Part 17 - Gradient theorem

Introduction

In this part we’ll cover the fundamental theorem of line integrals (also called Gradient theorem).

Recall the fundamental theorem of calculus:

\int_a^b f^\prime(x) \, dx = f(b) - f(a).

Theorem 1 (Fundamental theorem of line integrals)

For line integrals: Let $f: D \subset \mathbb{R}^2 \rarr \mathbb{R}$

Let $C \subset D$ be parameterized as $\vec{r}(t)$ , for $a \leq t \leq b$ .

\boxed{\int_C \nabla f \cdot dr = f(\vec{r}(b)) - f(\vec{r}(a))}

Notice how this doesn’t depend on the actual curve, but only of the start and endpoint.

Additionally, if we assume that $C$ is closed, meaning the curve starts and ends at the same point:

\int_C \nabla f \cdot dr = 0.

Theorem 2

Let $D \subset \mathbb{R}^2$ be open (every point in $D$ has a small disc around it, which contains $D$ ), connected (consists of one single “piece”).

Let $F: D \rarr \mathbb{R}^2$ be a vector field. Then $F$ is conservative (there exists a $f$ such that $\nabla f = F$ ), if and only if:

\int_C F \cdot dr = 0 \quad \text{for every closed curve } C \subset D.

Proposition 1

NB: This only applies for $\mathbb{R}^2$ .

Suppose $F(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a conservative field. Then:

\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}.

Proof

If $F$ is conservative, then there is a function, $f: D \subset \mathbb{R}^2 \rarr \mathbb{R}$ such that $\nabla f = F$ . In other words, $f_x = P$ and $f_y = Q$ .

\begin{align*} P_y & = \dfrac{\partial P}{\partial y} \newline & = \dfrac{\partial}{\partial y} \left(\dfrac{\partial f}{\partial x} \right) \newline & = \dfrac{\partial^2 f}{\partial y \partial x} \newline & = \dfrac{\partial^2 f}{\partial x \partial y} \newline & = \dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial y} \right) \newline & = \dfrac{\partial Q}{\partial x} \newline & = \boxed{Q_x} \end{align*}

Remark 1

If given a field, $F = P \vec{i} + Q\vec{j}$ and if it is conservative, you can check $P_y = Q_x$ , if this is not the case, this means that $F$ is not conservative.

Definition 1 (Simply connected)

$D \subset \mathbb{R}^2$ is called simply connected if no “holes” exists.

Theorem 3

Suppose $D \subset \mathbb{R}^2$ is open, connected and simply connected. If $F: D \rarr \mathbb{R}^2$ is a vector field, $F = P\vec{i} + Q\vec{j}$ and $P_y = Q_x$ , then $F$ is conservative.

Let’s now do a bunch of examples to understand all of this theory.

Example 1

Determine if the following vector fields are conservative:

\begin{align*} F(x, y) & = (x - y)\vec{i} + (x - 2)\vec{j}, \newline F(x, y) & = (3 + 2xy)\vec{i} + (x^2 - 3y^2)\vec{j}. \end{align*}

As we can see, both of these have the domain $D = \mathbb{R}^2$ . Which means:

In the first case:

P_y = -1, \quad Q_x = 1.

Which means this is not a conservative vector field.

In the second case:

P_y = 2x, \quad Q_x = 2x.

Which means this is a conservative vector field.

Example 2

$F(x, y) = (3 + 2xy)\vec{i} + (x^2 - 3y^2)\vec{j}$ is conservative. Find a potential for it. (Find $f$ , such that $\nabla f = F$ ).

Let, $f: \mathbb{R}^2 \rarr \mathbb{R}$ be a function with $\nabla f = F$ .

Therefore:

\begin{align*} f_x & = 3 + 2xy, \newline f_y & = x^2 - 3y^2. \end{align*}

We generally will use this approach (this doesn’t always work but, it’s a good strategy):

Integrate one of these with respect to “their” variable:

f(x, y) = \int f_x(x, y) \, dx = \int 3 + 2xy \, dx = 3x + x^2y + g(y).

Notice the $g(y)$ , since we can a function depending on y if we differentiate with respect to $x$ .

Let’s now differentiate this with respect to the other variable:

f_y(x, y) = 0 + x^2 + g^\prime(y).

Now we have two equations for $f_y$ :

\begin{align*} x^2 - 3y^2 & = x^2 + g^\prime(y), \newline g^\prime(y) & = -3y^2, \newline g(y) & = -y^3 + C. \end{align*}

Which means:

f(x, y) = 3x + x^2y - y^3 + C.

Let’s do the other way around as well:

f(x, y) = \int f_y(x, y) \, dy = \int x^2 - 3y^2 \, dy = x^2y - y^3 + g(x).

\begin{align*} f_x(x, y) & = 2xy + g^\prime(x), \newline 3 + 2xy & = 2xy + g^\prime(x), \newline g^\prime(x) & = 3, \newline g(x) & = 3x + C, \end{align*}

f(x, y) = x^2y - y^3 + 3x + C \quad \text{same!}

Example 3

Evaluate $\int_C F \cdot dr$ where, $F(x, y) = (3 + 2xy)\vec{i} + (x^2 - 3y^2)\vec{j}$ and $C = \vec{r}(t) = (e^t \sin(t))\vec{i} + (e^t \cos(t))\vec{j}$ .

For $0 \leq t \leq \pi$ .

Since we already know that $F$ is conservative, which means there exists a $\nabla f = F$ , which we calculated in the last question, since the domain of the potential is $\mathbb{R}^2$ as well, we can use the fundamental theorem of line integrals!

\int_C F \cdot dr = \int_C \nabla f \cdot dr = f(\vec{r}(\pi)) - f(\vec{r}(0))

Remember that:

f(x, y) = x^2y - y^3 + 3x + C.

Let’s calculate the start and end point:

\begin{align*} \vec{r}(0) & = \langle 0, 1 \rangle, \newline \vec{r}(\pi) & = \langle 0, -e^{\pi} \rangle, \end{align*}

\begin{align*} f(0, 1) & = -1 + C, \newline f(0, -e^{\pi}) & = -(-e^{\pi})^3 + C = e^{3\pi} + C. \end{align*}

Which means:

\int_C F \cdot dr = (e^{3\pi} + C) - (-1 + C) = \boxed{e^{3\pi} + 1}

Conservation of energy

(This is just a motivation of the name, nothing really important, but I found it cool so).

Suppose $F$ is a force field, suppose that it moves an object with mass, $m$ , along a curve $C$ , parameterized as $\vec{r}(t)$ for $a \leq t \leq b$ . According to Newton’s second law of motion:

F = ma

We can write this as:

F(\vec{r}(t)) = m \vec{r^{\prime\prime}}(t).

The work done by the force field:

W = \int_C F \cdot dr = \int_a^b F(\vec{r}(t)) \cdot \vec{r^\prime}(t) \, dt.

Using Newton’s second law of motion:

\begin{align*} \int_a^b m \vec{r^{\prime\prime}}(t) \cdot \vec{r^\prime}(t) \, dt & = m \int_a^b \vec{r^{\prime\prime}}(t) \cdot \vec{r^\prime}(t) \, dt \newline & = \dfrac{m}{2} \int_a^b \dfrac{d}{dt} \left| \vec{r^\prime}(t) \right|^2 \, dt \newline & = \dfrac{m}{2} \left( \left| \vec{r^\prime}(b) \right|^2 - \left| \vec{r^\prime}(a) \right|^2 \right) \newline & = \dfrac{m}{2} \left( \left| \vec{v}(b) \right|^2 - \left| \vec{v}(a) \right|^2 \right). \end{align*}

Using that $K_E = \dfrac{mv^2}{2}$ :

K_E(b) - K_E(a).

Suppose now that $\nabla f = F$ . The potential energy of the object is $P_E(t) = -f(\vec{r}(t))$

W = \int_C F \cdot dr = \int_C - \nabla P_E \cdot dr = -(P_E(b) - P_E(a)).

Which means:

K_E(b) - K_E(a) = -(P_E(b) - P_E(a)),

\boxed{K_E(b) + P_E(b) = K_E(a) + P_E(a)}

Which is the law of conservation of energy!