Motion with Vector-Valued Functions (College Board AP® Calculus BC): Revision Note

Motion with vector-valued functions

How can a displacement vector be used to model the motion of a particle in 2D?

• A particle moving along a two-dimensional path in the $xy$-plane can be modelled by a displacement vector $<x\left(t\right),y\left(t\right)>$ that depends on time, $t$

  • The first component is the $x$-coordinate of the particle at time $t$

  • The second component is the $y$-coordinate of the particle at time $t$

• These coordinates move with time and are measured relative to an origin, $O$

  • $x\left(t\right)$ and $y\left(t\right)$ are parametric equations of the path of the particle

  • Motion using vector-valued functions can be thought of as motion with parametric equations using vector notation

    • See the study guide on 'Motion with Parametric Equations'

• The magnitude of the displacement vector is $\sqrt{{\left(x\left(t\right)\right)}^2+{\left(y\left(t\right)\right)}^2}$

How do I find the velocity and acceleration vectors from a displacement vector?

• The velocity vector, $<x\text{'}\left(t\right),y\text{'}\left(t\right)>$, is the derivative of the displacement vector $<x\left(t\right),y\left(t\right)>$

  • The velocity vector always points in the direction of motion of the particle

  • The magnitude of the velocity vector, $\sqrt{{\left(x\text{'}\left(t\right)\right)}^2+{\left(y\text{'}\left(t\right)\right)}^2}$, is the speed of the particle

• The acceleration vector, $<x\text{'}\text{'}\left(t\right),y\text{'}\text{'}\left(t\right)>$, is the derivative of the velocity vector $<x\text{'}\left(t\right),y\text{'}\left(t\right)>$

  • This makes it the second derivative of the displacement vector, $<x\left(t\right),y\left(t\right)>$

  • The magnitude of the acceleration vector is $\sqrt{{\left(x\text{'}\text{'}\left(t\right)\right)}^2+{\left(y\text{'}\text{'}\left(t\right)\right)}^2}$

• If $x$ is horizontal and $y$ is vertical:

  • then if $x\text{'}\left(t\right)>0$, the particle moves to the right ( $x\text{'}\left(t\right)<0$ is to the left)

  • and if $y\text{'}\left(t\right)>0$, the particle moves upwards ( $y\text{'}\left(t\right)<0$ is downwards)

How do I find the displacement vector from the velocity or acceleration vector?

• To find the displacement vector from the velocity vector:

  • Use indefinite integration to integrate both components of the velocity vector

  • Add constants of integration to each component

  • Use information in the question to find these constants

    • This gives the velocity vector

• To find the displacement vector from the acceleration vector:

  • Use indefinite integration and add integration constants to each component

  • Use information in the question to find these constants

    • This gives the velocity vector

  • Then use indefinite integration a second time and add a new set of integration constants

  • Use additional information in the question to find these new constants

    • This gives the displacement vector

How do I find the speed, slope of path and total distance traveled?

• The formulas for the speed, slope of path and total distance traveled are explained in the study guide on 'Motion with Parametric Equations' and briefly recapped below

  • The speed of a particle at time $t$ is $\sqrt{{\left(x\text{'}\left(t\right)\right)}^2+{\left(y\text{'}\left(t\right)\right)}^2}$

  • The slope of the line tangent to the path of a particle at time $t$ is $\frac{y\text{'}\left(t\right)}{x\text{'}\left(t\right)}$

  • The total distance traveled between times $t_1$ and $t_2$ is ${\int }_{t_1}^{t_2}\sqrt{{\left(x\text{'}\left(t\right)\right)}^2+{\left(y\text{'}\left(t\right)\right)}^2}dt$