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Part 16 - Vector fields

LMA017 Date: October 3, 2023 4 min read

Introduction

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

Definition 1 (Vector field)

A vector field is a function, FF:

F:D⊂R2→R2 ∣ 2DF:D⊂R3→R3 ∣ 3DF: D \subset \mathbb{R}^2 \rarr \mathbb{R}^2 \ | \ \text{2D} \newline F: D \subset \mathbb{R}^3 \rarr \mathbb{R}^3 \ | \ \text{3D}
Example 1

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

The graph for this looks like:

Gradient field

By this definiton, we can say that.

Let f:D⊂R2→Rf: D \subset \mathbb{R}^2 \rarr \mathbb{R} be a differentiable function. The gradient is ∇f(x,y)=⟨fx(x,y),fy(x,y)⟩\nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle.

The gradient is therefore a vector field!

∇f:D→R2\nabla f: D \rarr \mathbb{R}^2

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

Example 2

Find ∇f\nabla f for f(x,y)=x2y−y3f(x, y) = x^2 y - y^3

∇f(x,y)=⟨fx(x,y),fy(x,y)⟩∇f(x,y)=⟨2xy,x2−3y2⟩\nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle \newline \nabla f(x, y) = \langle 2xy, x^2 - 3y^2 \rangle \newline

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

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

Definition 2 (Conservative vector field)

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

In this case, ff, is called a potential for FF.

Integrating over vector fields

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

If a particle moves along the curve, the work done by FF, to move it a distance:

∫CF⋅dr\int_C F \cdot dr

We call this the line integral of FF over CC.

Definition 3 (Line integral over a vector field)

If CC is parameterized as r⃗(t)=⟨x(t),y(t),z(t)⟩\vec{r}(t) = \langle x(t), y(t), z(t) \rangle in the interval, a≤t≤ba \leq t \leq b. Then:

∫CF⋅dr=∫abF(x(t),y(t),z(t))⋅r′⃗(t) dt\int_C F \cdot dr = \int_a^b F(x(t), y(t), z(t)) \cdot \vec{r^\prime}(t)\ dt
Example 3

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

Let’s first parameterize our curve.

r⃗(t)=⟨x0+ta,y0+tb,z0+tc⟩r⃗(t)=⟨2+t(3−2),0+t(4−0),0+t(5−0)⟩r⃗(t)=⟨2+t,4t,5t⟩ ∣ 0≤t≤1\vec{r}(t) = \langle x_0 + ta, y_0 + tb, z_0 + tc \rangle \newline \vec{r}(t) = \langle 2 + t(3 - 2), 0 + t(4 - 0), 0 + t(5 - 0) \rangle \newline \vec{r}(t) = \langle 2 + t, 4t, 5t \rangle \ | \ 0 \leq t \leq 1

Which means:

r′⃗(t)=⟨1,4,5⟩ ∣ 0≤t≤1\vec{r^\prime}(t) = \langle 1, 4, 5 \rangle \ | \ 0 \leq t \leq 1

Using our definition:

∫CF⋅dr=∫01⟨4t,5t,2+t⟩⋅⟨1,4,5⟩ dt\int_C F \cdot dr = \int_0^1 \langle 4t, 5t, 2 + t \rangle \cdot \langle 1, 4, 5 \rangle \ dt∫014t+20t+10+5t dt\int_0^1 4t + 20t + 10 + 5t\ dt∫0129t+10 dt\int_0^1 29t + 10\ dt292t2+10t∣t=0t=1\dfrac{29}{2} t^2 + 10t \bigg\rvert_{t = 0}^{t = 1}292+10\boxed{\dfrac{29}{2} + 10}
Example 4

Evaluate ∫CF⋅dr\int_C F \cdot dr for F(x,y,z)=xyi⃗+yzj⃗+zxk⃗F(x, y, z) = xy \vec{i} + yz \vec{j} + zx \vec{k}.

CC is given by, r⃗(t)=⟨t,t2,t3⟩\vec{r}(t) = \langle t, t^2, t^3 \rangle. For 0≤t≤10 \leq t \leq 1.

Let’s first rewrite FF:

F(x,y,z)=⟨xy,yz,zx⟩F(x, y, z) = \langle xy, yz, zx \rangle

Let’s differentiate:

r′⃗(t)⟨1,2t,3t2⟩\vec{r^\prime}(t) \langle 1, 2t, 3t^2 \rangle

Use our definition:

∫CF⋅dr=∫01⟨t3,t5,t4⟩⋅⟨1,2t,3t2⟩ dt\int_C F \cdot dr = \int_0^1 \langle t^3, t^5, t^4 \rangle \cdot \langle 1, 2t, 3t^2 \rangle \ dt∫01t3+2t6+3t6 dt\int_0^1 t^3 + 2t^6 + 3t^6\ dt∫015t6+t3 dt\int_0^1 5t^6 + t^3\ dt57t7+14t4∣t=0t=1\dfrac{5}{7} t^7 + \dfrac{1}{4} t^4 \bigg\rvert_{t = 0}^{t = 1}57+14\boxed{\dfrac{5}{7} + \dfrac{1}{4}}