# Multivariable Calculus

## Partial Differentiation

### Explicit Differentiation

#### Basic Example

The rule of thumb is to treat whatever variable we are not taking the derivative of as a constant:

f(x,y) = x^3 y+2x^2+9y^2+xy+10

We may wish to find the derivative of f with respect to x:

df/dx =

In this case we need to treat y as a constant. The best way to think of this is that dz/dx is the path that would be available if we were to walk in any direction keeping our y-coordinate fixed.

df/dx = 3x^2y + 4x + 0 + y + 0

df/dx = 3x^2y + 4x + y

To break this down a little, whenever we had an x multiplied by y we were not able to just get rid of the y as its treated as a constant. However, an isolated y or constant we just treat as 0.

#### Chain Rule Example

f(x,y) = (x+y^2)^3

In this case we apply the chain rule:

{insert chain rule here}

df/dx = 3(x + y^2)^2 * (1+0)

df/dx = 3(x + y^2)^2

#### Trigonometric Example

f(x,y) = x^2 y + sin(y)

df/dx = 2xy + 0

df/dx = 2xy

df/dy = x^2 + cos(y)

### Implicit Differentiation

Implicit differentiation includes some type of equality, as below:

z(x,y) = xyz = x – y + z

The trick to it is that we need to differentiate with respect to x, so we apply the typical derivative rule, but whenever we see the z term we need to multiply it by dz/dx as follows:

z(x,y) = xyz = x – y + z

dz/dx = yz*dz/dx = 1 + z*dz/dx

Why have we done this? the y on the right hand side disappears because it is treated as a constant, the y on the left remains because it is multiplied by x.

Let’s try an example:

z(x,y)= x^2+y^2+z^2=10

dz/dx =

−x/z,

## Gradient

The gradient points in the direction of steepest ascent.

The length of the gradient vector measures how steep the ascent is.

## Critical Points of a Function

Critical points occur when the partial derivative with respect to one of the variables equals 0. Example:

To find an evaluate critical points of the function:

x^3 – 15x^2 – 20y^2 + 5

**Step 1: Take the first-order and second-order partial derivatives with respect to each variable.**

fx (x, y) = 3x^2-30x

fy (x, y) = -40y

fxx (x, y) =

fyy (x, y) =

fxy (x, y) =

**Step 2: Find critical points by solving the system of equations:**

Critical points occur when the first derivatives equal 0:

3x^2-30x = 0

-40y = 0

**Step 3: Find the discriminant of f(x,y) at critical point (a,b)**

D(a,b) = fxx(a,b)*fyy(a,b) – [fxy(a,b)]^2

**Step 4: Classifying critical points using second order partial derivative test:**

## General Solutions to Differential Equations

### Initial Value Problems

## Fourier Series

We use the Fourier series when we have a periodic function:

The above simply means that when x is between negative pi and 0, y= -1 and when x is between 0 and pi, y = 1. The function changes sign at the origin, basically.

Let’s break down how the series works in practice:

a0 is the average level of the function, most often it is 0 for the origin but not always.

an is the

bn is the

The rule is: if the function is odd (meaning it alternates between negative and positive), only sins. if the function is even (meaning the values are all positive or all negative), only cosines.

### Fourier Series Examples

#### Even Function

**Step 1: Write the Fourier Equation in Full**

**Step 2: Determine whether it is an odd or even function**

**Step 3: Finding an**

#### Odd Function

**Step 1: Write the Fourier Equation in Full**

**Step 2: Determine whether it is an odd or even function**

The function is an odd function so we know that a0 and an will be 0, so remove those elements from consideration and find only bn.

**Step 3: Finding bn**

The above formula is designed for odd and even functions, but we can take a shortcut:

The 2 will be negative because the integral of sine is negative cosine:

Cos(0) is equal to 1, so the answer will be:

## Taylor Series

### Taylor Series Examples

f(x) = (sin x)^3

**Step 1: Write the Taylor Series Equation**

**Step 2: Find associated coefficients**

## Euler’s Method

### Euler’s Method Examples

**Step 1: Write Euler’s Method Formula**

**Step 2: Set out the question, defining it.**

Δx = 1

1 ≤ x ≤ 4

y(1) = 1

As the above figure shows, we take steps from 1 onwards, in steps of 1, to get to 4.

**Step 3: Calculate y values**

Our first y value is given by the initial condition:

y0 = 1

y1 = y0 + x0(1+y0) = 1 + 1(1+1) = 3

y2 = y1 + x1(1+y1) = 3 + 2(1+3) = 11

y3 = y2 + x2(1+y2) = 11 + 3(1+11) = 47

**Ans:**

y(4) = 47

## Identity Matrix

1*x = x

The 1 is the identity. When we apply this concept to matrix multiplication, becuase matrix multiplication is ordered row * column, we get the following:

IM = M

The identity matrix is thus:

How this works is that the first item in the matrix ‘M’, 1, is equal to one after being multiplied by the identity matrix because 1*1 + 0*4 + 0*7 = 1.

## Eigenvalues of a Matrix

### Calculating Eigenvalues of a Matrix Examples

## Eigenvectors of a Matrix

### Calculating Eigenvectors of a Matrix

## Reference

http://www.columbia.edu/itc/sipa/math/calc_rules_multivar.html

http://personal.maths.surrey.ac.uk/S.Zelik/teach/calculus/partial_derivatives.pdf

https://www.youtube.com/watch?v=mU9xb-j7cOI