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Week 7 Discussion Notes

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Conservative Vector Fields

A vector field F⃗\vec{F} is conservative on a domain D\mathscr{D} if there is a scalar-valued function ff such that ∇f=F⃗\nabla f = \vec{F} on all of D\mathscr{D}.

If F⃗\vec{F} is conservative with potential function ff, you can calculate line integrals really quickly. If C\mathscr{C} is a curve starting at a point Q⃗\vec{Q} and ending at P⃗\vec{P}, then

∫CF⃗⋅dr⃗=f(P⃗)−f(Q⃗).\int_{\mathscr{C}} \vec{F} \cdot \diff\vec{r} = f\p{\vec{P}} - f\p{\vec{Q}}.

In particular, if C\mathscr{C} is a closed curve (i.e., a curve that starts and ends at the same point), then ∫CF⃗⋅dr⃗=0\int_{\mathscr{C}} \vec{F} \cdot \diff\vec{r} = 0.

Example 1.

Calculate ∫C⟨x,y⟩⋅dr⃗\int_{\mathscr{C}} \ang{x, y} \cdot \diff\vec{r}, where C\mathscr{C} is the curve below:
Solution.

When you see a very complicated curve like the one above, you almost never want to try to parametrize and integrate over each segment. Instead, you'll hope that there's a potential function and integrate using that instead. You can eyeball a potential function for ⟨x,y⟩\ang{x, y} here:

f(x,y)=x2+y22.f\p{x, y} = \frac{x^2 + y^2}{2}.

So, ⟨x,y⟩\ang{x, y} is conservative, which means we can calculate the line integral using the endpoints:

∫C⟨x,y⟩⋅dr⃗=f(1,0)−f(0,0)=12.\int_{\mathscr{C}} \ang{x, y} \cdot \diff\vec{r} = f\p{1, 0} - f\p{0, 0} = \boxed{\frac{1}{2}}.

The domain is very important when determining whether a vector field is conservative or not.

Example 2.

Recall the vortex field:

F⃗(x,y)=∇arctan⁡(yx)=⟨−y,x⟩x2+y2\vec{F}\p{x, y} = \nabla\arctan\p{\frac{y}{x}} = \frac{\ang{-y, x}}{x^2 + y^2}

Be careful, though: arctan⁡(yx)\arctan\p{\frac{y}{x}} isn't even continuous on all of R2∖{(0,0)}\R^2 \setminus \set{\p{0, 0}}. arctan⁡(yx)\arctan\p{\frac{y}{x}} is the angle of (x,y)\p{x, y} measured from the positive real axis, so

lim⁡y→0+arctan⁡(y1)=0butlim⁡y→0−arctan⁡(y1)=2π.\lim_{y\to0^+} \arctan\p{\frac{y}{1}} = 0 \quad\text{but}\quad \lim_{y\to0^-} \arctan\p{\frac{y}{1}} = 2\pi.

F⃗\vec{F} is not conservative on R2∖{(0,0)}\R^2 \setminus \set{\p{0, 0}}, since if C\mathscr{C} is the unit circle oriented counter-clockwise, then parametrizing it via r⃗(θ)=⟨cos⁡θ,sin⁡θ⟩\vec{r}\p{\theta} = \ang{\cos\theta, \sin\theta},

∫CF⃗⋅dr⃗=∫02π⟨−sin⁡θ,cos⁡θ⟩cos⁡2θ+sin⁡2θ⋅⟨−sin⁡θ,cos⁡θ⟩ dθ=∫02πsin⁡2θ+cos⁡2θ dθ=∫02πdθ=2π.\begin{aligned} \int_\mathscr{C} \vec{F} \cdot \diff\vec{r} &= \int_0^{2\pi} \frac{\ang{-\sin\theta, \cos\theta}}{\cos^2\theta + \sin^2\theta} \cdot \ang{-\sin\theta, \cos\theta} \,\diff\theta \\ &= \int_0^{2\pi} \sin^2\theta + \cos^2\theta \,\diff\theta \\ &= \int_0^{2\pi} \diff\theta \\ &= 2\pi. \end{aligned}

(Remember that if you integrate a conservative vector field over a closed curve, you should get 00.)

Because arctan⁡(yx)\arctan\p{\frac{y}{x}} is the angle, though, you can calculate line integrals of the vortex field just by counting the number of times the curve loops around the origin: