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.
Contents
Velocity
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 $t = 2?$
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,
$v_{\text{total}} = \sqrt{ \dot{x}^2 + \dot{y}^2 } = \sqrt{(16)^2 + (36)^2} = 39.40.\ _\square$
Acceleration
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 $\vec{r} = \big(4t^2 -2t+1, -3t^2\big).$
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:
$a_{\text{total}} = \sqrt{ \ddot{x}^2 + \ddot{y}^2 } =\sqrt{ (8)^2 + (-6)^2 } =10.\ _\square$
References
- Ruryk, . Oscillating pendulum. Retrieved May 18, 2016, from https://commons.wikimedia.org/wiki/File:Oscillating_pendulum.gif