Part 16 - Vector fields

Introduction

In this part we’ll cover so-called vector fields.

Definition

A vector field is a function, $F$: $$ F: D \subset \mathbb{R}^2 \rarr \mathbb{R}^2 \ | \ \text{2D} \\ F: D \subset \mathbb{R}^3 \rarr \mathbb{R}^3 \ | \ \text{3D} $$

Example

Consider $F(x, y) = (-y, x)$ (which is the same as $-y \vec{i} + x\vec{j}$).

The graph for this looks like:

Gradient field

By this definiton, we can say that.

Let $f: D \subset \mathbb{R}^2 \rarr \mathbb{R}$ be a differentiable function. The gradient is $\nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle$.

The gradient is therefore a vector field!

$$ \nabla f: D \rarr \mathbb{R}^2 $$

Let’s quickly just do a simple gradient example to refresh our memory.

Example

Find $\nabla f$ for $f(x, y) = x^2 y - y^3$ $$ \nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle \\ \nabla f(x, y) = \langle 2xy, x^2 - 3y^2 \rangle \\ $$

Let’s also remember that $\nabla f(x, y)$ is perpendicular to the level curves of $f$.

Which means, $\nabla f$ has larger (vector) length when the level curves are tighter together.

Definition

A vector field, $F$, is called conservative, if there is a function, $f$, for which $\nabla f = F$.

In this case, $f$, is called a potential for $F$.

Integrating over vector fields

Suppose the vector field, $F$, represents some force on the plane/space and suppose there is a string, curved of the shape of a curve, $C$.

If a particle moves along the curve, the work done by $F$, to move it a distance: $$ \int_C F \cdot dr $$

We call this the line integral of $F$ over $C$.

Definition

If $C$ is parameterized as $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$ in the interval, $a \leq t \leq b$. Then:

$$ \int_C F \cdot dr = \int_a^b F(x(t), y(t), z(t)) \cdot \vec{r^\prime}(t)\ dt $$

Example

Evaluate $\int_C F \cdot dr$ for $F(x, y, z) = (y, z, x)$ on the straight line from $(2, 0, 0)$ to $(3, 4, 5)$.

Let’s first parameterize our curve.

$$ \vec{r}(t) = \langle x_0 + ta, y_0 + tb, z_0 + tc \rangle \\ \vec{r}(t) = \langle 2 + t(3 - 2), 0 + t(4 - 0), 0 + t(5 - 0) \rangle \\ \vec{r}(t) = \langle 2 + t, 4t, 5t \rangle \ | \ 0 \leq t \leq 1 $$

Which means: $$ \vec{r^\prime}(t) = \langle 1, 4, 5 \rangle \ | \ 0 \leq t \leq 1 $$

Using our definition: $$ \int_C F \cdot dr = \int_0^1 \langle 4t, 5t, 2 + t \rangle \cdot \langle 1, 4, 5 \rangle \ dt $$

$$ \int_0^1 4t + 20t + 10 + 5t\ dt $$

$$ \int_0^1 29t + 10\ dt $$

$$ \dfrac{29}{2} t^2 + 10t \bigg\rvert_{t = 0}^{t = 1} $$

$$ \boxed{\dfrac{29}{2} + 10} $$

Example

Evaluate $\int_C F \cdot dr$ for $F(x, y, z) = xy \vec{i} + yz \vec{j} + zx \vec{k}$.

$C$ is given by, $\vec{r}(t) = \langle t, t^2, t^3 \rangle$. For $0 \leq t \leq 1$.

Let’s first rewrite $F$: $$ F(x, y, z) = \langle xy, yz, zx \rangle $$

Let’s differentiate: $$ \vec{r^\prime}(t) \langle 1, 2t, 3t^2 \rangle $$

Use our definition: $$ \int_C F \cdot dr = \int_0^1 \langle t^3, t^5, t^4 \rangle \cdot \langle 1, 2t, 3t^2 \rangle \ dt $$

$$ \int_0^1 t^3 + 2t^6 + 3t^6\ dt $$

$$ \int_0^1 5t^6 + t^3\ dt $$

$$ \dfrac{5}{7} t^7 + \dfrac{1}{4} t^4 \bigg\rvert_{t = 0}^{t = 1} $$

$$ \boxed{\dfrac{5}{7} + \dfrac{1}{4}} $$