Our last topic!!

Suppose we shoot a 1kg object horizontally off of platform that's 64 feet hight with an initial speed of 10 feet per second.

What is the work done by gravity on this object?

In the preceding slide,

- Path: $\mathbf{r}(t) = \langle 10t, 64-16t^2 \rangle$
- Gravitational field: $\mathbf{F}(x,y) = \langle 0, -32 \rangle$.

The vector field is constant in this case but need not be in general.

We can express the work as a line integral:

$$ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^2 \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$- The left side is an abstract formulation where $C$ is the path, $\mathbf{F}$ is the vector field, and $\mathbf{r}$ is a parametrization of the path.
- The right side allows you to evaluate the line integral by translating it to a normal integral.

Using the line integral formulation, the work is

$$\begin{align} \int_C \mathbf{F} \cdot d\mathbf{r} &= \int_0^2 \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt \\ &= \int_0^2 \langle 0, -32 \rangle \cdot \langle 10,-32t \rangle \, dt \\ &= \int_0^2 1024t \, dt = 2048. \end{align}$$The work required to move the object from one point to the next is approximately $\mathbf{F}(\mathbf{r}(t_i))\cdot \mathbf {r}'(t_i)\Delta t$. If we add those all up, we get

$$ \sum_{i=1}^n \mathbf{F}(\mathbf{r}(t_i))\cdot \mathbf {r}'(t_i)\Delta t \to \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$- A vector field is a function $\mathbf{F}:\mathbb R^n \to \mathbb R^n$.
- Typically, we visualize a vector field by drawing a collection of vectors each emanating from the point where we evalaute the field.
- Often, we need to scale the vectors to keep the images fro being too cluttered.

An electrostatic field

The path of $\textbf{r}(t) = \langle 2 t, t^2, t^3 \rangle$ over the time interval $-1\leq t \leq 1$ through the vector field $\mathbf{F}(x,y,z) = \langle x,-z,y \rangle$.

To find the work done by that vector field on an object moving through that path, we evaluate

$$ \int_C \textbf{F}\cdot d\textbf{r} = \int_{-1}^1 \textbf{F}(\textbf{r}(t))\cdot\textbf{r}(t) \, dt. $$Since $\mathbf{F}(x,y,z) = \langle x,-z,y \rangle$ and $\textbf{r}(t) = \langle 2 t, t^2, t^3 \rangle$,

$$\textbf{F}(\textbf{r}(t))\cdot \textbf{r}(t) = \langle 2t, -t^3, t^2 \rangle\cdot\langle 2,2t,3t^2\rangle = 4t+t^4.$$Finishing the integral, we get

$$\begin{align} \int_C \textbf{F}\cdot d\textbf{r} &= \int_{-1}^1 \textbf{F}(\textbf{r}(t))\cdot\textbf{r}(t) \, dt\\ &= \int_{-1}^1 \left(4t+t^4\right) \, dt = 2/5. \end{align}$$
A vector field $\mathbf{F}$ is called conservative if it arises as the gradient of a real valued function.

That is, there is a function $f:\mathbb R^n\to \mathbb R$ such that $\mathbf{F} = \nabla f$.

The function $-f$ is sometimes called the potential of the vector field.

Let $$\mathbf{F}(x,y,z) = - \frac{\langle x,y,z \rangle}{(x^2+y^2+z^2)^{3/2}}.$$ Then $\mathbf{F}$ is conservative because $\mathbf{F} = \nabla f,$ where $$f(x,y,z) = 1/\sqrt{x^2+y^2+z^2},$$ as you can check. This example is is essentially the gravitational field.

- Vector fields might or might not be conservative.
- If $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$ is a conservative 2D field, then $$P = \frac{\partial f}{\partial x} \: \text{ and } \: Q = \frac{\partial f}{\partial y},$$ for some $f$. Thus (equating the mixed partials of $f$), $$\frac{\partial P}{\partial y}= \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial Q}{\partial x}.$$ This gives us a test to see if a vector field is conservative.

- $\mathbf{F}(x,y) = \langle xy, x^2y^2 \rangle$ is not conservative since $$\frac{\partial}{\partial y}xy = x \neq 2xy^2 = \frac{\partial}{\partial x}x^2y^2.$$
- $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$ is conservative since $$\frac{\partial}{\partial y}2xy^3 = 6xy^2 = \frac{\partial}{\partial x}(3x^2y^2+1).$$

Since $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$ is conservative, there must be a function $f$ such that $\mathbf{F}=\nabla f$.

- Surely, $\frac{\partial f}{\partial x} = 2xy^3$. Thus, $$f(x,y) = x^2y^3 + g(y).$$
- Note that $g(y)$ is a constant of integration; it is constant as far as $x$ is concerned. To find it more precisely, note that $$\frac{\partial f}{\partial y} = 3x^2y^2 + g'(y) = 3x^2y^2 + 1.$$ Thus, $g'(y)=1$ so $g(y)=y$ and $f(x,y) = x^2y^3+y$.

If $\mathbf{F} = \nabla f$ is a conservative vector field and $C$ is a path starting at $A$ and ending at $B$, then

$$\int_C \mathbf{F}\cdot d\mathbf{r} = f(B)-f(A).$$This makes it easy to compute line integrals!

(for conservative vector fields)

Let $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$, let $\mathbf{r}(t) = \langle t^2,t \rangle$, and $C$ denote the portion of the path parametrized by $\mathbf{r}$ over the time interval $[0,1]$.

Let's evaluate $\displaystyle \int_C \mathbf{F}\cdot d\mathbf{r}.$

Solution: We already know that $\mathbf{F}$ is the gradient of $f(x,y)=x^2y^3+y$. We can also see that $C$ moves from $(0,0)$ to $(1,1)$ over the given time interval. Thus, $$\int_C \mathbf{F}\cdot d\mathbf{r} = f(1,1)-f(0,0) = 2.$$

We already know that $\mathbf{F}$ is the gradient of $f(x,y)=x^2y^3+y$. We can also see that $C$ moves from $(0,0)$ to $(1,1)$ over the given time interval. Thus, $$\int_C \mathbf{F}\cdot d\mathbf{r} = f(1,1)-f(0,0) = 2.$$

Our vector field is $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$ and our path is $\mathbf{r}(t) = (x(t),y(t)) = (t^2,t)$. We can substitute in directly to get

$$\begin{align} \int_C \mathbf{F}\cdot d\mathbf{r} &= \int_0^1 \langle 2(t^2)(t)^3,3(t^2)^2(t)^2 + 1\rangle \cdot \langle 2t,1 \rangle \, dt \\ &= \int_0^1 \langle 2t^5,3t^6 + 1\rangle \cdot \langle 2t,1 \rangle \, dt \\ &= \int_0^1 (4t^6 + 3t^6 + 1)\,dt \\ & = \int_0^1 (7t^6 + 1) \, dt = 2. \end{align}$$