First off, it's important to keep in mind that this unit stands apart from all others in structure, style, and tone.

All our other units focus on a specific aspect of differential equations and draw on real-world problems for motivation. They're pieces of a whole, and very much like short stories themselves where each page connects to the ones before and the ones that follow.

This unit is different in that it provides a roadmap to other areas of math that'll be important to us later. The unit can't cover these topics in great detail, but it does provide links to other courses and wikis that can help fill in any gaps.

The good news is that we don't need to borrow much from other branches of math. The first two units hinted that we'll need a little bit of multivariable calculus as well as comfort with vectors. We'll expand on what's needed from linear algebra, multivariable calculus, and vector calculus with a few practice problems, starting with partial derivatives.

The vibrating I-beam equation of the opening unit uses partial derivatives because the unknown function depends on more than one variable. Partial differential equations are certainly important to us and loom on the distant horizon, but we'll actually need partial derivatives much earlier to tackle certain first-order problems.

If \( f(x,y) \) is a function of two variables, then \[ \begin{align} \frac{\partial f}{\partial x} & =f_{x}(x,y) = \text{ rate of change of} \ f \ \text{as} \ x \ \text{varies,} \ y \ \text{ held fixed} \\ \frac{\partial f}{\partial y} & = f_{y}(x,y) = \text{ rate of change of} \ f \ \text{as} \ y \ \text{varies,} \ x \ \text{ held fixed}. \end{align}\] For example, if \( f(x,y) = \cos\big( x^2 - y^2\big),\) then \[ \begin{align} \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \bigg( \cos\big( \underbrace{x^2 - y^2}_{u} \big)\bigg) = \frac{d}{du} \cos(u) \underbrace{\frac{\partial}{\partial x} \big( x^2 - y^2\big) }_{\frac{\partial u}{\partial x}} = -2 x \sin( x^2 - y^2). \end{align}\] What is \( f_{y} = \frac{\partial f}{\partial y}?\)

One of the most important things we can do with partial derivatives is assemble them into the **gradient vector**; if \( f\) depends only on \(x\) and \(y,\) then the gradient is defined to be
\[ \nabla f = \left \langle \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} \right \rangle.\]
For example, the function \( f(x,y) = \cos\big( x^2 - y^2\big)\) from the last problem has gradient \[ \nabla f = \Big\langle \underbrace{-2 x \sin\big( x^2 - y^2\big)}_{f_{x}},\ \underbrace{2y \sin\big( x^2 - y^2\big)}_{f_{y}} \Big\rangle. \]
We care about gradients in this course since they help us find vectors perpendicular to **level sets** of a function \( f(x,y).\) A level set, or **level curve**, of a function of two variables is the collection of pairs \( (x,y) \) making \( f(x,y) = c \) true from some fixed constant \( c.\)

The visualization below shows part of the graph of \( f(x,y) = \cos\big( x^2 - y^2\big)\) (blue) together with a few level sets \( f = c \) (red), which are seen to be the curves that result from slicing the surface horizontally by the plane \( z = c\) (yellow).

What shape describes these level curves?

The level sets of the function \( f(x,y) = a x^2 + b y^2 \) are displayed below in blue. The sliders \( a\) and \(b\) (red) control the constants defining \(f.\) Any given arrow (red) is perpendicular, or **normal**, to the level curve on which it sits. This means the vector and a very small segment of the level curve around its tail form a right angle where they meet.

**Hint:** What was the reason we gave for caring about gradients in the last problem?

**Vector fields** like the gradient in this unit and the slope field in the previous one are great visual tools for some types of differential equations. We'll see this many times in the future.

Given the formula for a vector field, we can lay down a grid and draw the vector at each grid point with tail pinned there. This can be terribly tedious, though it's something a computer can easily do.

For our work, we won't care too much about the vector lengths; we'll mostly be interested in the *directions* of the field arrows. We can quickly sketch a direction field using the signs of the components.

Consider \( \langle y-1,x+2 \rangle. \) The first component is positive when \( y > 1 \) and the second component is negative when \( x < - 2.\) If we split the \(xy\)-plane up into four sections using the lines \(x=-2\) and \(y = 1,\) in one sector the vectors looks schematically like \( \langle +, - \rangle;\) the arrows point down and to the right.

We can sketch the rest of the vector field this way. From the options below, select the one that correctly fills in the other sectors and completes the graph of \( \langle y-1,x+2 \rangle. \)

When we investigate first-order equations, we'll need to know if a given vector field is actually the gradient of some function.

For example, we know that \( \left\langle -2 x \sin\big( x^2 - y^2\big) ,2y \sin\big( x^2 - y^2\big) \right\rangle \) is a gradient, but that's cheating: we built it *from* the function \( f(x,y) = \cos\big(x^2-y^2\big)\) by *taking* the gradient! How can we know if the vector field \( \langle y-1, x+2 \rangle \) from the last problem has such an \( f,\) too?

Vector calculus gives us the perfect tool for sorting vector fields that are gradients and those that are not: the curl. The curl measures how much “swirl” a field has near a given point.

Here's a good analogy: the flow of a river is like a 2D vector field; the strength and direction of current at a point is assigned an arrow there.

If we dump a few X-shaped paddle wheels into the stream, they'll move with the flow and also rotate as fluid strikes the paddles. Each X is assigned a vector (the **curl**) pointing along the \(z\)-axis and providing the axis of rotation. Its length is related to the spin rate.

Concretely, the curl vector of a 2D field \( \big\langle M(x,y), N(x,y) \big\rangle \) has a single component: \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}.\) Under pretty mild conditions (which will always be true for us), the field is the gradient of some \( f\) if and only if the curl is zero.

Getting back to the question at hand, is there an \(f(x,y)\) where \[ \nabla f = \langle y-1, x+2 \rangle ? \]

Using the curl, we can show there has to be an \( f \) with \( \nabla f = \langle y-1, x+2 \rangle. \) We can't just call it a day, though; for our work with first-order equations later, we need a process to *find* this **potential function**.

In terms of components, \( \nabla f = \left \langle \frac{\partial f}{\partial x} ,\frac{\partial f}{\partial y} \right \rangle,\) so \( f_{x} = y -1 \) and \( f_{y} = x+2 \) are two equations we need to solve. (It's funny that, to solve a first-order equation with just a *single* variable, we need to go through a *pair* of *partial* differential equations... such is life.)

It doesn't matter where we begin, so let's start with \( f_{y} = x + 2.\) Antiderivatives can remove partial derivatives as well; we just have to be careful about the “constant” of integration:
\[ \begin{align}
\frac{\partial f}{\partial y} = x+2 \implies \int \frac{\partial f}{\partial y} dy
&= f(x,y) \\
& = \int [x+2]\, dy \\
&= (x+2) \int 1\, dy \\
&= (x+2) y + h(x).
\end{align}\]
Since \(x\) is held fixed during a \(y\)-partial differentiation, it's also constant during \(y\)-*anti*differentiation! That's why we can pull \( x+2\) out of the integral and the “constant” of integration depends on \(x.\)

The other equation, \( f_{x} = y - 1,\) may be used to determine \( h(x).\) Ignoring any additional constant, what is \( h(x)? \)

So far, we've used angle brackets \( \langle \ \ , \ \ \rangle\) exclusively to represent vectors; later, it'll be equally important to write them as column **matrices**. A **matrix** is just a table of numbers; a column matrix has a single column.

For example, \( \langle 1, -2\rangle \) and \( \begin{pmatrix} 1 \\ -2 \end{pmatrix} \) represent exactly the same vector, the latter being a matrix.

If we go from a column matrix to angle bracket notation or vice versa, we have to be careful about the order of the numbers since it *does* matter: the first (second) component in the angle brackets is the top (bottom) entry in the column matrix.

Given the dot product of two vectors \( \vec{a} \cdot \vec{b} = \langle a_{x}, a_{y} \rangle \cdot \langle b_{x} , b_{y} \rangle = a_{x} b_{x} + a_{y} b_{y},\)

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \langle a, b \rangle \cdot \langle x, y\rangle \\ \langle c, d \rangle \cdot \langle x, y\rangle \end{pmatrix} = \begin{pmatrix} a x + b y \\ c x + d y \end{pmatrix}\]
is the **matrix-vector** product we'll use (almost) exclusively throughout the entire course\(^{\dagger}.\)

This is expressed as “multiplying row against column,” which makes sense after staring at it for a while. To get a little practice with this vector and matrix arithmetic, calculate \[ \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix}. \]

\(^{\dagger}\) **Footnote:** There are a few sections where we need to multiply an \( n \times n \) matrix by a column with \(n \geq 2\) components, but the generalization from this formula is straightforward.

Usually, a matrix-vector product scrambles up the original vector components so the new arrow points somewhere entirely else and has a different length. The vector in the last problem, though, is special: the matrix only changed its length, not its direction: \[ \underbrace{\begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix}}_{A} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} = 2\begin{pmatrix} 2 \\ 1 \end{pmatrix}.\]

A nonzero vector that satisfies \( A \vec{v} = \lambda \vec{v} \) is called an **eigenvector**, and \(\lambda\) is the corresponding **eigenvalue**. So \( \langle 2 , 1 \rangle \) is an eigenvector of the matrix \( A \) with eigenvalue 2. If you play around with the visualization above, you may stumble onto others...

Eigenvalue and eigenvectors are absolutely essential for our work with linear systems and higher-order equations, so we better know how to find them!

If we define \( I_{2} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) to be the \( 2 \times 2 \) **identity matrix**, the eigenvalues are given by
\[ \det( A - \lambda I_{2} ) = 0, \]
where \( A \) is also \( 2 \times 2.\) The determinant, in turn, is defined by the rather weird formula
\[ \det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = a d - b c.\]
Why all of this works is too complicated to go into here, so check out our linear algebra course for more information.

What are the eigenvalues of \( \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix}? \) Select *all* that apply.

Finding \( \lambda =2 \) isn't much of a surprise since we already knew that \[\underbrace{\begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix}}_{A} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} = 2\begin{pmatrix} 2 \\ 1 \end{pmatrix}.\] What is interesting is that another eigenvector lurks somewhere in the shadows. We need to bring it into the light.

We start with the only piece of information we have about it: it's eigenvalue is \( \lambda = 1.\) So,
\[ A \vec{v} = 1 \vec{v} \implies \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} v_{x} \\ v_{y} \end{pmatrix} = \begin{pmatrix} 3 v_{x} - 2 v_{y} \\ v_{x} \end{pmatrix} = \begin{pmatrix} v_{x} \\ v_{y} \end{pmatrix} .\]
From the options, select *every* vector obeying this equation.

We saved the best for last! To wrap up this unit, let's briefly talk about Euler's formula \( e^{i \theta} = \cos(\theta) + i \sin(\theta), \) which we'll use many times in the second half of the course.

Here and throughout our course, \( i = \sqrt{-1}.\)

This rather deep and mysterious formula deserves an entire course of its own. We'll only use it to relate trig functions to complex exponentials and vice versa.

For example, integrating \( e^{x} \sin(x) \) is hard, but strangely enough, if we can write \( \sin(x) \) in terms of complex exponentials, the integration is a piece of cake.

Use Euler's formula to write \( \sin(x)\) in terms of \( e^{i x } \) and \( e^{- i x}.\)

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