Lecture 05      

Motion in Space

APMA E2000

Drew Youngren dcy2@columbia.edu

$\gdef\RR{\mathbb{R}}$ $\gdef\vec{\mathbf}$ $\gdef\bv#1{\begin{bmatrix} #1 \end{bmatrix}}$ $\gdef\proj{\operatorname{proj}}$ $\gdef\comp{\operatorname{comp}}$

Announcements

  • Recitation 03 this week.
  • No quiz this week.
  • HW3 due next Tues

1-minute review

Curves
\[\lim_{t \to a} \vec r(t) = \bv{\lim\limits_{t \to a} x(t) \\\lim\limits_{t \to a} y(t) \\ \lim\limits_{t \to a} z(t)} \qquad \text{and} \qquad \vec r'(t) = \bv{x'(t) \\ y'(t) \\ z'(t) } \]

Properties

\[\begin{align*} &\frac{d}{dt} (c\vec r(t)) = c \vec r'(t) \\ &\frac{d}{dt} (\vec r(t) + \vec q(t)) = \vec r'(t) + \vec q'(t) \\\end{align*}\]
Demo

Properties

\[\begin{align*} &\frac{d}{dt} (f(t) \vec r(t)) = f'(t)\vec r(t) + f(t) \vec r'(t) \\ &\frac{d}{dt} (\vec q(t) \cdot \vec r(t)) = \vec q'(t)\cdot\vec r(t) + \vec q(t) \cdot \vec r'(t) \\ &\frac{d}{dt} (\vec q(t) \times \vec r(t)) = \vec q'(t)\times\vec r(t) + \vec q(t) \times \vec r'(t) \\ \end{align*}\]

Properties

\[\begin{align*} &\frac{d}{dt} \vec r(f(t)) = f'(t)\vec r'(f(t)) \\ \end{align*}\]
Example

Let $\vec r(t)$ parameterize a curve on the surface of a sphere centered at the origin. Show that the tangent vector is orthogonal to the position vector at every point.

Integration

Integrals are limits of sums. Vectors can be added and scaled, so this definition also goes right through, though the interpretation is less clear. \[\int_a^b \vec r(t)\,dt = \lim_{N\to\infty} \sum_{i=1}^N \vec r(t_i^*) \,\Delta t\]
\[ = \left\langle \int_a^b x(t)\,dt, \int_a^b y(t)\,dt ,\int_a^b z(t)\,dt \right\rangle. \]

Note: this is not an area.

The Fundamental Theorem

\[f(b) - f(a) = \int_a^b f'(x)\,dx\]
\[\vec r(t) - \vec r(t_0) = \int_{t_0}^t \vec r'(u)\,du \]
\[\vec r(t) = \vec r(t_0) + \int_{t_0}^t \vec r'(u)\,du \]

Motion

Position, Velocity, Acceleration

Suppose the position of a particle in space at time $t$ is given by $\vec r(t)$. Its derivative $\vec r'(t) = \vec v(t)$ is called the velocity, and its second derivative $\vec r''(t) = \vec a(t)$ is the acceleration. \[\vec a(t) = \vec v'(t) = \vec r''(t) \]
Speed is the magnitude of velocity $v = |\vec v(t)|$.
You are likely familiar with Newton's Second Law of motion \[\vec F = m \,\vec a.\]
Quick exercises

1. Show that a particle that changes direction has nonzero acceleration.

Use contradiction. Assume $\vec a (t) = \vec 0$. Then $\vec v(t) = \vec c$, a constant, so the particle does not change direction.

Quick exercises

2. Is the converse true?

No. Consider the straight-line path (for $t > 0$) $ \vec r(t) = \langle t^2, t^2, t^2 \rangle $. It has direction \[ \frac{\vec r'(t)}{|\vec r'(t)|} = \frac{\langle 1,1,1 \rangle}{\sqrt{3}} \] which does not change, but $\vec r''(t) = \langle 2,2,2 \rangle \neq \vec 0$.

Quick exercises

3. Find an expression for the range formula, i.e., the distance a projectile fired from the ground with initial speed $v_0$ at angle $\alpha$ will travel before landing.

Example

A particle initially at rest at the origin is subjected to an acceleration \[\vec a(t) = \begin{cases} \vec i - t\,\vec j, & t\leq 6 \\ \vec 0, & t > 6 \end{cases}. \] Find its position at $t=10$.

Solution

Learning Outcomes

You should be able to...
  • Formulate the Fundamental Theorem of Calculus for vector-valued functions.
  • Identify position, velocity, speed, and acceleration using the machinery of vector-valued functions.
  • Solve initial-value motion problems with piecewise-defined acceleration functions.
  • Try your hand at the silly rocket game.