$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Motion in Space and Curvature

## Velocity and Acceleration

If the position of an object in space is described by a vector-valued function $$\vec{r}(t) = \langle x(t),y(t),z(t) \rangle$$, the velocity of the object is the first derivative of the position function, and its acceleration is the second derivative:

\begin{align} \textrm{Velocity: } &\vec{v}(t) = \vec{r}\ '(t)\\ \textrm{Speed: } &|\vec{v}(t)| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\\ \textrm{Acceleration: } &\vec{a}(t) = \vec{v}\ '(t) = \vec{r}\ ''(t) \end{align}

Note that the velocity is a vector, and the speed is a scalar (the magnitude of the velocity vector).

If $$\vec{r}(t)=t\ \hat{i} + t^3\ \hat{j} + t^2\ \hat{k}$$, find $$\vec{v}(t)$$ and $$\vec{a}(t)$$.

#### Solution

Find the first and second derivatives of $$\vec{r}(t)$$:

\ans{\begin{align} \vec{v}(t) &= \hat{i} + 3t^2\ \hat{j} + 2t\ \hat{k}\\ \vec{a}(t) &= 6t\ \hat{j} + 2\ \hat{k} \end{align}}

If $$\vec{r}(t)=3\cos t\ \hat{i} + 3\sin t\ \hat{j}$$ for $$0\leq t\leq 2\pi$$, find the velocity $$\vec{v}(t)$$, the speed $$|\vec{v}(t)|$$, and the acceleration $$\vec{a}(t)$$.

#### Solution

Note that $$x(t)^2 + y(t)^2 = 9$$, so this describes a circle in the xy plane centered at the origin with radius 3.

\ans{\begin{align} \vec{v}(t) &= -3\sin t\ \hat{i} + 3\cos t\ \hat{j} \longrightarrow |\vec{v}(t)| = 3\\ \vec{a}(t) &= -3\cos t\ \hat{i} - 3\sin t\ \hat{j} \end{align}}

## Curvature

Suppose an object is traveling along a curving path through space (think of a car driving along a mountain path). We can use the unit tangent and normal vectors at each point to describe this motion.

Note that $$\vec{T} \perp \vec{N} \longrightarrow \vec{T} \cdot \vec{N} = 0$$.

1. Unit Tangent Vector: we've already seen this $\ans{\vec{T}(t) = \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|}}$
2. Curvature: how quickly the direction (tangent vector) changes $\ans{\kappa = \dfrac{|\vec{T}\ '(t)|}{|\vec{r}\ '(t)|} = \dfrac{|\vec{v} \times \vec{a}|}{|\vec{v}|^3}}$ (note that the curvature is a scalar)
3. Unit Normal Vector: as long as $$\kappa \neq 0$$ $\ans{\vec{N}(t) = \dfrac{\vec{T}\ '(t)}{|\vec{T}\ '(t)|}}$
###### Curvature of a Circle

If $$\vec{r}(t)=\langle R\cos t, R\sin t \rangle$$ (a circle of radius $$R$$ centered at the origin), find an expression for the unit tangent vector and the curvature.

#### Solution

\begin{align} \vec{T}(t) &= \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} = \dfrac{1}{R} \langle -R\sin t, R\cos t \rangle\\ &= \ans{\langle -\sin t,\cos t \rangle}\\ \\ \kappa &= \dfrac{|\vec{T}\ '(t)|}{|\vec{r}\ '(t)|} = \ans{\dfrac{1}{R}} \textrm{ (note: constant curvature)} \end{align}
###### Helix

A helix is defined by $$\vec{r}(t)=\langle 2\cos t, 2\sin t, 3t \rangle$$.

Find the curvature, unit tangent, and unit normal.

#### Solution

1. Curvature: $$\kappa = \dfrac{|\vec{v} \times \vec{a}|}{|\vec{v}|^3}$$ \begin{align} \vec{v} &= \langle -2\sin t,2\cos t,3 \rangle\\ \vec{a} &= \langle -2\cos t,-2\sin t,0 \rangle\\ \\ \vec{v} \times \vec{a} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ -2\sin t & 2\cos t & 3\\ -2\cos t & -2\sin t & 0 \end{vmatrix}\\ \\ &= 6\sin t\ \hat{i} - 6\cos t\ \hat{j} + 4\ \hat{k}\\ \\ |\vec{v} \times \vec{a}| &= \sqrt{36\sin^2 t + 36\cos^2 t + 16} = \sqrt{36+16} = 2\sqrt{13}\\ \\ |\vec{v}|^3 &= \sqrt{4\sin^2 t + 4\cos^2 t + 9} = \sqrt{4+9} = \sqrt{13}\\ \\ &\implies \ans{\kappa = \dfrac{|\vec{v} \times \vec{a}|}{|\vec{v}|^3} = \dfrac{2\sqrt{13}}{13\sqrt{13}} = \dfrac{2}{13}} \end{align}
2. Unit Tangent: $$\vec{T}(t) = \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|}$$ \begin{align} \vec{r}\ '(t) &= \langle -2\sin t,2\cos t,3 \rangle\\ |\vec{r}\ '(t)| &= \sqrt{13}\\ \\ &\implies \ans{\vec{T}(t) = \left\langle -\dfrac{2\sqrt{13}}{13}\sin t, \dfrac{2\sqrt{13}}{13}\cos t,\dfrac{3\sqrt{13}}{13} \right\rangle} \end{align}
3. Unit Normal: $$\vec{N}(t) = \dfrac{\vec{T}\ '(t)}{|\vec{T}\ '(t)|}$$ \begin{align} \vec{T}(t) &= \left\langle -\dfrac{2\sqrt{13}}{13}\sin t, \dfrac{2\sqrt{13}}{13}\cos t,\dfrac{3\sqrt{13}}{13} \right\rangle\\ \vec{T}\ '(t) &= \left\langle -\dfrac{2\sqrt{13}}{13}\cos t, -\dfrac{2\sqrt{13}}{13}\sin t,0 \right\rangle\\ |\vec{T}\ '(t)| &= \dfrac{2\sqrt{13}}{13}\\ \\ &\implies \ans{\vec{N}(t) = \left\langle -\cos t, -\sin t,0 \right\rangle} \end{align}

## Components of Acceleration

Consider a car driving along a road; the acceleration of the car includes changes in speed (speeding up or slowing down) and changes in direction (turning), so we can talk about the tangential component of acceleration (speed changes) and the normal component of acceleration (direction changes).

\begin{align} \vec{a} &= a_N \vec{N} + a_T \vec{T}\\ \textrm{where } a_N &= \vec{a} \cdot \vec{N} = \dfrac{|\vec{a} \times \vec{v}|}{|\vec{v}|}\\ \textrm{and } a_T &= \vec{a} \cdot \vec{T} = \dfrac{\vec{a} \cdot \vec{v}}{|\vec{v}|} \end{align}

###### Car Traveling

If $$\vec{r}(t)=\langle t,t^2 \rangle$$ for $$-2 \leq t \leq 2$$, find the components of the acceleration vector.

#### Solution

First, find the acceleration vector:

$\vec{r}(t) = \langle t,t^2 \rangle \longrightarrow \vec{v}(t) = \langle 1,2t \rangle \longrightarrow \vec{a}(t) = \langle 0,2 \rangle$

Next, find the unit tangent and unit normal vectors:

\begin{align} \vec{T}(t) &= \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} = \dfrac{\langle 1,2t \rangle}{\sqrt{1+4t^2}}\\ \\ \vec{T}\ '(t) &= \left\langle \dfrac{-4t}{(1+4t^2)^{3/2}}, \dfrac{2}{\sqrt{1+4t^2}}-\dfrac{8t^2}{(1+4t^2)^{3/2}} \right\rangle\\ &= \dfrac{1}{(1+4t^2)^{3/2}} \langle -4t,2 \rangle\\ \\ \vec{N}(t) &= \dfrac{\vec{T}\ '(t)}{|\vec{T}\ '(t)|} = \dfrac{1}{\sqrt{4t^2+1}} \langle -2t,1 \rangle\\ \\ \end{align}\\ \ans{\begin{align} a_N &= \vec{a} \cdot \vec{N} = \dfrac{2}{\sqrt{1+4t^2}}\\ a_T &= \vec{a} \cdot \vec{T} = \dfrac{4t}{\sqrt{1+4t^2}} \end{align}} (note: we could have also calculated $$a_N$$ and $$a_T$$ using the alternate formulas above)

## Projectile Motion

### In Two Dimensions: no wind

Suppose we launch a projectile and neglect air resistance, so the only force on the projectile is gravity. Suppose the components of the initial velocity vector are

\begin{align} v_{0x} &= |\vec{v}_0| \cos \theta\\ v_{0y} &= |\vec{v}_0| \sin \theta \end{align}

Since $$\vec{a} = \vec{r}\ ''$$, we can we can integrate to work backwards to the position function $$\vec{r}$$:

\begin{align} \vec{a} &= -g\ \hat{j}\\ \\ \vec{v} &= \int \vec{a} \ dt = -gt\ \hat{j} + \vec{C}\\ &\longrightarrow \vec{C} = \vec{v}_0\\ \vec{v} &= -gt\ \hat{j} + \vec{v}_0\\ \\ \vec{r} &= \int \vec{v}\ dt = -\dfrac{1}{2}gt^2\ \hat{j} + \vec{v}_0 t + \vec{C}\\ &\longrightarrow \vec{C} = \vec{r}_0\\ \\ \vec{r} &= -\dfrac{1}{2}gt^2\ \hat{j} + \vec{v}_0 t + \vec{r}_0 \end{align}

Summary:

\begin{align} \vec{a} &= \langle 0,-g \rangle\\ \\ \vec{v} &= \langle v_{0x},-gt+v_{0y} \rangle\\ \\ \vec{r} &= \langle v_{0x}t + r_{0x}, -\dfrac{1}{2}gt^2+v_{0y}t+r_{0y} \rangle \end{align}

(note: the gravitational constant is $$g=9.81\ m/s^2 = 32.2 \ ft/s^2$$)

###### Projectile Motion

A projectile is launched at 80 ft/s at 30 degrees from the horizontal.

1. How long is the projectile in the air?
2. What is the range?
3. What is the maximum height?

#### Solution

\begin{align} v_{0x} &= 80 \cos 30^{\circ} = 40\sqrt{3}\\ v_{0y} &= 80 \sin 30^{\circ} = 40\\ \\ r_{0x} &= r_{0y} = 0 \ \ \textrm{(by assumption)}\\ \\ \vec{a} &= \langle 0,-32 \rangle\\ \\ \vec{v} &= \langle v_{0x},-32t+v_{0y} \rangle\\ &= \langle 40\sqrt{3},-32t+40 \rangle\\ \\ \vec{r} &= \langle 40\sqrt{3}t, -16t^2+40t \rangle \end{align}
1. Time in the air: find $$t$$ when the $$\hat{j}$$ component is 0 \begin{align} -16t^2+40t &= 0 \longrightarrow t(40-16t) = 0 \longrightarrow t = 0,\dfrac{40}{16}\\ &\implies \ans{t=\dfrac{5}{2}\ s} \end{align}
2. Range: find the $$\hat{i}$$ component when $$t=\dfrac{5}{2}$$ \begin{align} 40\sqrt{3}\left(\dfrac{5}{2}\right) &= 100\sqrt{3}\\ &\implies \ans{\textrm{Range } \approx 173.21 \ ft} \end{align}
3. Max Height: find where the $$\hat{j}$$ component of velocity is 0 \begin{align} -32t+40 &= 0 \longrightarrow t = \dfrac{40}{32} = \dfrac{5}{4}\\ \textrm{at } t &= \dfrac{5}{4}, \ \vec{r} = -16\left(\dfrac{5}{4}\right)^2+40\left(\dfrac{5}{4}\right) &\implies \ans{\textrm{Max Height } = 25 \ ft} \end{align}

### In Three Dimensions: crosswind

Suppose we launch a projectile due east at 300 m/s at 45 degrees above the horizontal, and there is a crosswind from south to north that adds 0.36 m/s$$^2$$ to the projectile. Where does this projectile land?

In three dimensions, the acceleration vector is

$\vec{a} = \langle 0,0.36,-9.81 \rangle$

Thus, the velocity vector is

\begin{align} \vec{v} &= \langle v_{0x},0.36t+v_{0y},-9.81t+v_{oz} \rangle\\ &= \langle 150\sqrt{2}, 0.36t,-9.81t+150\sqrt{2} \rangle \end{align}

Finally, the position vector is

$\vec{r} = \langle 150\sqrt{2}t,0.18t^2,-4.9t^2+150\sqrt{2}t \rangle$

The projectile lands when the $$\hat{k}$$ component of $$\vec{r}$$ is 0:

\begin{align} -4.9t^2+150\sqrt{2}t &= 0 \longrightarrow t(150\sqrt{2}-4.9t) = 0\\ t &= 0, 43.3\\ \\ \vec{r}(43.3) &= \ans{\langle 9183.7,337.4 \rangle} \end{align}

The projectile lands 9184 ft east and 337.4 ft north of where it was launched.