notes/uni/mmme/1xxx/1026_maths_for_engineering/calculus.md

---
author: Alvie Rahman
date: \today
title: MMME1026 // Calculus
tags:
- uni
- nottingham
- mechanical
- engineering
- mmme1026
- maths
- calculus
uuid: 126b21f8-e188-48f6-9151-5407f2b2b644
---

# Calculus of One Variable Functions

## Key Terms

<details>
<summary>

### Function

A function is a rule that assigns a **unique** value $f(x)$ to each value $x$ in a given *domain*.

</summary>

The set of value taken by $f(x)$ when $x$ takes all possible value in the domain is the *range* of
$f(x)$.

</details>

<details>
<summary>

### Rational Functions

A function of the type

$$ \frac{f(x)}{g(x)} $$

</summary>

where $f$ and $g$ are polynomials, is called a rational function.

Its range has to exclude all those values of $x$ where $g(x) = 0$.

</details>

<details>
<summary>

### Inverse Functions

Consider the function $f(x) = y$.
If $f$ is such that for each $y$ in the range there is exactly one $x$ in the domain,
we can define the inverse $f^{-1}$ as:

$$f^{-1}(y) = f^{-1}(f(x)) = x$$

</summary>
</details>

<details>
<summary>

### Limits

Consider the following:

$$f(x) = \frac{\sin x}{x}$$

The value of the function can be easily calculated when $x \neq 0$, but when $x=0$, we get the
expression $\frac{\sin 0 }{0}$.
However, when we evaluate $f(x)$ for values that approach 0, those values of $f(x)$ approach 1.

This suggests defining the limit of a function

$$\lim_{x \rightarrow a} f(x)$$

to be the limiting value, if it exists, of $f(x)$ as $x$ gets approaches $a$.

</summary>

#### Limits from Above and Below

Sometimes approaching 0 with small positive values of $x$ gives you a different limit from
approaching with small negative values of $x$.

The limit you get from approaching 0 with positive values is known as the limit from above:

$$\lim_{x \rightarrow a^+} f(x)$$

and with negative values is known as the limit from below:

$$\lim_{x \rightarrow a^-} f(x)$$

If the two limits are equal, we simply refer to the *limit*.


</details>

## Important Functions

<details>
<summary>

### Exponential Functions

$$f(x) = e^x = \exp x$$

</summary>

It can also be written as an infinite series:

$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$

The two important limits to know are:

- as $x \rightarrow + \infty$, $\exp x \rightarrow +\infty$ ($e^x \rightarrow +\infty$)
- as $x \rightarrow -\infty$, $\exp x \rightarrow 0$ ($e^x \rightarrow 0$)

Note that $e^x > 0$ for all real values of $x$.

</details>

<details>
<summary>

### Hyperbolic Functions (sinh and cosh)

The hyperbolic sine ($\sinh$) and hyperbolic cosine function ($\cosh$) are defined by:

$$\sinh x = \frac 1 2 (e^x - e^{-x}) \text{ and } \cosh x = \frac 1 2 (e^x + e^{-x})$$
$$\tanh = \frac{\sinh x}{\cosh x}$$

</summary>

![[Fylwind at English Wikipedia, Public domain, via Wikimedia Commons](https://commons.wikimedia.org/wiki/File:Sinh_cosh_tanh.svg)](./images/Sinh_cosh_tanh.svg)

Some key facts about these functions:

- $\cosh$ has even symmetry and $\sinh$ and $\tanh$ have odd symmetry
- as $x \rightarrow + \infty$, $\cosh x \rightarrow +\infty$ and $\sinh x \rightarrow +\infty$
- $\cosh^2x - \sinh^2x = 1$
- $\tanh$'s limits are -1 and +1
- Derivatives:
  - $\frac{\mathrm{d}}{\mathrm{d}x} \sinh x = \cosh x$
  - $\frac{\mathrm{d}}{\mathrm{d}x} \cosh x = \sinh x$
  - $\frac{\mathrm{d}}{\mathrm{d}x} \tanh x = \frac{1}{\cosh^2x}$

</details>

<details>
<summary>

### Natural Logarithm

$$\ln{e^y} = \ln{\exp y} = y$$

</summary>

Since the exponential of any real number is positive, the domain of $\ln$ is $x > 0$.

</details>

<details>
<summary>

### Implicit Functions

An implicit function takes the form

$$f(x, y) = 0$$

</summary>

To draw the curve of an implicit function you have to rewrite it in the form $y = f(x)$.
There may be more than one $y$ value for each $x$ value.

</details>

# Differentiation

The derivative of the function $f(x)$ is denoted by:

$$f'(x) \text{ or } \frac{\mathrm{d}}{\mathrm dx} f(x)$$

Geometrically, the derivative is the gradient of the curve $y = f(x)$.

It is a measure of the rate of change of $f(x)$ as $x$ varies.

For example, velocity, $v$, is the rate of change of displacement, $s$, with respect to time, $t$,
or:

$$v = \frac{\mathrm ds}{dt}$$


<details>
<summary>

#### Formal Definition

</summary>

![](./images/vimscrot-2021-12-27T14:33:20,836330991+00:00.png)

As $h\rightarrow 0$, the clospe of the cord $\rightarrow$ slope of the tangent, or:

$$f'(x_0) = \lim_{h\rightarrow0}\frac{f(x_0+h) - f(x_0)}{h}$$

whenever this limit exists.

</details>

## Rules for Differentiation

### Powers

$$\frac{\mathrm d}{\mathrm dx} x^n = nx^{-1}$$

### Trigonometric Functions

$$\frac{\mathrm d}{\mathrm dx} \sin x = \cos x$$
$$\frac{\mathrm d}{\mathrm dx} \cos x = \sin x$$

### Exponential Functions

$$\frac{\mathrm d}{\mathrm dx} e^{kx} = ke^{kx}$$

$$\frac{\mathrm d}{\mathrm dx} \ln kx^n = \frac n x$$

where $n$ and $k$ are constant.

### Linearity

$$\frac{\mathrm d}{\mathrm dx} (f + g) = \frac{\mathrm d}{\mathrm dx} f + \frac{\mathrm d}{\mathrm dx} g$$

### Product Rule

$$\frac{\mathrm d}{\mathrm dx} (fg) = \frac{\mathrm df}{\mathrm dx}g + \frac{\mathrm dg}{\mathrm dx}f$$

### Quotient Rule

$$ \frac{\mathrm d}{\mathrm dx} \frac f g = \frac 1 {g^2} \left( \frac{\mathrm df}{\mathrm dx} g - f \frac{\mathrm dg}{\mathrm dx} \right) $$

$$ \left( \frac f g \right)' = \frac 1 {g^2} (gf' - fg')$$

### Chain Rule

Let

$$f(x) = F(u(x))$$

$$ \frac{\mathrm df}{\mathrm dx} = \frac{\mathrm{d}F}{\mathrm du} \frac{\mathrm du}{\mathrm dx} $$

<details>
<summary>

#### Example 1

Differentiate $f(x) = \cos{x^2}$.

</summary>

Let $u(x) = x^2$, $F(u) = \cos u$

$$ \frac{\mathrm df}{\mathrm dx} = -\sin u \cdot 2x = 2x\sin{x^2} $$

</details>

## L'Hôpital's Rule

l'Hôpital's rule provides a systematic way of dealing with limits of functions like
$\frac{\sin x} x$.

Suppose

$$\lim_{x\rightarrow{a}} f(x) = 0$$

and

$$\lim_{x\rightarrow{a}} g(x) = 0$$

and we want $\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)}$.

If

$$\lim_{x\rightarrow{a}} \frac{f'(x)}{g'(x)} = L $$

where any $L$ is any real number or $\pm \infty$, then

$$\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)} = L$$

You can keep applying the rule until you get a sensible answer.

# Graphs

## Stationary Points

An important application of calculus is to find where a function is a maximum or minimum.

![](./images/vimscrot-2021-12-27T15:30:26,494800477+00:00.png)

when these occur the gradient of the tangent to the curve, $f'(x) = 0$.
The condition $f'(x) = 0$ alone however does not guarantee a minimum or maximum.
It only means that point is a *stationary point*.

There are three main types of stationary points:

- maximum
- minimum
- point of inflection

### Local Maximum

The point $x = a$ is a local maximum if:

$$f'(a) = 0 \text{ and } f''(a) < 0$$

This is because $f'(x)$ is a decreasing function of $x$ near $x=a$.

### Local Minimum

The point $x = a$ is a local minimum if:

$$f'(a) = 0 \text{ and } f''(a) > 0$$

This is because $f'(x)$ is a increasing function of $x$ near $x=a$.

### Point of Inflection

$$f'(a) = 0 \text{ and } f''(a) = 0 \text { and } f'''(a) \ne 0$$

#### $f'''(a) > 0$

![](./images/vimscrot-2021-12-27T15:38:11,125781274+00:00.png)

#### $f'''(a) < 0$

![](./images/vimscrot-2021-12-27T15:38:29,395666506+00:00.png)

# Approximating with the Taylor series

The expansion

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

is an example of a *Taylor series*.
These enable us to approximate a given function f(x) using a series which is often easier to
calculate.
Among other uses, they help us:

- calculate complicated function using simple arithmetic operations
- find useful analytical approximations which work for $x$ near a given value
  (e.g. $e^x  \approx 1 + x$ for $x$ near 0)
- Understand the behaviour of a function near a stationary point

## Strategy

Suppose we know information about $f(x)$ only at the point $x=0$.
How can we find out about $f$ for other values of $x$?

We could approximate the function by successive polynomials,
each time matching more derivatives at $x=0$.

\begin{align*}
  g(x) = a_0 &\text{ using } f(0) \\
  g(x) = a_0 + a_1x &\text{ using } f(0), f'(0) \\
  g(x) = a_0 + a_1x + a_2x^2 &\text{ using } f(0), f'(0), f''(0) \\
  &\text{and so on...}
\end{align*}

<details>
<summary>

#### Example 1

For $x$ near 0, approximate $f(x) = \cos x$ by a quadratic.

</summary>

1. Set $f(0) = g(0$:

   $$f(0) = 1 \rightarrow g(0) = a_0 = 1$$

2. Set $f'(0) = g'(0$:

   $$f'(0) = -\sin0 = 0 \rightarrow g'(0) = a_1 = 0$$

3. Set $f''(0) = g''(0$:

   $$f''(0) = -\cos = -1 \rightarrow g''(0) = 2a_2 = -1 \rightarrow a_2 = -0.5$$

So for $x$ near 0,

$$\cos x \approx 1 - \frac 1 2 x^2$$

Check: 

$x$ | $\cos x$ | $1 - 0.5x^2$
--- | -------- | ------------
0.4 | 0.921061 | 0.920
0.2 | 0.960066 | 0.980
0.1 | 0.995004 | 0.995

</details>

## General Case

### Maclaurin Series

A Maclaurin series is a Taylor series expansion of a function about 0.

Any function $f(x)$ can be written as an infinite *Maclaurin Series*

$$f(x) = a_0 + a_1x + a_2x^2 + a_3x^2 + \cdots$$

where

$$a_0 = f(0) \qquad a_n = \frac 1 {n!} \frac{\mathrm d^nf}{\mathrm dx^n} \bigg|_{x=0}$$

($|_{x=0}$ means evaluated at $x=0$)

### Taylor Series

We may alternatively expand about any point $x=a$ to give a Taylor series:

\begin{align*}
f(x) = &f(a) + (x-a)f'(a) \\
       & + \frac 1 {2!}(x-a)^2f''(a) \\
       & + \frac 1 {3!}(x-a)^3f'''(a) \\
       & + \cdots + \frac 1 {n!}(x-a)^nf^{(n)}(a)
\end{align*}

a generalisation of a Maclaurin series.

An alternative form of Taylor series is given by setting $x = a+h$ where $h$ is small:

$$f(a+h) = f(a) + hf'(a) + \cdots + \frac 1 {n!}h^nf^{(n)}(a) + \cdots$$


## Taylor Series at a Stationary Point

If f(x) has a stationary point at $x=a$, then $f'(a) = 0$ and the Taylor series begins

$$f(x) = f(a) + \frac 1 2 f''(a)(x-a)^2 + \cdots$$

- If $f''(a) > 0$ then the quadratic part makes the function increase going away from $x=a$ and we
  have a minimum
- If $f''(a) < 0$ then the quadratic part makes the function decrease going away from $x=a$ and we
  have a maximum
- If $f''(a) = 0$ then we must include a higer order terms to determine what happens
  have a minimum

# Integration

Integration is the reverse of [differentiation](#differentiation).

Take velocity and displacement as an example:

$$\int\! v \mathrm dt = s + c$$

where $c$ is the constant of integration, which is required for
[indefinite integrals](#indefinite-integrals).A

## Definite Integrals

The definite integral of a function $f(x)$ in the range $a \le x \le b$ is denoted be:

$$\int^b_a \! f(x) \,\mathrm dx$$

If $f(x) = F'(x)$ ($f(x)$ is the derivative of $F(x)$) then

$$\int^b_a \! f(x) \,\mathrm dx = \left[F(x)\right]^b_a = F(b) - F(a)$$

## Area and Integration

Approximate the area under a smooth curve using a large number of narrow rectangles.

![](./images/vimscrot-2021-12-28T15:18:59,911868873+00:00.png)

Area under curve $\approx \sum_{n} f(x_n)\Delta x_n$.

As the rectangles get more numerous and narrow, the approximation approaches the real area.

The limiting value is denoted

$$\approx \sum_{n} f(x_n)\Delta x_n \rightarrow \int^b_a\! f(x) \mathrm dx$$

This explains the notation used for integrals.

<details>
<summary>

#### Example 1

Calculate the area between these two curves:

\begin{align*}
y &= f_1(x) = 2 - x^2 \\
y &= f_2(x) = x
\end{align*}

</summary>

![](./images/vimscrot-2021-12-28T15:25:12,556743251+00:00.png)

1. Find the crossing points $P$ and $Q$

   \begin{align*}
   f_1(x) &= f_2(x) \\
   x &= 2-x^2 \\
   x &= 1 \\
   x &= -2
   \end{align*}

2. Since $f_1(x) \ge f_2(x)$ between $P$ and $Q$

  \begin{align*}
  A &= \int^1_{-2}\! (f_1(x) - f_2(x)) \mathrm dx \\
    &= \int^1_{-2}\! (2 - x^2  - x) \mathrm dx \\
    &= \left[ 2x - \frac 13 x^3 - \frac 12 x^2 \right]^1_{-2} \\
    &= \left(2 - \frac 13 - \frac 12 \right) - \left( -4 + \frac 83 - \frac 42 \right) \\
    &= \frac 92
  \end{align*}

</details>

## Techniques for Integration

Integration requires multiple techniques and methods to do correctly because it is a PITA.

These are best explained by examples so try to follow those rather than expect and explanation.

### Integration by Substitution

Integration but substitution lets us integrate functions of functions.

<details>
<summary>

#### Example 1

Find

$$I = \int\!(5x - 1)^3 \mathrm dx$$

</summary>

1. Let $w(x) = 5x - 1$
2. 
   \begin{align*}
   \frac{\mathrm d}{\mathrm dx} w &= 5 \\
   \frac 15 \mathrm dw &= \mathrm dx
   \end{align*}

3. The integral is then

   \begin{align*}
   I &= \int\! w^3 \frac 15 \mathrm dw \\
     &= \frac 15 \cdot \frac 14 \cdot w^4 + c \\
     &= \frac{1}{20}w^4 + c
   \end{align*}

4. Finally substitute $w$ out

  $$I = \frac{(5x-1)^4}{20} + c$$

</details>

<details>
<summary>

#### Example 2

Find

$$I = \int\! \cos x \sqrt{\sin x + 1} \mathrm dx$$

</summary>

1. Let

   $$w(x) = \sin x + 1$$

2. Then

   \begin{align*}
   \frac{\mathrm d}{\mathrm dx} w = \cos x \\
   \mathrm dw = \cos x \mathrm dx \\
   \end{align*}

3. The integral is now

   \begin{align*}
   I &= \int\! \sqrt w \,\mathrm dw \\
     &= \int\! w^{\frac12} \,\mathrm dw \\
     &= \frac23w^{\frac32} + c
   \end{align*}

4. Finally substitute $w$ out to get:

   $$I = \frac23 (\sin x + 1)^{\frac32} + c$$

</details>

<details>
<summary>

#### Example 3

Find

$$I = \int^{\frac\pi2}_0\! \cos x \sqrt{\sin x + 1} \,\mathrm dx$$

</summary>

1. Use the previous example to get to

   $$I = \int^2_1\! \sqrt w \,\mathrm dw = \frac23w^{\frac32} + c$$

2. Since $w(x) = \sin x + 1$ the limits are:

   \begin{align*}
   x = 0 &\rightarrow w = 1\\
   x = \frac\pi2 &\rightarrow w = 2
   \end{align*}

3. This gives us

   $$I = \left[ \frac23w^{\frac32} \right]^2_1 = \frac23 (2^{\frac23} = 1)$$

</details>

<details>
<summary>

#### Example 4

Find

$$I = \int^1_0\! \sqrt{1 - x^2} \,\mathrm dx$$

</summary>

1. Try a trigonmetrical substitution:

  \begin{align*}
  x &= \sin w \\
  \\
  \frac{\mathrm dx}{\mathrm dw} = \cos w \\
  \mathrm dx = \cos 2 \,\mathrm dw \\
  \end{align*}

2.

   \begin{align*}
   x=0 &\rightarrow w=0 \\
   x=1 &\rightarrow w=\frac\pi2
   \end{align*}

3. Therefore

   \begin{align*}
   I &= \int^{\frac\pi2}_0\! \sqrt{1 - \sin^2 w} \cos w \,\mathrm dw \\
     &= \int^{\frac\pi2}_0\! \cos^w w \,\mathrm dw
   \end{align*}

   But $\cos(2w) = 2\cos^2w - 1$ so:

   $$\cos^2w = \frac12 \cos(2w) + \frac12$$


   Hence

   \begin{align*}
   I &= \int^{\frac\pi2}_0\! \frac12 \cos(2w) + \frac12 \,\mathrm dw \\
     &= \left[ \frac14 \sin(2w) + \frac w2 \right]^{\frac\pi2}_0 \\
     &= \left( \frac14 \sin\pi + \frac\pi4 \right) - 0 \\
     &= \frac\pi4
   \end{align*}

### Integration by Parts

$$uv = \int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx + \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$

or 

$$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$

This technique is derived from integrating the product rule.

</details>

<details>
<summary>

#### Example 1

Find

$$I = \int\! \ln x \,\mathrm dx$$

</summary>

1. Use

   $$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$

2. Set $u = \ln x$  
   and $v' = 1$.

3. This means that $u' = \frac1x$ and $v = x$.
4. 

   \begin{align*}
   I &= x\ln x - \int\! x\cdot\frac1x \,\mathrm dx + c \\
     &= x\ln x - \int\! \,\mathrm dx + c \\
     &= x\ln x - x + c \\
   \end{align*}

</details>

# Application of Integration

## Differential Equations

Consider the equation

$$\frac{\mathrm dy}{\mathrm dx} = y^2$$

To find $y$, is not a straightforward integration:

$$y = \int\!y^2 \,\mathrm dx$$

The equation above does not solve for $y$ as we can't integrate the right until we know $y$...
which is what we're trying to find.

This is an example of a first order differential equation.
The general form is:

$$\frac{\mathrm dy}{\mathrm dx} = F(x, y)$$

### Separable Differential Equations

A first order diferential equation is called *separable* if it is of the form

$$\frac{\mathrm dy}{\mathrm dx} = f(x)g(y)$$

We can solve these by rearranging:

$$\frac1{g(y} \cdot \frac{\mathrm dy}{\mathrm dx} = f(x)$$

$$\int\! \frac1{g(y)} \,\mathrm dy = \int\! f(x) \,\mathrm dx + c$$


<details>
<summary>

#### Example 1

Find $y$ such that

$$\frac{\mathrm dy}{\mathrm dx} = ky$$

where $k$ is a constant.

</summary>

Rearrange to get

\begin{align*}
\int\! \frac1y \,\mathrm dy &= \int\! k \mathrm dx + c \\
\ln y &= kx + c \\
y &= e^{kx + c} = e^ce^{kx} \\
  &= Ae^{kx}
\end{align*}

where $A = e^c$ is an arbitrary constant.

</details>