Differentiate the x and y components of the position vector with respect to time: v(t) = <x'(t), y'(t)>.

Calculate the magnitude of the velocity vector: |v(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}.

Evaluate x'(t) at that time. If x'(t) > 0, the particle is moving to the right.

Evaluate y'(t) at that time. If y'(t) > 0, the particle is moving upwards.

Integrate the velocity vector to find the displacement vector, then add the displacement vector to the initial position vector.

Set y'(t) = 0 and solve for t. This gives the times when the vertical component of velocity is zero.

Set x'(t) = 0 and solve for t. This gives the times when the horizontal component of velocity is zero.

Calculate $\int_{a}^{b} |x'(t)| dt$.

Calculate $\int_{a}^{b} |y'(t)| dt$.

Differentiate the x and y components of the velocity vector with respect to time: a(t) = <x''(t), y''(t)>

p(t) = <x(t), y(t)> = x(t)i + y(t)j

v(t) = <x'(t), y'(t)>

|v(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}

Integrate the absolute value of the horizontal velocity: $\int |x'(t)| dt$

Integrate the absolute value of the vertical velocity: $\int |y'(t)| dt$

Displacement = $\int_{t_1}^{t_2} v(t) dt = <x(t_2) - x(t_1), y(t_2) - y(t_1)>$

a(t) = <x''(t), y''(t)>

T(t) = v(t) / |v(t)|

N(t) = T'(t) / |T'(t)|

Arc Length = $\int_{a}^{b} |v(t)| dt = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} dt$

The sign of x'(t) indicates the direction of horizontal movement: positive means right, negative means left.

The sign of y'(t) indicates the direction of vertical movement: positive means up, negative means down.

Speed is the magnitude (or absolute value) of the velocity vector; it represents how fast the particle is moving, regardless of direction.

The derivative of the velocity vector is the acceleration vector, which describes the rate of change of velocity.

Velocity is the derivative of position, and acceleration is the derivative of velocity. Integration reverses this relationship.

A particle is at rest when both the horizontal and vertical components of its velocity are zero, i.e., x'(t) = 0 and y'(t) = 0.

Integrating the velocity vector gives the displacement vector, representing the change in position of the particle.

Arc length represents the total distance traveled by the particle along its path over a given time interval. It's found by integrating the speed.

A particle changes direction when either x'(t) or y'(t) changes sign. Solve for when x'(t) = 0 or y'(t) = 0 and check for sign changes.

Displacement is the change in position from start to end, while total distance traveled considers the entire path, including any changes in direction.