WIP

quarto/differentiable_vector_calculus/vector_fields.qmd

@@ -24,7 +24,7 @@ For a scalar function $f: R^n \rightarrow R$, the gradient of $f$, $\nabla{f}$,
 |   $f: R\rightarrow R$   |  univariate   | familiar graph of function  |         $f$          |
 |  $f: R\rightarrow R^m$  | vector-valued |  space curve when n=2 or 3  | $\vec{r}$, $\vec{N}$ |
 |  $f: R^n\rightarrow R$  |    scalar     |     a surface when n=2      |         $f$          |
-| $F: R^n\rightarrow R^n$ | vector field  |   a vector field when n=2   |         $F$          |
+| $F: R^n\rightarrow R^n$ | vector field  |   a vector field when n=2, 3|         $F$          |
 | $F: R^n\rightarrow R^m$ | multivariable | n=2,m=3 describes a surface |     $F$, $\Phi$      |
 
 
@@ -34,7 +34,9 @@ After an example where the use of a multivariable function is of necessity, we d
 ## Vector fields
 
 
-We have seen that the gradient of a scalar function, $f:R^2 \rightarrow R$, takes a point in $R^2$ and associates a vector in $R^2$. As such $\nabla{f}:R^2 \rightarrow R^2$ is a vector field. A vector field  can be visualized by sampling a region and representing the field at those points. The details, as previously mentioned, are in the `vectorfieldplot` function of `CalculusWithJulia`.
+We have seen that the gradient of a scalar function, $f:R^2 \rightarrow R$, takes a point in $R^2$ and associates a vector in $R^2$. As such $\nabla{f}:R^2 \rightarrow R^2$ is a vector field. A vector field is a vector-valued function from $R^n \rightarrow R^n$ for $n \geq 2$.
+
+An input/output pair  can be visualized by identifying the input values as a point, and the output as a vector visualized by anchoring the vector at the point. A vector field is a sampling of such pairs, usually taken over some ordered grid. The details, as previously mentioned, are in the `vectorfieldplot` function of `CalculusWithJulia`.
 
 
 ```{julia}
@@ -1071,7 +1073,7 @@ Adjusting $f$ to have a vanishing second -- but not third -- derivative at $c_0$
 As for $g_1$, we have by construction $g_1(b_0) = 0$. By differentiation we get a pattern for some constants $c_j = (j+1)\cdot(j+2)\cdots \cdot k$ with $c_k = 1$.
 
 $$
-g^{(k)}(b) = k! \cdot \frac{f(a_0) - f(b)}{(a_0-b)^{k+1}} - \sum_{j=1}^k c_j \frac{f^{(j)}(b)}{(a_0 - b)^{k-j+1}}. 
+g^{(k)}(b) = k! \cdot \frac{f(a_0) - f(b)}{(a_0-b)^{k+1}} - \sum_{j=1}^k c_j \frac{f^{(j)}(b)}{(a_0 - b)^{k-j+1}}.
 $$
 
 Of note that when $f(a_0) = f(b_0) = 0$ that if $f^{(k)}(b_0)$ is the first non-vanishing derivative of $f$ at $b_0$ that $g^{(k)}(b_0) = f^{(k)}(b_0)/(b_0 - a_0)$ (they have the same sign).
@@ -1146,6 +1148,30 @@ This handles most cases, but leaves the possibility that a function with infinit
 
 ## Questions
 
+##### Question
+
+```{julia}
+#| echo: false
+gr()
+p1 = vectorfieldplot((x,y) -> [x,y],   xlim=(-4,4), ylim=(-4,4), nx=9, ny=9, title="A");
+p2 = vectorfieldplot((x,y) -> [x-y,x],  xlim=(-4,4), ylim=(-4,4), nx=9, ny=9,title="B");
+p3 = vectorfieldplot((x,y) -> [y,0],    xlim=(-4,4), ylim=(-4,4), nx=9, ny=9, title="C");
+p4 = vectorfieldplot((x,y) -> [-y,x],   xlim=(-4,4), ylim=(-4,4), nx=9, ny=9, title="D");
+plot(p1, p2, p3, p4; layout=[2,2])
+```
+
+In the above figure, match the function with the vector field plot.
+
+```{julia}
+#| echo: false
+plotly()
+matchq(("`F(x,y)=[-y ,x]`",  "`F(x,y)=[y,0]`",
+        "`F(x,y)=[x-y,x]`",  "`F(x,y)=[x,y]`"),
+       ("A", "B", "C", "D"),
+       (4,3,2,1);
+       label="For each function mark the correct vector field plot"
+       )
+```
 
 ###### Question