Calculus with vectors

Differentiating position vectors to find velocity and acceleration vectors, speed, integrating acceleration and velocity vectors using initial conditions, and analysing motion in the plane.

Differentiating

Motion vectors

$$\underset{\sim}{r} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} \qquad \underset{\sim}{v} = \dot{\underset{\sim}{r}} = \begin{pmatrix} \dot x \\ \dot y \end{pmatrix} \qquad \underset{\sim}{a} = \dot{\underset{\sim}{v}} = \begin{pmatrix} \ddot x \\ \ddot y \end{pmatrix}$$

Differentiate each component separately. A dot means differentiation with respect to time.

Example 1●●

$\underset{\sim}{r} = \begin{pmatrix} 3t \\ 4t - t^2 \end{pmatrix}$. Find the velocity and acceleration, the speed at $t = 2$, and the highest point.

$\underset{\sim}{v} = \begin{pmatrix} 3 \\ 4 - 2t \end{pmatrix}$ and $\underset{\sim}{a} = \begin{pmatrix} 0 \\ -2 \end{pmatrix}$.

At $t = 2$, $\underset{\sim}{v} = \begin{pmatrix} 3 \\ 0 \end{pmatrix}$, so the speed is $3$.

The vertical velocity is $0$ at $t = 2$, so the highest point is $(6, 4)$.

Integrating

Going the other way, integrate each component and use the initial conditions to find the constants.

Example 2●●

A particle has $\underset{\sim}{a} = \begin{pmatrix} 0 \\ -10 \end{pmatrix}$, starts at the origin, and has initial velocity $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$. Find $\underset{\sim}{v}$ and $\underset{\sim}{r}$.

$\underset{\sim}{v} = \begin{pmatrix} C_1 \\ -10t + C_2 \end{pmatrix}$, and $t = 0$ gives $C_1 = 6$, $C_2 = 8$. So $\underset{\sim}{v} = \begin{pmatrix} 6 \\ 8 - 10t \end{pmatrix}$.

$\underset{\sim}{r} = \begin{pmatrix} 6t + D_1 \\ 8t - 5t^2 + D_2 \end{pmatrix}$, and $\underset{\sim}{r}(0) = \underset{\sim}{0}$ gives $D_1 = D_2 = 0$. So $\underset{\sim}{r} = \begin{pmatrix} 6t \\ 8t - 5t^2 \end{pmatrix}$.

Flashcards

Velocity vector

$\underset{\sim}{v} = \dfrac{d\underset{\sim}{r}}{dt}$, component by component.

Acceleration vector

$\underset{\sim}{a} = \dfrac{d\underset{\sim}{v}}{dt}$

Speed at time $t$

$|\underset{\sim}{v}(t)| = \sqrt{\dot x^2 + \dot y^2}$

Practice

$\underset{\sim}{r} = \begin{pmatrix} 3t \\ 4t - t^2 \end{pmatrix}$. Find the velocity at $t = 2$. Enter it as $x, y$.
For the same $\underset{\sim}{r}$, find the acceleration. Enter it as $x, y$.
For the same $\underset{\sim}{r}$, find the speed at $t = 0$.
For the same $\underset{\sim}{r}$, find the greatest height.
In Example 2, find the velocity at $t = 2$. Enter it as $x, y$.
In Example 2, find the position at $t = 2$. Enter it as $x, y$.
In Example 2, when does the particle stop rising?
If $\underset{\sim}{r} = \begin{pmatrix} \cos t \\ \sin t \end{pmatrix}$, the speed is:
When integrating $\underset{\sim}{a}$ to find $\underset{\sim}{v}$, the constants come from: