

s(t) = ( ∫(t + sin(t) + cos(t))dt, ∫(2t - 6)dt )

= ( t²/2 - cos(t) + sin(t) + C, t² - 6t + D )

for any constants C and D.

Those two constants are determined from the initial condition.

(-1, -1) = s(0) = (0²/2 - cos(0) + sin(0) + C, 0² - 6×0 + D) = (-1 + C, D)

Therefore C = 0, D = -1.

So the function of the position is

s(t) = (t²/2 - cos(t) + sin(t), t² - 6t - 1)

b)

The particle will be on the x-axis whenever the y-coordinate is 0.

t² - 6t - 1 = 0

t = (6 ± √(6² + 4))/2 = 3 ± √10

The particle will be on the x-axis at two times, first at time 3-√10 then 3+√10

c)

v(0)) = (0 + sin(0) + cos(0), 2×0 - 6) = (1, -6)

That means it is moving in the positive x-direction and negative y-direction.

That is, right and down.

d)

Let s_x(t) and s_y(t) be the x and y-coordinates of s(t).

Then the tangent to the curve s(t) is the function

s'(t) = ds_y/ds_x = (ds_y/dt)/(ds_x/dt) = v_y/v_x = (t + sin(t) + cos(t))/(2t - 6)

The tangent line at time t = 0 going through the point (-1,-1) will be of the form

y + 1 = s'(0)(x + 1)

We can evaluate

s'(0) = (0 + sin(0) + cos(0))/(2×0 - 6) = -1/6