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:
\[\begin{align} \dot{x} &= 8t -2\\ \ddot{x} &= 8\\ \dot{y} &= -6t\\ \ddot{y} &= -6. \end{align}\]
Now plug these into the acceleration magnitude equation:
\[a_{\text{total}} = \sqrt{ \ddot{x}^2 + \ddot{y}^2 } =\sqrt{ (8)^2 + (-6)^2 } =\sqrt{113}.\ _\square\]
References
- Ruryk, . Oscillating pendulum. Retrieved May 18, 2016, from https://commons.wikimedia.org/wiki/File:Oscillating_pendulum.gif