A particle moves along the curve defined by the parametric equations x(t) = 2t a

Question

A particle moves along the curve defined by the parametric equations x(t) = 2t and y(t) = 36 - t^2 for time t, 0 lessthanorequalto t lessthanorequalto 6. A laser light on the particle points in the direction of motion and shines on the x-axis. (a) What is the velocity vector of the particle? (b) In terms of t. Write an equation of the line tangent to the graph of the curve at the point (2t, 36 - t^2). (c) Express the x-coordinate of the point on the x-axis that the laser light hits as a function of t. (d) At what speed is the laser light moving along the x-axis at lime t = 3 ? Justify your answer.

Explanation / Answer

x(t) = 2t
y(t) = 36 - t²

1)

v = <dx/dt, dy/dt>
v = <2, -2t>

2)

dy/dx = (dy/dt) / (dx/dt) = -2t/2 = -t

At time t, tangent line has slope -t and passes through point (2t, 36-t²)
Using point slope form:
y - (36-t²) = -t (x - 2t)
y = -tx + 2t² + 36 - t²
y = -tx + (t² + 36)

3)

Since the laser light points in the direction of motion, then laser light shines along the tangent line found in part 2)
Laser light will hit x-axis at same point that tangent intersects x-axis.

Set y = 0 and solve for x:

0 = -tx + (t² + 36)
tx = (t² + 36)
x = (t² + 36)/t

4)

Position of light on x-axis at time t: x(t) = (t² + 36)/t = t + 36/t

Velocity of light on x-axis at time t: v(t) = 1 - 36/t²

velocity at time t=3 is : v(3) = 1 - 36/9 = 1 – 4 = -3

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