Parametric Equations - Velocity and Acceleration

The speed of a particle whose motion is described by a parametric equation is given in terms of the time derivatives of the $x$-coordinate, $\dot{x},$ and $y$-coordinate, $\dot{y}:$

$$v_{\text{total}} = \sqrt{ \dot{x}^2 + \dot{y}^2}.$$

The magnitude of the acceleration of a particle whose motion is described by a parametric function is given in terms of the second time derivatives of the $x$-coordinate, $\ddot{x},$ and $y$-coordinate, $\ddot{y}:$

$$a_{\text{total}} = \sqrt{ \ddot{x}^2 + \ddot{y}^2 }.$$

Both of these relations fall out of the definitions of one-dimensional kinematics and vector addition, and can be used to compute these quantities for any particle whose position is known.

The motion of this pendulum is complex mathematically, but the acceleration vector is always the rate of change of the velocity vector. ^[1]

For a vector quantity with two components, like velocity, the resultant magnitude (speed) is

$$v_{\text{total}} = \sqrt{ v_x^2 + v_y^2}.$$

Since velocity is defined to be $\vec{v} = \frac{d\vec{r}}{dt},$ the total speed is

$$v_{total} = \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}.$$

This equation is less headache-inducing if written using Newton's dot notation, by which $\dot{u}$ to represent the first derivative of $u$ with respect to $t$ and $\ddot{u}$ to represent the second derivative of $u$ with respect to $t$.

$$v_{\text{total}} = \sqrt{ \dot{x}^2 + \dot{y}^2}$$

A particle has a position given by $\vec{r} = \big(4t^2, 3t^3 -1\big).$ What is the particle's speed at $t = 2?$

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Write down the derivatives:

$$\dot{x} = 8t,\quad \dot{y} = 9t^2.$$

Evaluate each at $t=2:$

$$\dot{x} = 16,\quad \dot{y} = 36.$$

Finally,

$$v_{\text{total}} = \sqrt{ \dot{x}^2 + \dot{y}^2 } = \sqrt{(16)^2 + (36)^2} = 39.40.\ _\square$$

Since acceleration is a vector, its magnitude is found in quadrature:

$$a_{total}^2 = a_x^2 +a_y^2.$$

In terms of the position, the acceleration is $\vec{a} = \frac{d^2\vec{r}}{dt^2},$ so the total magnitude of the acceleration is

$$a_{total}^2 = \left(\frac{d^2x}{dt^2}\right)^2 + \left(\frac{d^2y}{dt^2}\right)^2.$$

$$a_{\text{total}} = \sqrt{ \ddot{x}^2 + \ddot{y}^2 }$$

Find the magnitude of the acceleration of a particle with position $\vec{r} = \big(4t^2 -2t+1, -3t^2\big).$

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Write down the necessary derivatives:

$$\begin{aligned} \dot{x} &= 8t -2\\ \ddot{x} &= 8\\ \dot{y} &= -6t\\ \ddot{y} &= -6. \end{aligned}$$

Now plug these into the acceleration magnitude equation:

$$a_{\text{total}} = \sqrt{ \ddot{x}^2 + \ddot{y}^2 } =\sqrt{ (8)^2 + (-6)^2 } =10.\ _\square$$

1. Ruryk, . Oscillating pendulum. Retrieved May 18, 2016, from https://commons.wikimedia.org/wiki/File:Oscillating_pendulum.gif