Motion with Parametric Equations (College Board AP® Calculus BC): Revision Note

Motion with parametric equations

How do I model a particle moving in 2D using parametric equations?

• A particle moving in two-dimensions follows a path (curve) in the $xy$-plane

  • This path can be given parametrically as $x\left(t\right)$ and $y\left(t\right)$

    • $x\left(t\right)$ is the displacement in the $x$-direction

    • $y\left(t\right)$ is the displacement in the $y$-direction

  • Displacement is measured relative to a fixed origin, $O$

  • The parameter, $t$, represents time, where $t\ge 0$

  • The coordinates of the particle's position at time $t$ are $\left(x\left(t\right),y\left(t\right)\right)$

• The direction of motion is the direction along the curve as $t$ increases

• The initial position of the particle is the point at which $t=0$

  • The coordinates of the initial position are $\left(x\left(0\right),y\left(0\right)\right)$

    • This may not be at the origin

• Sometimes a time interval for the motion is given, $t_1\le t\le t_2$

How do I find the velocity and acceleration in 2D?

• Recall the ideas in the study guide on 'Motion in a Straight Line'

  • Velocity, $v$, is the rate of change of displacement, $s$, with respect to time

    • $v=\frac{ds}{dt}$

  • Acceleration, $a$, is the rate of change of velocity, $v$, with respect to time

    • $a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$

• In two dimensions, the displacement, velocity and acceleration of a particle each have two components in perpendicular directions

  • Displacement splits into $x$ and $y$ components

    • Often horizontal and vertical

  • Derivatives of these are as follows:

How do I calculate the speed of a particle in 2D?

• The speed of a particle at time $t$ is given by the formula:

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

• This is because speed is the magnitude (hypotenuse) of a velocity triangle formed out of:

  • the velocity of the particle in the $x$-direction, $\frac{dx}{dt}$, at time $t$

  • and the velocity of the particle in the $y$-direction , $\frac{dy}{dt}$, at time $t$

    • By Pythagoras' theorem, this gives $\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}$

• The velocity triangle also gives the direction of motion of the particle at that time

• The particle is at rest if the speed is zero

How do I calculate the slope of a line that is tangent to the path of the particle?

• The slope of the line that is tangent to the path of the particle (a tangent to the curve) at time $t$ is given by the formula:

$\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{y\text{'}\left(t\right)}{x\text{'}\left(t\right)}$

• See the study guide on 'Derivatives of Parametric Equations'

• This is the same slope as the velocity triangle at that time (see above)

  • Note that the direction of motion could be either way along the tangent

    • It will follow the direction in which $t$ is increasing

How do I use indefinite integration for particles moving in 2D?

• Indefinite integration can be used to find

  • displacement, $x\left(t\right)$ or $y\left(t\right)$, by integrating velocity, $x\text{'}\left(t\right)$ or $y\text{'}\left(t\right)$

  • velocity, $x\text{'}\left(t\right)$ or $y\text{'}\left(t\right)$, by integrating acceleration, $x\text{'}\text{'}\left(t\right)$ or $y\text{'}\text{'}\left(t\right)$

• Always remember to add a constant of integration each time you integrate

  • Then use information in the question to find the constant

    • e.g. you may be told that $x=5$ when $t=0$

• Indefinite integration is good when you need an algebraic expression in terms of time $t$

  • You can then evaluate it at different times

How do I use definite integration for particles moving in 2D?

• If an algebraic expression in terms of $t$ is not required but a specific value of a quantity is required (e.g. find $x$ at time $t=2$) then definite integration can be used

  • Below are the commonly used definite integrals in motion questions:

• Definite integrals often need rearranging to find the specific value of a quantity at either end of the time interval

  • e.g. ${\int }_{t_1}^{t_2}x\text{'}\left(t\right)dt=x\left(t_2\right)-x\left(t_1\right)$ rearranges to either

    • $x\left(t_2\right)=x\left(t_1\right)+{\int }_{t_1}^{t_2}x\text{'}\left(t\right)dt$

    • or $x\left(t_1\right)=x\left(t_2\right)-{\int }_{t_1}^{t_2}x\text{'}\left(t\right)dt$

How do I calculate the distance traveled by a particle in 2D?

• The distance traveled by a particle between time $t_1$and time $t_2$ is the arc length of the path traveled, given by the definite integral:

  • ${\int }_{t_1}^{t_2}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

    • See the study guide on 'Arc Lengths of Parametric Equations'

  • This is always positive