mmme2046 lecture 1
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---
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author: Alvie Rahman
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date: \today
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title: MMME1026 // Calculus
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tags:
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- uni
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- nottingham
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- mechanical
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- engineering
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- mmme1026
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- maths
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- calculus
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uuid: 126b21f8-e188-48f6-9151-5407f2b2b644
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---
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# Calculus of One Variable Functions
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## Key Terms
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<details>
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<summary>
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### Function
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A function is a rule that assigns a **unique** value $f(x)$ to each value $x$ in a given *domain*.
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</summary>
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The set of value taken by $f(x)$ when $x$ takes all possible value in the domain is the *range* of
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$f(x)$.
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</details>
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<details>
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<summary>
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### Rational Functions
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A function of the type
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$$ \frac{f(x)}{g(x)} $$
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</summary>
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where $f$ and $g$ are polynomials, is called a rational function.
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Its range has to exclude all those values of $x$ where $g(x) = 0$.
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</details>
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<details>
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<summary>
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### Inverse Functions
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Consider the function $f(x) = y$.
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If $f$ is such that for each $y$ in the range there is exactly one $x$ in the domain,
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we can define the inverse $f^{-1}$ as:
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$$f^{-1}(y) = f^{-1}(f(x)) = x$$
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</summary>
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</details>
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<details>
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<summary>
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### Limits
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Consider the following:
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$$f(x) = \frac{\sin x}{x}$$
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The value of the function can be easily calculated when $x \neq 0$, but when $x=0$, we get the
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expression $\frac{\sin 0 }{0}$.
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However, when we evaluate $f(x)$ for values that approach 0, those values of $f(x)$ approach 1.
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This suggests defining the limit of a function
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$$\lim_{x \rightarrow a} f(x)$$
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to be the limiting value, if it exists, of $f(x)$ as $x$ gets approaches $a$.
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</summary>
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#### Limits from Above and Below
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Sometimes approaching 0 with small positive values of $x$ gives you a different limit from
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approaching with small negative values of $x$.
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The limit you get from approaching 0 with positive values is known as the limit from above:
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$$\lim_{x \rightarrow a^+} f(x)$$
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and with negative values is known as the limit from below:
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$$\lim_{x \rightarrow a^-} f(x)$$
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If the two limits are equal, we simply refer to the *limit*.
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</details>
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## Important Functions
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<details>
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<summary>
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### Exponential Functions
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$$f(x) = e^x = \exp x$$
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</summary>
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It can also be written as an infinite series:
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$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
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The two important limits to know are:
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- as $x \rightarrow + \infty$, $\exp x \rightarrow +\infty$ ($e^x \rightarrow +\infty$)
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- as $x \rightarrow -\infty$, $\exp x \rightarrow 0$ ($e^x \rightarrow 0$)
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Note that $e^x > 0$ for all real values of $x$.
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</details>
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<details>
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<summary>
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### Hyperbolic Functions (sinh and cosh)
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The hyperbolic sine ($\sinh$) and hyperbolic cosine function ($\cosh$) are defined by:
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$$\sinh x = \frac 1 2 (e^x - e^{-x}) \text{ and } \cosh x = \frac 1 2 (e^x + e^{-x})$$
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$$\tanh = \frac{\sinh x}{\cosh x}$$
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</summary>
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](./images/Sinh_cosh_tanh.svg)
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Some key facts about these functions:
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- $\cosh$ has even symmetry and $\sinh$ and $\tanh$ have odd symmetry
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- as $x \rightarrow + \infty$, $\cosh x \rightarrow +\infty$ and $\sinh x \rightarrow +\infty$
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- $\cosh^2x - \sinh^2x = 1$
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- $\tanh$'s limits are -1 and +1
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- Derivatives:
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- $\frac{\mathrm{d}}{\mathrm{d}x} \sinh x = \cosh x$
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- $\frac{\mathrm{d}}{\mathrm{d}x} \cosh x = \sinh x$
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- $\frac{\mathrm{d}}{\mathrm{d}x} \tanh x = \frac{1}{\cosh^2x}$
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</details>
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<details>
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<summary>
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### Natural Logarithm
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$$\ln{e^y} = \ln{\exp y} = y$$
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</summary>
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Since the exponential of any real number is positive, the domain of $\ln$ is $x > 0$.
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</details>
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<details>
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<summary>
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### Implicit Functions
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An implicit function takes the form
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$$f(x, y) = 0$$
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</summary>
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To draw the curve of an implicit function you have to rewrite it in the form $y = f(x)$.
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There may be more than one $y$ value for each $x$ value.
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</details>
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# Differentiation
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The derivative of the function $f(x)$ is denoted by:
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$$f'(x) \text{ or } \frac{\mathrm{d}}{\mathrm dx} f(x)$$
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Geometrically, the derivative is the gradient of the curve $y = f(x)$.
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It is a measure of the rate of change of $f(x)$ as $x$ varies.
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For example, velocity, $v$, is the rate of change of displacement, $s$, with respect to time, $t$,
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or:
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$$v = \frac{\mathrm ds}{dt}$$
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<details>
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<summary>
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#### Formal Definition
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</summary>
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As $h\rightarrow 0$, the clospe of the cord $\rightarrow$ slope of the tangent, or:
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$$f'(x_0) = \lim_{h\rightarrow0}\frac{f(x_0+h) - f(x_0)}{h}$$
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whenever this limit exists.
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</details>
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## Rules for Differentiation
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### Powers
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$$\frac{\mathrm d}{\mathrm dx} x^n = nx^{-1}$$
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### Trigonometric Functions
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$$\frac{\mathrm d}{\mathrm dx} \sin x = \cos x$$
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$$\frac{\mathrm d}{\mathrm dx} \cos x = \sin x$$
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### Exponential Functions
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$$\frac{\mathrm d}{\mathrm dx} e^{kx} = ke^{kx}$$
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$$\frac{\mathrm d}{\mathrm dx} \ln kx^n = \frac n x$$
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where $n$ and $k$ are constant.
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### Linearity
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$$\frac{\mathrm d}{\mathrm dx} (f + g) = \frac{\mathrm d}{\mathrm dx} f + \frac{\mathrm d}{\mathrm dx} g$$
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### Product Rule
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$$\frac{\mathrm d}{\mathrm dx} (fg) = \frac{\mathrm df}{\mathrm dx}g + \frac{\mathrm dg}{\mathrm dx}f$$
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### Quotient Rule
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$$ \frac{\mathrm d}{\mathrm dx} \frac f g = \frac 1 {g^2} \left( \frac{\mathrm df}{\mathrm dx} g - f \frac{\mathrm dg}{\mathrm dx} \right) $$
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$$ \left( \frac f g \right)' = \frac 1 {g^2} (gf' - fg')$$
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### Chain Rule
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Let
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$$f(x) = F(u(x))$$
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$$ \frac{\mathrm df}{\mathrm dx} = \frac{\mathrm{d}F}{\mathrm du} \frac{\mathrm du}{\mathrm dx} $$
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<details>
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<summary>
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#### Example 1
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Differentiate $f(x) = \cos{x^2}$.
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</summary>
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Let $u(x) = x^2$, $F(u) = \cos u$
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$$ \frac{\mathrm df}{\mathrm dx} = -\sin u \cdot 2x = 2x\sin{x^2} $$
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</details>
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## L'Hôpital's Rule
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l'Hôpital's rule provides a systematic way of dealing with limits of functions like
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$\frac{\sin x} x$.
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Suppose
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$$\lim_{x\rightarrow{a}} f(x) = 0$$
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and
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$$\lim_{x\rightarrow{a}} g(x) = 0$$
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and we want $\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)}$.
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If
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$$\lim_{x\rightarrow{a}} \frac{f'(x)}{g'(x)} = L $$
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where any $L$ is any real number or $\pm \infty$, then
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$$\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)} = L$$
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You can keep applying the rule until you get a sensible answer.
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# Graphs
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## Stationary Points
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An important application of calculus is to find where a function is a maximum or minimum.
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when these occur the gradient of the tangent to the curve, $f'(x) = 0$.
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The condition $f'(x) = 0$ alone however does not guarantee a minimum or maximum.
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It only means that point is a *stationary point*.
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There are three main types of stationary points:
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- maximum
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- minimum
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- point of inflection
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### Local Maximum
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The point $x = a$ is a local maximum if:
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$$f'(a) = 0 \text{ and } f''(a) < 0$$
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This is because $f'(x)$ is a decreasing function of $x$ near $x=a$.
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### Local Minimum
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The point $x = a$ is a local minimum if:
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$$f'(a) = 0 \text{ and } f''(a) > 0$$
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This is because $f'(x)$ is a increasing function of $x$ near $x=a$.
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### Point of Inflection
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$$f'(a) = 0 \text{ and } f''(a) = 0 \text { and } f'''(a) \ne 0$$
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#### $f'''(a) > 0$
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#### $f'''(a) < 0$
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# Approximating with the Taylor series
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The expansion
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$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
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is an example of a *Taylor series*.
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These enable us to approximate a given function f(x) using a series which is often easier to
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calculate.
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Among other uses, they help us:
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- calculate complicated function using simple arithmetic operations
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- find useful analytical approximations which work for $x$ near a given value
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(e.g. $e^x \approx 1 + x$ for $x$ near 0)
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- Understand the behaviour of a function near a stationary point
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## Strategy
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Suppose we know information about $f(x)$ only at the point $x=0$.
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How can we find out about $f$ for other values of $x$?
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We could approximate the function by successive polynomials,
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each time matching more derivatives at $x=0$.
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\begin{align*}
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g(x) = a_0 &\text{ using } f(0) \\
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g(x) = a_0 + a_1x &\text{ using } f(0), f'(0) \\
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g(x) = a_0 + a_1x + a_2x^2 &\text{ using } f(0), f'(0), f''(0) \\
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&\text{and so on...}
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\end{align*}
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<details>
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<summary>
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#### Example 1
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For $x$ near 0, approximate $f(x) = \cos x$ by a quadratic.
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</summary>
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1. Set $f(0) = g(0$:
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$$f(0) = 1 \rightarrow g(0) = a_0 = 1$$
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2. Set $f'(0) = g'(0$:
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$$f'(0) = -\sin0 = 0 \rightarrow g'(0) = a_1 = 0$$
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3. Set $f''(0) = g''(0$:
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$$f''(0) = -\cos = -1 \rightarrow g''(0) = 2a_2 = -1 \rightarrow a_2 = -0.5$$
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So for $x$ near 0,
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$$\cos x \approx 1 - \frac 1 2 x^2$$
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Check:
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$x$ | $\cos x$ | $1 - 0.5x^2$
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--- | -------- | ------------
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0.4 | 0.921061 | 0.920
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0.2 | 0.960066 | 0.980
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0.1 | 0.995004 | 0.995
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</details>
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## General Case
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### Maclaurin Series
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A Maclaurin series is a Taylor series expansion of a function about 0.
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Any function $f(x)$ can be written as an infinite *Maclaurin Series*
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$$f(x) = a_0 + a_1x + a_2x^2 + a_3x^2 + \cdots$$
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where
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$$a_0 = f(0) \qquad a_n = \frac 1 {n!} \frac{\mathrm d^nf}{\mathrm dx^n} \bigg|_{x=0}$$
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($|_{x=0}$ means evaluated at $x=0$)
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### Taylor Series
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We may alternatively expand about any point $x=a$ to give a Taylor series:
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\begin{align*}
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f(x) = &f(a) + (x-a)f'(a) \\
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& + \frac 1 {2!}(x-a)^2f''(a) \\
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& + \frac 1 {3!}(x-a)^3f'''(a) \\
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& + \cdots + \frac 1 {n!}(x-a)^nf^{(n)}(a)
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\end{align*}
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a generalisation of a Maclaurin series.
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An alternative form of Taylor series is given by setting $x = a+h$ where $h$ is small:
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$$f(a+h) = f(a) + hf'(a) + \cdots + \frac 1 {n!}h^nf^{(n)}(a) + \cdots$$
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## Taylor Series at a Stationary Point
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If f(x) has a stationary point at $x=a$, then $f'(a) = 0$ and the Taylor series begins
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$$f(x) = f(a) + \frac 1 2 f''(a)(x-a)^2 + \cdots$$
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- If $f''(a) > 0$ then the quadratic part makes the function increase going away from $x=a$ and we
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have a minimum
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- If $f''(a) < 0$ then the quadratic part makes the function decrease going away from $x=a$ and we
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have a maximum
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- If $f''(a) = 0$ then we must include a higer order terms to determine what happens
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have a minimum
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# Integration
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Integration is the reverse of [differentiation](#differentiation).
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Take velocity and displacement as an example:
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$$\int\! v \mathrm dt = s + c$$
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where $c$ is the constant of integration, which is required for
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[indefinite integrals](#indefinite-integrals).A
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## Definite Integrals
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The definite integral of a function $f(x)$ in the range $a \le x \le b$ is denoted be:
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$$\int^b_a \! f(x) \,\mathrm dx$$
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If $f(x) = F'(x)$ ($f(x)$ is the derivative of $F(x)$) then
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$$\int^b_a \! f(x) \,\mathrm dx = \left[F(x)\right]^b_a = F(b) - F(a)$$
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## Area and Integration
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Approximate the area under a smooth curve using a large number of narrow rectangles.
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Area under curve $\approx \sum_{n} f(x_n)\Delta x_n$.
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As the rectangles get more numerous and narrow, the approximation approaches the real area.
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The limiting value is denoted
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$$\approx \sum_{n} f(x_n)\Delta x_n \rightarrow \int^b_a\! f(x) \mathrm dx$$
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This explains the notation used for integrals.
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<details>
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<summary>
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#### Example 1
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Calculate the area between these two curves:
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\begin{align*}
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y &= f_1(x) = 2 - x^2 \\
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y &= f_2(x) = x
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\end{align*}
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</summary>
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1. Find the crossing points $P$ and $Q$
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\begin{align*}
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f_1(x) &= f_2(x) \\
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x &= 2-x^2 \\
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x &= 1 \\
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x &= -2
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\end{align*}
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2. Since $f_1(x) \ge f_2(x)$ between $P$ and $Q$
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\begin{align*}
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A &= \int^1_{-2}\! (f_1(x) - f_2(x)) \mathrm dx \\
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&= \int^1_{-2}\! (2 - x^2 - x) \mathrm dx \\
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&= \left[ 2x - \frac 13 x^3 - \frac 12 x^2 \right]^1_{-2} \\
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&= \left(2 - \frac 13 - \frac 12 \right) - \left( -4 + \frac 83 - \frac 42 \right) \\
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&= \frac 92
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\end{align*}
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</details>
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## Techniques for Integration
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Integration requires multiple techniques and methods to do correctly because it is a PITA.
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These are best explained by examples so try to follow those rather than expect and explanation.
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### Integration by Substitution
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Integration but substitution lets us integrate functions of functions.
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|
||||
<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>
|
||||
@@ -1,398 +0,0 @@
|
||||
---
|
||||
author: Alvie Rahman
|
||||
date: \today
|
||||
title: MMME1026 // Complex Numbers
|
||||
tags:
|
||||
- uni
|
||||
- nottingham
|
||||
- mechanical
|
||||
- engineering
|
||||
- mmme1026
|
||||
- maths
|
||||
- complex_numbers
|
||||
uuid: b9be8780-1ab7-402f-9c67-8cc74a74f7a9
|
||||
---
|
||||
|
||||
# Complex Numbers
|
||||
|
||||
## What is a Complex Number?
|
||||
|
||||
- $i$ is the unit imaginary number, which is defined by:
|
||||
|
||||
$$ i^2 = -1 $$
|
||||
|
||||
- An arbritary complex number is written in the form
|
||||
|
||||
$$z = x + iy$$
|
||||
|
||||
Where:
|
||||
|
||||
- $x$ is the real part of $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$)
|
||||
- $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$)
|
||||
|
||||
- Two complex numbers are equal if both their real and imaginary parts are equal
|
||||
|
||||
e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$
|
||||
|
||||
### The Complex Conjugate
|
||||
|
||||
Given complex number $z$:
|
||||
|
||||
$$z = x + iy$$
|
||||
|
||||
The complex conjugate of z, $\bar z$ is:
|
||||
|
||||
$$\bar{z} = x -iy$$
|
||||
|
||||
### Division of Complex Numbers
|
||||
|
||||
- Multiply numerator and denominator by the conjugate of the denominator
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example
|
||||
|
||||
</summary>
|
||||
|
||||
> \begin{align*}
|
||||
z_1 &= 5 + i \\
|
||||
z_2 &= 1 -i \\
|
||||
\\
|
||||
\frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
|
||||
&= \frac{5 + i + 5i -1}{1 + 1} \\
|
||||
&= \frac{4 + 6i}{2} = 2 + 3i
|
||||
> \end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
### Algebra and Conjugation
|
||||
|
||||
When taking complex conjugate of an algebraic expresion, we can replace $i$ by $-i$ before or after
|
||||
doing the algebraic operations:
|
||||
|
||||
\begin{align*}
|
||||
\overline{z_1+z_2} &= \bar{z_1}+\bar{z_2} \\
|
||||
\overline{z_1z_2} &= \bar{z_1}\bar{z_2} \\
|
||||
\overline{z_1/z_2} &= \frac{\bar{z_1}}{\bar{z_2}}
|
||||
\end{align*}
|
||||
|
||||
The conjugate of a real number is the same as that number.
|
||||
|
||||
#### Application
|
||||
|
||||
If $z$ is a root of the polynomial equation
|
||||
|
||||
$$0 = az^2 + bz + c$$
|
||||
|
||||
with **real** coefficients $a$, $b$, and $c$, then $\bar{z}$ is also a root because
|
||||
|
||||
\begin{align*}
|
||||
0 &= \overline{az^2 + bz + c} \\
|
||||
&= \bar{a}\bar{z}^2 + \bar{b}\bar{z} + \bar{c} \\
|
||||
&= a\bar{z}^2 + b\bar{z} + c
|
||||
\end{align*}
|
||||
|
||||
### The Argand Diagram
|
||||
|
||||
A general complex number $z = x + iy$ has two components so it can can be represented as a point in
|
||||
the plane with Cartesion coordinates $(x, y)$.
|
||||
|
||||
\begin{align*}
|
||||
4-2i &\leftrightarrow (4, -2) \\
|
||||
-i &\leftrightarrow (0, -1) \\
|
||||
z &\leftrightarrow (x, y) \\
|
||||
\bar z &\leftrightarrow (x, -y)
|
||||
\end{align*}
|
||||
|
||||
### Plotting on a Polar Graph
|
||||
|
||||
We can also describe points in the complex plain with polar coordinates $(r, \theta)$:
|
||||
|
||||
\begin{align*}
|
||||
z &= r(\cos\theta + i\sin\theta) &\text{ polar form of $z$} \\
|
||||
r &= \sqrt{x^2+y^2} &\text{(modulus)}\\
|
||||
\theta &= \arg z,\text{ where} \tan \theta = \frac y x &\text{(argument)} \\
|
||||
x &= r\cos \theta \\
|
||||
y &= r\sin \theta
|
||||
\end{align*}
|
||||
|
||||
Be careful when turning $(x, y)$ into $(r, \theta)$ form as $\tan^{-1} \frac y x = \theta$ does not
|
||||
always hold true as there are many solutions.
|
||||
|
||||
#### Choosing $\theta$ Correctly
|
||||
|
||||
1. Determine which quadrant the point is in (draw a picture).
|
||||
2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent.
|
||||
If it puts you in the wrong quadrant, add or subtract $\pi$.
|
||||
|
||||
## Exponential Functions
|
||||
|
||||
- The exponential function $f(x) = \exp x$ may be wirtten as an infinite series:
|
||||
|
||||
$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
|
||||
|
||||
- The function $f(x) = e^{-x}$ is just $\frac 1 {e^x}$
|
||||
- Note the important properties:
|
||||
|
||||
\begin{align*}
|
||||
e^{a+b} &= e^a e^b \\
|
||||
(e^a)^b &= e^{ab}
|
||||
\end{align*}
|
||||
|
||||
## Euler's Formula
|
||||
|
||||
$$e^{i\theta} = \cos\theta + i\sin\theta$$
|
||||
|
||||
- Properties of $e^{i\theta}$: For any real angle $\theta$ we have
|
||||
|
||||
$$|e^{i\theta}| = |\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$$
|
||||
|
||||
and
|
||||
|
||||
$$ \arg {e^{i\theta}} = \theta $$
|
||||
|
||||
- A complex number in *polar form*, where $r = |z|$, and $\theta = \arg z$, may alternatively be
|
||||
written in its *exponential form*:
|
||||
|
||||
$$z = re^{i\theta}$$
|
||||
|
||||
**Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 1
|
||||
|
||||
Write $z = -1 + i$ in exponential form
|
||||
|
||||
</summary>
|
||||
|
||||
> $\arg z = \frac {3\pi} 4$
|
||||
> $|z| = \sqrt 2$
|
||||
>
|
||||
> So $z = \sqrt2e^{i\frac{3\pi} 4}$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 2
|
||||
|
||||
The equations for a mechanical vibration problem are found to have the following mathematical
|
||||
solution:
|
||||
|
||||
$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
|
||||
|
||||
</summary>
|
||||
|
||||
where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
|
||||
constants.
|
||||
Although $z(t)$
|
||||
is complex and cannot directly represent a physical solution, it turns out that the real and
|
||||
imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
|
||||
this physical information efficiently as follows:
|
||||
|
||||
a. Put the denominator in the form
|
||||
|
||||
$$ae^{i\delta}$$
|
||||
|
||||
where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
|
||||
and $\gamma$.
|
||||
|
||||
|
||||
> \begin{align*}
|
||||
a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
|
||||
\delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
|
||||
> \end{align*}
|
||||
|
||||
b. Hence find the constants $b$ and $\varphi$ such that
|
||||
|
||||
$$x(t) = b\cos(\omega t + \varphi)$$
|
||||
|
||||
and write a similar expression for $y(t)$.
|
||||
|
||||
> \begin{align*}
|
||||
z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
|
||||
x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
|
||||
\therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
|
||||
\Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
|
||||
\\
|
||||
b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
|
||||
\varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
|
||||
\\
|
||||
y(t) &= \frac 1 a \sin(\omega t - \delta) \\
|
||||
> \end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
## Products of Complex Numbers
|
||||
|
||||
Suppose we have 2 complex numbers:
|
||||
|
||||
$$z_1 = x_1 + iy_1 = r_1e^{i\theta_1}$$
|
||||
$$z_2 = x_2 + iy_2 = r_2e^{i\theta_2}$$
|
||||
|
||||
Using $e^a e^b = e^{a+b}$, the product is:
|
||||
|
||||
\begin{align*}
|
||||
z_3 = z_1 z_2 &= (r_1e^{i\theta_1})(r_2e^{i\theta_2}) \\
|
||||
&= r_1r_2e^{i\theta_1}e^{i\theta_2} \\
|
||||
&= r_1r_2e^{i(\theta_1+\theta_2)} \\
|
||||
\\
|
||||
|z_1z_2| &= |z_1|\times|z_2| \\
|
||||
\arg z_1z_2 &= \arg z_1 \times \arg z_2
|
||||
\end{align*}
|
||||
|
||||
## de Moivre's Theorem
|
||||
|
||||
Let $z = re^{i\theta}$. Consider $z^n$.
|
||||
|
||||
Since $z = r(\cos\theta + i\sin\theta)$,
|
||||
\begin{align*}
|
||||
z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
|
||||
\end{align*}
|
||||
|
||||
But also
|
||||
|
||||
\begin{align*}
|
||||
z^n &= (re^{i\theta})^n \\
|
||||
&= r^n(e^{i\theta})^n \\
|
||||
&= r^ne^{in\theta} \\
|
||||
&= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
|
||||
\end{align*}
|
||||
|
||||
By equating (1) and (2), we find de Moivre's theorem:
|
||||
|
||||
\begin{align*}
|
||||
r^n(\cos\theta +i\sin\theta)^n &= r^n(\cos{n\theta} + i\sin{n\theta}) \\
|
||||
(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
|
||||
\end{align*}
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 1
|
||||
|
||||
Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
|
||||
|
||||
</summary>
|
||||
|
||||
> \begin{align*}
|
||||
r &= |1+i| = \sqrt2 \\
|
||||
\theta &= \arg{1+i} = \frac \pi 4 \\
|
||||
\\
|
||||
\text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
|
||||
(i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
|
||||
&= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
|
||||
&= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
|
||||
&= 2^7 (1 - i) \\
|
||||
&= 128 - 128i
|
||||
> \end{align*}
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 2
|
||||
|
||||
Use de Moivre's theorem to show that
|
||||
|
||||
\begin{align*}
|
||||
\cos{2\theta} &= \cos^2\theta-\sin^2\theta \\
|
||||
\text{and} \\
|
||||
\sin{2\theta} &= 2\sin\theta\cos\theta
|
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\end{align*}
|
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|
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</summary>
|
||||
|
||||
> Let $n=2$:
|
||||
|
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> \begin{align*}
|
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(\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
|
||||
\text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{2\theta}\\
|
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\text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
|
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> \end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 3
|
||||
|
||||
Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
|
||||
$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
|
||||
|
||||
</summary>
|
||||
|
||||
> \begin{align*}
|
||||
r &= \sqrt{2} \\
|
||||
\arg \omega = \theta &= \frac 3 4 \pi \\
|
||||
\\
|
||||
\omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
|
||||
\bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
|
||||
\omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
|
||||
&= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
|
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> \end{align*}
|
||||
|
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</details>
|
||||
|
||||
## Complex Roots of Polynomials
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example
|
||||
|
||||
Find which complex numbers $z$ satisfy
|
||||
|
||||
$$z^3 = 8i$$
|
||||
|
||||
</summary>
|
||||
|
||||
> 1. Write $8i$ in exponential form,
|
||||
>
|
||||
> $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
|
||||
>
|
||||
> $\therefore 8i = 8e^{i\frac \pi 2}$
|
||||
>
|
||||
>
|
||||
> 2. Let the solution be $r = re^{i\theta}$.
|
||||
>
|
||||
> Then $z^3 = r^3e^{3i\theta}$.
|
||||
>
|
||||
> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
|
||||
>
|
||||
> i. Compare modulus:
|
||||
>
|
||||
> $r^3 = 8 \rightarrow r = 2$
|
||||
>
|
||||
> ii. Compare argument:
|
||||
>
|
||||
> $$3\theta = \frac \pi 2$$
|
||||
>
|
||||
> is a solution but there are others since
|
||||
>
|
||||
> $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
|
||||
>
|
||||
> so we get a solution whenever
|
||||
>
|
||||
> $$3\theta = \frac \pi 2 + 2n\pi$$
|
||||
>
|
||||
> for any integer `n`
|
||||
>
|
||||
> - $n = 0 \rightarrow z = \sqrt3 + i$
|
||||
> - $n = 1 \rightarrow z = -\sqrt3 + i$
|
||||
> - $n = 2 \rightarrow z = -2i$
|
||||
> - $n = 3 \rightarrow z = \sqrt3 + i$
|
||||
> - $n = 4 \rightarrow z = -\sqrt3 + i$
|
||||
> - The solutions start repeating as you can see
|
||||
>
|
||||
> In general, an $n$-th order polynomial has exactly $n$ complex roots.
|
||||
> Some of these complex roots may be real numbers.
|
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>
|
||||
> 4. There are three solutions
|
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|
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</details>
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@@ -1,753 +0,0 @@
|
||||
---
|
||||
author: Alvie Rahman
|
||||
date: \today
|
||||
title: MMME1026 // Systems of Equations and Matrices
|
||||
tags:
|
||||
- uni
|
||||
- nottingham
|
||||
- mechanical
|
||||
- engineering
|
||||
- mmme1026
|
||||
- maths
|
||||
- systems_of_equations
|
||||
- matrices
|
||||
uuid: 16edb140-9946-4759-93df-50cad510fe31
|
||||
---
|
||||
# Systems of Equations (Simultaneous Equations)
|
||||
|
||||
## Gaussian Elimination
|
||||
|
||||
Gaussian eliminiation can be used when the number of unknown variables you have is equal to the
|
||||
number of equations you are given.
|
||||
|
||||
I'm pretty sure it's the name for the method you use to solve simultaneous equations in school.
|
||||
|
||||
For example if you have 1 equation and 1 unknown:
|
||||
|
||||
\begin{align*}
|
||||
2x &= 6 \\
|
||||
x &= 3
|
||||
\end{align*}
|
||||
|
||||
### Number of Solutions
|
||||
|
||||
Let's generalise the example above to
|
||||
|
||||
$$ax = b$$
|
||||
|
||||
There are 3 possible cases:
|
||||
|
||||
\begin{align*}
|
||||
a \ne 0 &\rightarrow x = \frac b a \\
|
||||
a = 0, b \ne 0 &\rightarrow \text{no solution for $x$} \\
|
||||
a = 0, b = 0 &\rightarrow \text{infinite solutions for $x$}
|
||||
\end{align*}
|
||||
|
||||
### 2x2 Systems
|
||||
|
||||
A 2x2 system is one with 2 equations and 2 unknown variables.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
\begin{align*}
|
||||
3x_1 + 4x_2 &= 2 &\text{(1)} \\
|
||||
x_1 + 2x_2 &= 0 &\text{(2)} \\
|
||||
\end{align*}
|
||||
|
||||
</summary>
|
||||
|
||||
\begin{align*}
|
||||
3\times\text{(2)} = 3x_1 + 6x_2 &= 0 &\text{(3)} \\
|
||||
\text{(3)} - \text{(1)} = 0x_1 + 2x_2 &= -2 \\
|
||||
x_2 &= -1
|
||||
\end{align*}
|
||||
|
||||
We've essentially created a 1x1 system for $x_2$ and now that's solved we can back substitute it
|
||||
into equation (1) (or equation (2), it doesn't matter) to work out the value of $x_1$:
|
||||
|
||||
\begin{align*}
|
||||
3x_1 + 4x_2 &= 2 \\
|
||||
3x_1 - 1 &= 2 \\
|
||||
3x_1 &= 6 \\
|
||||
x_1 &= 2
|
||||
\end{align*}
|
||||
|
||||
You can check the values for $x_1$ and $x_2$ are correct by substituting them into equation (2).
|
||||
|
||||
</details>
|
||||
|
||||
### 3x3 Systems
|
||||
|
||||
A 3x3 system is one with 3 equations and 3 unknown variables.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
\begin{align*}
|
||||
2x_1 + 3x_2 - x_3 &= 5 &\text{(1)} \\
|
||||
4x_1 + 4x_2 - 3x_3 &= 5 &\text{(2)} \\
|
||||
2x_1 - 3x_2 + x_3 &= 5 &\text{(3)} \\
|
||||
\end{align*}
|
||||
|
||||
</summary>
|
||||
|
||||
The first step is to eliminate $x_1$ from (2) and (3) using (1):
|
||||
|
||||
\begin{align*}
|
||||
\text{(2)}-2\times\text{(1)} = -2x_2 -x_3 &= -1 &\text{(4)} \\
|
||||
\text{(3)}-\text{(1)} = -6x_2 + 3x_3 &= -6 &\text{(5)} \\
|
||||
\end{align*}
|
||||
|
||||
This has created a 2x2 system of $x_2$ and $x_3$ which can be solved as any other 2x2 system.
|
||||
I'm too lazy to type up the working, but it is solved like any other 2x2 system.
|
||||
|
||||
\begin{align*}
|
||||
x_2 &= -2
|
||||
x_3 &= 5
|
||||
\end{align*}
|
||||
|
||||
These values can be back-substituted into any of the first 3 equations to find out $x_1$:
|
||||
|
||||
\begin{align*}
|
||||
-2x_1 + 3x_2 - x_3 = 2x_1 + 6 - 3 = 5 \rightarrow x_1 = 1
|
||||
\end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 2
|
||||
|
||||
\begin{align*}
|
||||
x_1 + x_2 - 2x_3 &= 1 &R_1 \\
|
||||
2x_1 - x_2 - x_3 &= 1 &R_2 \\
|
||||
x_1 + 4x_2 + 7x_3 &= 2 &R_3 \\
|
||||
\end{align*}
|
||||
|
||||
</summary>
|
||||
|
||||
1. Eliminate $x_1$ from $R_2$, $R_3$:
|
||||
|
||||
\begin{align*}
|
||||
x_1 + x_2 - 2x_3 &= 1 &R_1' = R_1\\
|
||||
- 3x_2 - 5x_3 &= -1 &R_2' = R_2 - 2R_1 \\
|
||||
3x_2 + 5x_3 &= 1 &R_3' = R_3 - R_1 \\
|
||||
\end{align*}
|
||||
|
||||
We've created another 2x2 system of $R_2'$ and $R_3'$
|
||||
|
||||
2. Eliminate $x_2$ from $R_3''$
|
||||
|
||||
\begin{align*}
|
||||
x_1 + x_2 - 2x_3 &= 1 &R_1'' = R_1' = R_1\\
|
||||
- 3x_2 - 5x_3 &= -1 &R_2'' = R_2' = R_2 - 2R_1 \\
|
||||
0x_3 &= 0 &R_3'' = R_3 '+ R_2' \\
|
||||
\end{align*}
|
||||
|
||||
We can see that $x_3$ can be any number, so there are infinite solutions. Let:
|
||||
|
||||
$$x_3 = t$$
|
||||
|
||||
where $t$ can be any number
|
||||
|
||||
3. Substitute $x_3$ into $R_2''$:
|
||||
|
||||
$$R_2'' = -3x_2 - 5t = -1 \rightarrow x_2 = \frac 1 3 - \frac{5t} 3$$
|
||||
|
||||
4. Substitute $x_2$ and $x_3$ into $R_1''$:
|
||||
|
||||
$$R_1'' = x_1 + \frac 1 3 - \frac{5t} 3 + 2t = 1 \rightarrow x_1 = \frac 2 3 - \frac t 3$$
|
||||
|
||||
</details>
|
||||
|
||||
## Systems of Equations and Matrices
|
||||
|
||||
Many problems in engineering have a very large number of unknowns and equations to solve
|
||||
simultaneously.
|
||||
We can use matrices to solve these efficiently.
|
||||
|
||||
Take the following simultaneous equations::
|
||||
|
||||
\begin{align*}
|
||||
3x_1 + 4x_2 &= 2 &\text{(1)} \\
|
||||
x_1 + 2x_2 &= 0 &\text{(2)}
|
||||
\end{align*}
|
||||
|
||||
They can be represented by the following matrices:
|
||||
|
||||
\begin{align*}
|
||||
A &= \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix} \\
|
||||
\pmb x &= \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \\
|
||||
\pmb b &= \begin{pmatrix} 2 \\ 0 \end{pmatrix} \\
|
||||
\end{align*}
|
||||
|
||||
You can then express the system as:
|
||||
|
||||
$$A\pmb x = \pmb b$$
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### A 3x3 System as a Matrix
|
||||
|
||||
</summary>
|
||||
|
||||
\begin{align*}
|
||||
2x_1 + 3x_2 - x_3 &= 5 \\
|
||||
4x_1 + 4x_2 - 3x_3 &= 3 \\
|
||||
2x_1 - 3x_2 + x_3 &= -1
|
||||
\end{align*}
|
||||
|
||||
Could be expressed in the form $A\pmb x = \pmb b$ where:
|
||||
|
||||
\begin{align*}
|
||||
A &= \begin{pmatrix} 2 & 3 & -1 \\ 4 & 4 & -3 \\ 2 & -3 & -1 \end{pmatrix} \\
|
||||
\pmb x &= \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \\
|
||||
\pmb b &= \begin{pmatrix} 5 \\ 3 \\ -1 \end{pmatrix} \\
|
||||
\end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### An $m\times n$ System as a Matrix
|
||||
|
||||
</summary>
|
||||
|
||||
\begin{align*}
|
||||
a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\
|
||||
a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\
|
||||
\cdots \\
|
||||
a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \\
|
||||
\end{align*}
|
||||
|
||||
Could be expressed in the form $A\pmb x = \pmb b$ where:
|
||||
|
||||
\begin{align*}
|
||||
A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\
|
||||
a_{21} & a_{22} & \cdots & a_{2n} \\
|
||||
\vdots & & & \vdots \\
|
||||
a_{m1} & a_{m2} & \cdots & a_{mn}
|
||||
\end{pmatrix},
|
||||
\pmb x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix},
|
||||
\pmb b = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix}
|
||||
\end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
# Matrices
|
||||
|
||||
## Order of a Matrix
|
||||
|
||||
The order of a matrix is its size e.g. $3\times2$ or $m\times n$
|
||||
|
||||
## Column Vectors
|
||||
|
||||
- Column vectors are matrices with only one column:
|
||||
|
||||
$$ \begin{pmatrix} 1 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 45 \\ 12 \end{pmatrix} $$
|
||||
|
||||
- Column vector variables typed up or printed are expressed in $\pmb{bold}$ and when it is
|
||||
handwritten it is \underline{underlined}:
|
||||
|
||||
$$ \pmb x = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$$
|
||||
|
||||
## Matrix Algebra
|
||||
|
||||
### Equality
|
||||
|
||||
Two matrices are the same if:
|
||||
|
||||
- Their order is the same
|
||||
- Their corresponding elements are the same
|
||||
|
||||
### Addition and Subtraction
|
||||
|
||||
Only possible if their order is the same.
|
||||
\begin{align*}
|
||||
A + B&= \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} & \cdots & a_{1n} + b_{1n} \\
|
||||
a_{21} + b_{21} & a_{22} + b_{22} & \cdots & a_{2n} + b_{2n} \\
|
||||
\vdots & & & \vdots \\
|
||||
a_{m1} + b_{m1} & a_{m2} + b_{m2} & \cdots & a_{mn} + b_{mn}
|
||||
\end{pmatrix} \\
|
||||
A - B&= \begin{pmatrix} a_{11} - b_{11} & a_{12} - b_{12} & \cdots & a_{1n} - b_{1n} \\
|
||||
a_{21} - b_{21} & a_{22} - b_{22} & \cdots & a_{2n} - b_{2n} \\
|
||||
\vdots & & & \vdots \\
|
||||
a_{m1} - b_{m1} & a_{m2} - b_{m2} & \cdots & a_{mn} - b_{mn}
|
||||
\end{pmatrix},
|
||||
\end{align*}
|
||||
|
||||
### Zero Matrix
|
||||
|
||||
This is a matrix whose elements are all zeros.
|
||||
For any matrix $A$,
|
||||
|
||||
$$A + 0 =A$$
|
||||
|
||||
We can only add matrices of the same order, therefore 0 must be of the same order as $A$.
|
||||
|
||||
### Multiplication
|
||||
|
||||
Let
|
||||
|
||||
$$
|
||||
\begin{matrix}
|
||||
A & m\times n \\
|
||||
B & p\times q
|
||||
\end{matrix}
|
||||
$$
|
||||
|
||||
To be able to multiply $A$ by $B$, $n = p$.
|
||||
|
||||
If $n \ne p$, then $AB$ does not exist.
|
||||
|
||||
$$
|
||||
\begin{matrix}
|
||||
A & B & = & C \\
|
||||
m\times n & p \times q & & m\times q
|
||||
\end{matrix}
|
||||
$$
|
||||
|
||||
When $C = AB$ exists,
|
||||
|
||||
$$C_{ij} = \sum_r\! a_{ir}b_{rj}$$
|
||||
|
||||
That is, $C_{ij}$ is the 'product' of the $i$th row of $A$ and $j$th column of $B$.
|
||||
|
||||
#### Multiplication of a Matrix by a Scalar
|
||||
|
||||
If $\lambda$ is a scalar, we define
|
||||
|
||||
$$
|
||||
\lambda a = \begin{pmatrix} \lambda a_{11} & \lambda a_{12} & \cdots & \lambda a_{1n} \\
|
||||
\lambda a_{21} & \lambda a_{22} & \cdots & \lambda a_{2n} \\
|
||||
\vdots & & & \vdots \\
|
||||
\lambda a_{m1} & \lambda a_{m2} & \cdots & \lambda a_{mn}
|
||||
\end{pmatrix},
|
||||
$$
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
</summary>
|
||||
|
||||
$$
|
||||
\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}
|
||||
\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix} =
|
||||
\begin{pmatrix} -3 & -1 \\ 3 & 4 \end{pmatrix}
|
||||
$$
|
||||
$$
|
||||
\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix}
|
||||
\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} =
|
||||
\begin{pmatrix} 2 & 1 \\ 7 & -1 \end{pmatrix}
|
||||
$$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 2
|
||||
|
||||
</summary>
|
||||
|
||||
$$
|
||||
A = \begin{pmatrix} 4 & 1 & 6 \\ 3 & 2 & 1 \end{pmatrix},\,
|
||||
B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 0 \end{pmatrix}
|
||||
$$
|
||||
$$
|
||||
AB = \begin{pmatrix} 11 & 6 \\ 6 & 7 \end{pmatrix},\,
|
||||
BA = \begin{pmatrix} 7 & 3 & 7 \\ 10 & 5 & 8 \\ 4 & 1 & 6 \end{pmatrix}
|
||||
$$
|
||||
|
||||
</details>
|
||||
|
||||
### Other Properties of Matrix Algebra
|
||||
|
||||
- $(\lambda A)B = \lambda(AB) = A(\lambda B)$
|
||||
- $A(BC) = (AB)C = ABC$
|
||||
- $(A+B)C = AC + BC$
|
||||
- $C(A+B) = CA + CB$
|
||||
- In general, $AB \ne BA$ even if both exist
|
||||
- $AB = 0$ does not always mean $A = 0$ or $B = 0$:
|
||||
|
||||
$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}3 & 0 \\ 0 & 0 \end{pmatrix} =
|
||||
\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0$$
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
It follows that $AB = AC$ does not imply that $B=C$ as
|
||||
|
||||
$$AB = AC \leftrightarrow A(B + C) = 0$$
|
||||
|
||||
and as $A$ and $(B-C)$ are not necessarily 0, $B$ is not necessarily equal to $C$:
|
||||
|
||||
</summary>
|
||||
|
||||
$$AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix} =
|
||||
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$
|
||||
|
||||
and
|
||||
|
||||
$$AC = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}1 & 2 \\ 1 & 0 \end{pmatrix} =
|
||||
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = AB$$
|
||||
|
||||
but $B \ne C$
|
||||
|
||||
</details>
|
||||
|
||||
## Special Matrices
|
||||
|
||||
### Square Matrix
|
||||
|
||||
Where $m = n$
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
A $3\times3$ matrix.
|
||||
|
||||
</summary>
|
||||
|
||||
$$\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}$$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 2
|
||||
|
||||
A $2\times2$ matrix.
|
||||
|
||||
</summary>
|
||||
|
||||
$$\begin{pmatrix}1 & 2 \\ 4 & 5 \end{pmatrix}$$
|
||||
|
||||
</details>
|
||||
|
||||
### Identity Matrix
|
||||
|
||||
The identity matrix is a square matrix whose eleements are all 0, except the leading diagonal which
|
||||
is 1s.
|
||||
The leading diagonal is the top left to bottom right corner.
|
||||
|
||||
It is usually denoted by $I$ or $I_n$.
|
||||
|
||||
The identity matrix has the properties that
|
||||
|
||||
$$AI = IA = A$$
|
||||
|
||||
for any square matrix $A$ of the same order as I, and
|
||||
|
||||
$$Ix = x$$
|
||||
|
||||
for any vector $x$.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
The $3\times3$ identity matrix.
|
||||
|
||||
</summary>
|
||||
|
||||
$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 2
|
||||
|
||||
The $2\times2$ identity matrix.
|
||||
|
||||
</summary>
|
||||
|
||||
$$\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$$
|
||||
|
||||
</details>
|
||||
|
||||
### Transposed Matrix
|
||||
|
||||
The transpose of matrix $A$ of order $m\times n$ is matrix $A^T$ which has the order $n\times m$.
|
||||
It is found by reflecting it along the leading diagonal, or interchanging the rows and columns of
|
||||
$A$.
|
||||
|
||||
](./images/Matrix_transpose.gif)
|
||||
|
||||
Let matrix $D = EF$, then $D^T = (EF)^T = E^TF^T$
|
||||
|
||||
#### Example 1
|
||||
|
||||
$$
|
||||
A = \begin{pmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \end{pmatrix},\,
|
||||
A^T = \begin{pmatrix}3 & 4 \\ 2 & 5 \\ 1 & 6\end{pmatrix}
|
||||
$$
|
||||
|
||||
#### Example 2
|
||||
|
||||
$$
|
||||
B = \begin{pmatrix}1 \\ 4\end{pmatrix},\,
|
||||
B^T = \begin{pmatrix}1 & 4\end{pmatrix}
|
||||
$$
|
||||
|
||||
#### Example 3
|
||||
|
||||
$$
|
||||
C = \begin{pmatrix}1 & 2 & 3 \\ 0 & 5 & 1 \\ 2 & 3 & 7\end{pmatrix},\,
|
||||
C^T = \begin{pmatrix}1 & 0 & 2 \\ 2 & 5 & 4 \\ 3 & 1 & 7\end{pmatrix}
|
||||
$$
|
||||
|
||||
### Orthogonal Matrices
|
||||
|
||||
A matrix, $A$, such that
|
||||
|
||||
$$A^{-1} = A^T$$
|
||||
|
||||
is said to be orthogonal.
|
||||
|
||||
Another way to say this is
|
||||
|
||||
$$AA^T = A^TA = I$$
|
||||
|
||||
### Symmetric Matrices
|
||||
|
||||
A square matrix which is symmetric about its leading diagonal:
|
||||
|
||||
$$A = A^T$$
|
||||
|
||||
You can also express this as the matrix $A$, where
|
||||
|
||||
$$a_{ij} = a_{ji}$$
|
||||
|
||||
is satisfied to all elements.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
</summary>
|
||||
|
||||
$$\begin{pmatrix}
|
||||
1 & 0 & -1 & 3 \\
|
||||
0 & 3 & 4 & -1 \\
|
||||
-2 & 4 & -1 & 6 \\
|
||||
3 & -7 & 6 & 2
|
||||
\end{pmatrix}$$
|
||||
|
||||
</details>
|
||||
|
||||
### Anti-Symmetric
|
||||
|
||||
A square matrix is anti-symmetric if
|
||||
|
||||
$$A = -A^T$$
|
||||
|
||||
This can also be expressed as
|
||||
|
||||
$$a_{ij} = -a_{ji}$$
|
||||
|
||||
This means that all elements on the leading diagonal must be 0.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
</summary>
|
||||
|
||||
$$\begin{pmatrix}
|
||||
0 & -1 & 5 \\
|
||||
1 & 0 & 1 \\
|
||||
-5 & -1 & 0
|
||||
\end{pmatrix}$$
|
||||
|
||||
</details>
|
||||
|
||||
## The Determinant
|
||||
|
||||
### Determinant of a 2x2 System
|
||||
|
||||
The determinant of a $2x2$ system is
|
||||
|
||||
$$D = a_{11}a_{22} - a_{12}a_{21}$$
|
||||
|
||||
It is denoted by
|
||||
|
||||
$$
|
||||
\begin{vmatrix}
|
||||
a_{11} & a_{12} \\
|
||||
a_{21} & a_{22}
|
||||
\end{vmatrix}
|
||||
\text{ or }
|
||||
\det
|
||||
\begin{pmatrix}
|
||||
a_{11} & a_{12} \\
|
||||
a_{21} & a_{22}
|
||||
\end{pmatrix}
|
||||
$$
|
||||
|
||||
- A system of equations has a unique solution if $D \ne 0$
|
||||
- If $D = 0$, then there are either
|
||||
|
||||
- no solutions (the equations are inconsistent)
|
||||
- intinitely many solutions
|
||||
|
||||
### Determinant of a 3x3 System
|
||||
|
||||
Let
|
||||
|
||||
$$
|
||||
A = \begin{pmatrix}
|
||||
a_{11} & a_{12} & a_{13} \\
|
||||
a_{21} & a_{22} & a_{23} \\
|
||||
a_{31} & a_{32} & a_{33}
|
||||
\end{pmatrix}
|
||||
$$
|
||||
|
||||
\begin{align*}
|
||||
\det A = &a_{11} \times \det \begin{pmatrix}a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \\
|
||||
&-a_{12} \times \det \begin{pmatrix}a_{21} & a_{23} \\ a_{31} & a_{33} \end{pmatrix} \\
|
||||
&+a_{13} \times \det \begin{pmatrix}a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix}
|
||||
\end{align*}
|
||||
|
||||
The $2x2$ matrices above are created by removing any elements on the same row or column as its corresponding
|
||||
coefficient:
|
||||
|
||||
|
||||
|
||||
### Chessboard Determinant
|
||||
|
||||
$\det A$ may be obtained by expanding out any row or column.
|
||||
To figure out which coefficients should be subtracted and which ones added use the chessboard
|
||||
pattern of signs:
|
||||
|
||||
$$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$$
|
||||
|
||||
### Properties of Determinants
|
||||
|
||||
- $$\det A = \det A^T$$
|
||||
- If all elements of one row of a matrix are multiplied by a constant $z$, the determinant of the
|
||||
new matrix is $z$ times the determinant of the original matrix:
|
||||
|
||||
\begin{align*}
|
||||
\begin{vmatrix} za & zb \\ c & d \end{vmatrix} &= zad - zbc \\
|
||||
&= z(ad-bc) \\
|
||||
&= z\begin{vmatrix} a & b \\ c & d \end{vmatrix}
|
||||
\end{align*}
|
||||
|
||||
This is also true if a column of a matrix is mutiplied by a constant.
|
||||
|
||||
**Application** if the fator $z$ appears in each elements of a row or column of a determinant it
|
||||
can be factored out
|
||||
|
||||
$$\begin{vmatrix}2 & 12 \\ 1 & 3 \end{vmatrix} = 2\begin{vmatrix}1 & 6 \\ 1 & 3 \end{vmatrix} = 2 \times 3
|
||||
\begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix}$$
|
||||
|
||||
**Application** if all elements in one row or column of a matrix are zero, the value of the
|
||||
determinant is 0.
|
||||
|
||||
$$\begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix} = 0\times d - 0\times c = 0$$
|
||||
|
||||
|
||||
**Application** if $A$ is an $n\times n$ matrix,
|
||||
|
||||
$$\det(zA) = z^n\det A$$
|
||||
|
||||
- Swapping any two rows or columns of a matrix changes the sign of the determinant
|
||||
|
||||
\begin{align*}
|
||||
\begin{vmatrix} c & d \\ a & b \end{vmatrix} &= cb - ad \\
|
||||
&= -(ad - bc) \\
|
||||
&= -\begin{vmatrix} a & b \\ c & d \end{vmatrix}
|
||||
\end{align*}
|
||||
|
||||
**Application** If any two rows or two columns are identical, the determinant is zero.
|
||||
|
||||
**Application** If any row is a mutiple of another, or a column a multiple of another column, the
|
||||
determinant is zero.
|
||||
|
||||
- The value of a determinant is unchanged by adding to any row a constant multiple of another row,
|
||||
or adding to any column a constant multiple of another column
|
||||
|
||||
- If $A$ and $B$ are square matrices of the same order then
|
||||
|
||||
$$\det(AB) = \det A \times \det B $$
|
||||
|
||||
## Inverse of a Matrix
|
||||
|
||||
If $A$ is a square matrix, then its inverse matrix is $A^{-1}$ and is defined by the property that:
|
||||
|
||||
$$A^{-1}A = AA^{-1} = I$$
|
||||
|
||||
- Not every matrix has an inverse
|
||||
- If the inverse exists, then it is very useful for solving systems of equations:
|
||||
|
||||
\begin{align*}
|
||||
A\pmb{x} = \pmb b \rightarrow A^{-1}A\pmb x &= A^{-1}\pmb b \\
|
||||
I\pmb x &= A^{-1}\pmb b \\
|
||||
\pmb x &= A^{-1}\pmb b
|
||||
\end{align*}
|
||||
|
||||
Therefore there must be a unique solution to $A\pmb x = \pmb b$: $\pmb x = A^{-1}\pmb b$.
|
||||
|
||||
- If $D = EF$ then
|
||||
|
||||
$$D^-1 = (EF)^{-1} = F^{-1}E^{-1}$$
|
||||
|
||||
### Inverse of a 2x2 Matrix
|
||||
|
||||
If $A$ is the $2x2$ matrix
|
||||
|
||||
$$
|
||||
A = \begin{pmatrix}
|
||||
a_{11} & a_{12} \\
|
||||
a_{21} & a_{22}
|
||||
\end{pmatrix}
|
||||
$$
|
||||
|
||||
and its determinant, $D$, satisfies $D \ne 0$, $A$ has the inverse $A^{-1}$ given by
|
||||
|
||||
$$
|
||||
A^{-1} = \frac 1 D \begin{pmatrix}
|
||||
a_{22} & -a_{12} \\
|
||||
-a_{21} & a_{11}
|
||||
\end{pmatrix}
|
||||
$$
|
||||
|
||||
If $D = 0$, then matrix $A$ has no inverse.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
Find the inverse of matrix $A = \begin{pmatrix} -1 & 5 \\ 2 & 3 \end{pmatrix}$.
|
||||
|
||||
</summary>
|
||||
|
||||
1. Calculate the determinant
|
||||
|
||||
$$\det A = -1 \times 3 - 5 \times 2 = -13$$
|
||||
|
||||
Since $\det A \ne 0$, the inverse exists.
|
||||
|
||||
2. Calculate $A^{-1}$
|
||||
|
||||
$$ A^{-1} = \frac 1 {-13} \begin{pmatrix} 3 & -5 \\ -2 & -1\end{pmatrix}$$
|
||||
@@ -1,322 +0,0 @@
|
||||
---
|
||||
author: Akbar Rahman
|
||||
date: \today
|
||||
title: MMME1026 // Vectors
|
||||
tags:
|
||||
- uni
|
||||
- nottingham
|
||||
- mechanical
|
||||
- engineering
|
||||
- mmme1026
|
||||
- maths
|
||||
- vectors
|
||||
uuid: ef1669d9-b8c3-4813-8b33-5764a2f57a55
|
||||
---
|
||||
|
||||
Vectors have a *magnitude* (size) and *direction*.
|
||||
Examples of vectors include force, velocity, and acceleration.
|
||||
|
||||
In type, vectors are notated in **bold**: $\pmb{a}$.
|
||||
In handwriting is is \underline{underlined}.
|
||||
|
||||
# Vector Algebra
|
||||
|
||||
## Equality
|
||||
|
||||
Two vectors are said to be equal if their magnitudes and directions are equal.
|
||||
|
||||
You can also do this by checking if their vertical and horizontal components are equal.
|
||||
|
||||
## Addition
|
||||
|
||||
Two vectors, $\pmb{a}$ and $\pmb{b}$, can be summed together by summing their components.
|
||||
|
||||
You can also do this graphically by drawing $\pmb{a}$ and then $\pmb{b}$ by putting its tail on
|
||||
the tip of $\pmb{a}$.
|
||||
The sum of the vectors is from the tail of $\pmb{a}$ to the tip of $\pmb{b}$:
|
||||
|
||||
|
||||
|
||||
Vector addition is associative[^d_associative] and commutative[^d_commutative].
|
||||
|
||||
## Zero Vector
|
||||
|
||||
The *zero vector* is denoted by $\pmb{0}$ and has zero magnitude and arbitrary direction.
|
||||
|
||||
$$\pmb{a} + \pmb 0 = \pmb a$$
|
||||
|
||||
If $\pmb a + \pmb b = 0$ then it is normal to write
|
||||
|
||||
|
||||
$$\pmb b = -\pmb a$$
|
||||
|
||||
$-\pmb a$ is a vector with the same magnitude to $\pmb a$ but opposite direction.
|
||||
|
||||
## Multiplication
|
||||
|
||||
### Multiplication by a Scalar
|
||||
|
||||
Let $k$, an arbitrary scalar and $\pmb a$, an arbitrary vector.
|
||||
|
||||
- $k\pmb a$ is a vector of magnitude $|k|$ times that of $\pmb a$ and is parralel to it
|
||||
- $0\pmb a = 0$
|
||||
- $1\pmb a = a$
|
||||
- $(-k)\pmb a = -(k\pmb a)$
|
||||
- $(-1)\pmb a = -\pmb a$
|
||||
- $k(\pmb a + \pmb b) = k\pmb a + k\pmb b$
|
||||
- $(k_1 + k_2)\pmb a = k_1\pmb a + k_2\pmb a$
|
||||
- $(k_1k_2)\pmb a = k_1(k_2\pmb a)$
|
||||
|
||||
### The Scalar Product (Inner Product, Dot Product)
|
||||
|
||||
The scalar product of two vectors $\pmb a$ and $\pmb b$ is a scalar defined by
|
||||
|
||||
$$\pmb a \cdot \pmb b = |\pmb a||\pmb b|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3$$
|
||||
|
||||
where $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$
|
||||
|
||||
where $\theta$ is the angle between the two vectors (note that $\cos\theta = \cos(2\pi - \theta)$).
|
||||
This definition does not depend on a coordinate system.
|
||||
|
||||
- The dot product is commutative[^d_commutative]
|
||||
- The dot product is distributive[^d_distributive]
|
||||
- If $\pmb a$ is perpendicular to $\pmb b$, then $\pmb a \cdot \pmb b = 0$ and they are said to be
|
||||
orthogonal
|
||||
- If $\pmb a \cdot \pmb b = 0$ then either
|
||||
|
||||
i. The vectors are orthogonal
|
||||
ii. One or both of the vectorse are zero vectors
|
||||
|
||||
- $\pmb a \cdot \pmb a = |\pmb a|^2 = a^2$
|
||||
|
||||
The base vectors are said to be *orthonormal* when $\pmb i^2 = \pmb j^2 = \pmb k^2 = 1$ and
|
||||
$i\cdot j = i\cdot k = j\cdot k = 0$.
|
||||
|
||||
### The Vector Product (Cross Product)
|
||||
|
||||
The vector product between two vectors is defined by:
|
||||
|
||||
$$\pmb a \times \pmb b = |\pmb a||\pmb b|\sin\theta \hat{\pmb n}$$
|
||||
|
||||
where $0 \le \theta \le \pi$ is the angle between $\pmb a$ and $\pmb b$ and $\pmb n$ is a unit
|
||||
vector such that the three vectors from a right handed system:
|
||||
|
||||
|
||||
|
||||
- $\pmb a \times \pmb b = -\pmb b \times \pmb a$ (the vector product is anti-commutative[^d_commutative])
|
||||
- If $\pmb a \times \pmb b = 0$ then either
|
||||
|
||||
i. The vectors are parallel
|
||||
ii. One or both of the vectors are a zero vector
|
||||
|
||||
- $(k_1\pmb a)\times(k_2\pmb b) = (k_1k_2)(\pmb a \times \pmb b)$ where $k_1$, $k_2$ are scalars
|
||||
- If $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$ then
|
||||
|
||||
$$\pmb a \times \pmb b = (a_2b_3 - a_3b_2, a_1b_3-a_3b_1, a_1b_2-a_2b_1)$$
|
||||
|
||||
- In the notation of determinants, provided we **expand by row 1**:
|
||||
|
||||
$$\pmb a \times \pmb b = \begin{vmatrix} \pmb i & \pmb j & \pmb k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$
|
||||
|
||||
This is technically not a determinant because not all the elements are numbers but shhhhhh...
|
||||
|
||||
|
||||
### Scalar Triple Product (and co-planar vectors)
|
||||
|
||||
\begin{align*}
|
||||
[ \pmb a, \pmb b, \pmb c ] &= \pmb a \cdot (\pmb b \times \pmb c) \\
|
||||
&= \pmb b \cdot (\pmb c \times \pmb a) \\
|
||||
&= \pmb c \cdot (\pmb a \times \pmb b) \\
|
||||
&= (\pmb b \times \pmb c) \cdot \pmb a \\
|
||||
&= (\pmb c \times \pmb a) \cdot \pmb b \\
|
||||
&= (\pmb a \times \pmb b) \cdot \pmb c
|
||||
\end{align*}
|
||||
|
||||
In terms of determinants:
|
||||
|
||||
$$
|
||||
[ \pmb a, \pmb b, \pmb c ] = \begin{vmatrix}
|
||||
a_1 & a_2 & a_3 \\
|
||||
b_1 & b_2 & b_3 \\
|
||||
c_1 & c_2 & c_3
|
||||
\end{vmatrix}
|
||||
$$
|
||||
|
||||
If $[\pmb a, \pmb b, \pmb c] = 0$ then the vectors are coplanar.
|
||||
|
||||
The absolute value of the scalar triple product reperesents the volume of the parallelepiped defined
|
||||
by those vectors:
|
||||
|
||||
|
||||
|
||||
|
||||
## The Unit Vector
|
||||
|
||||
$$\hat{\pmb a}= \frac{\pmb a}{|\pmb a|}$$
|
||||
|
||||
## Components of a Vector
|
||||
|
||||
The component of a vector $\pmb a$ in the direction of the **unit vector** $\hat{\pmb n}$ is
|
||||
|
||||
$$\pmb a \cdot \hat{\pmb n}$$
|
||||
|
||||
|
||||
|
||||
Vectors are often written in terms of base vectors, such as the Cartesian system's $\pmb i$,
|
||||
$\pmb j$, and $\pmb k$ in three dimensions.
|
||||
|
||||
|
||||
|
||||
These vectors have unit magnitude, are perpendicular to each other, and are right handed.
|
||||
|
||||
If $\pmb a = a_1\pmb i + a_2\pmb j + a_3\pmb k$ then the scalars $a_1$, $a_2$, and $a_3$ are the
|
||||
*components* of the vector (relative to the base vectors).
|
||||
|
||||
### Vector Projections
|
||||
|
||||
The *vector projection* of $\pmb a$ onto $\hat{\pmb n}$ is given by
|
||||
|
||||
$$(\pmb a \cdot \hat{\pmb n})\hat{\pmb n}$$
|
||||
|
||||
|
||||
|
||||
They look like the same as [vector components](#components-of-a-vector) to me...
|
||||
no idea what the difference is but uh StackExchange says
|
||||
([permalink](https://physics.stackexchange.com/a/537690)):
|
||||
|
||||
> As pointed out, the projection and component actually refers to the same thing.
|
||||
> To solve a problem like this it useful to introduce a coordinate system, as you mentioned yourself
|
||||
> you project onto the x-axis.
|
||||
> As soon as you introduce a coordinate system you can talk about the components of some vector.
|
||||
|
||||
## Position Vectors
|
||||
|
||||
If an origin $O$ is fixed, then any point $P$ in space may be represented by the vector $\pmb r$
|
||||
which has a magnitude and direction given by the line $\overrightarrow{OP}$.
|
||||
|
||||
A point $(x, y, z)$ in Cartesian space has the position vector $r = x\pmb i + y\pmb j + z\pmb k$.
|
||||
|
||||
## Angle Between Vectors
|
||||
|
||||
By rearranging the [dot product equation](#the-scalar-product-inner-product-dot-product) you can get
|
||||
an equation to find the angle, $\theta$, between two vectors:
|
||||
|
||||
$$\cos\theta = \frac{\pmb a \cdot \pmb b}{|\pmb a||\pmb b|} = \frac{a_1b_1 + a_2b_2 + a_3b_3}{|\pmb a ||\pmb b|}$$
|
||||
|
||||
# Applications of Vectors
|
||||
|
||||
## Application of Vectors to Geometry
|
||||
|
||||
### Area of a Parallelogram
|
||||
|
||||
$$area = |\pmb a||\pmb b|\sin\theta = |\pmb a \times \pmb b|$$
|
||||
|
||||
### Volume of a Parallelepiped
|
||||
|
||||
$$volume = | [ \pmb a, \pmb, b, \pmb c ] |$$
|
||||
|
||||
### Equation of a Straight Line
|
||||
|
||||
A straight line can be specified by
|
||||
|
||||
- two points it passes
|
||||
- one point it passes and a direction
|
||||
|
||||
If $\pmb a$ and $\pmb b$ are the position vectors of two distinct points, then the position vectors
|
||||
of an arbitrary point on the line joining these points is:
|
||||
|
||||
$$\pmb r = \pmb a + \lambda(\pmb b - \pmb a)$$
|
||||
|
||||
where $\lambda \in \Re$ is a parameter.
|
||||
|
||||
|
||||
|
||||
Suppose $O$ is an origin and $\pmb a$, $\pmb b$, and $\pmb r$ are position vectors on the line such
|
||||
that
|
||||
|
||||
\begin{align*}
|
||||
\pmb a &= (x_0, y_0, z_0) \\
|
||||
\pmb b &= (x_1, y_1, z_1) \\
|
||||
\pmb r &= (x, y, z)\\
|
||||
\\
|
||||
(x, y, z) &= (x_0, y_0, z_0) + \lambda((x_1, y_1, z_1) - (x_0, y_0, z_0)) \\
|
||||
\\
|
||||
x &= x_0 + \lambda(x_1-x_0) \\
|
||||
y &= y_0 + \lambda(y_1-y_0) \\
|
||||
z &= z_0 + \lambda(z_1-z_0) \\
|
||||
\\
|
||||
\lambda &= \frac{x-x_0}{x_1-x_0} = \frac{y-y_0}{y_1-y_0} = \frac{z-z_0}{z_1-z_0}
|
||||
\end{align*}
|
||||
|
||||
In the above, the vector $\pmb b - \pmb a$ is in the direction of the line.
|
||||
Thus the equation of a line can be specified by giving a point it passes through ($\pmb a$, say) and
|
||||
the direction of the line ($\pmb d = (d_1, d_2, d_3)$, say).
|
||||
|
||||
The vector equation is then
|
||||
|
||||
$$\pmb r = \pmb a + \lambda\pmb d$$
|
||||
|
||||
#### The Cartesian Equation
|
||||
|
||||
$$\frac{x-x_0}{d_1} = \frac{y-y_0}{d_2} = \frac{z-z_0}{d_3}$$
|
||||
|
||||
### Equation of a Plane
|
||||
|
||||
A *plane* can be defined by specifying either:
|
||||
|
||||
- three points (as long as they're not in a straight line)
|
||||
- a point on the plane and two directions (useful for a parametric form)
|
||||
- specifying a point on the plane and the normal vector to the plane
|
||||
|
||||
#### Specifying a Point and a Normal Vector
|
||||
|
||||
Let $\pmb a$ be the position vector of a point on the plane, and $\pmb n$ a normal vector to the
|
||||
plane.
|
||||
|
||||
If $\pmb r$ is the position vector of an arbitrary point on the plane, then $\pmb r - \pmb a$ is a
|
||||
vector lying *in* the plane, so
|
||||
|
||||
$$(\pmb r - \pmb a) \cdot \pmb n = 0$$
|
||||
|
||||
So the *vector equation* of the plane is
|
||||
|
||||
$$\pmb r \cdot \pmb n = \pmb a \cdot \pmb n = D$$
|
||||
|
||||
where $\pmb r = (x, y, z)$ and the vectors $\pmb a$ and $\pmb n$ are known.
|
||||
|
||||
Suppose $\pmb a$, $\pmb n$, and $\pmb r$ are given by
|
||||
|
||||
\begin{align*}
|
||||
\pmb a &= (x_0, y_0, z_0) \\
|
||||
\pmb n &= (l, m, p) \\
|
||||
\pmb r &= (x, y, z)\\
|
||||
\text{then } 0 &= ((x, y, z) - (x_0, y_0, z_0))\cdot(l, m, p)
|
||||
\end{align*}
|
||||
|
||||
#### Specifying Three Points on a Plane
|
||||
|
||||
If we specify three points on a plane with position vectors $\pmb a$, $\pmb b$, and $\pmb c$ the
|
||||
vectors $\pmb c - \pmb a$ and $\pmb c - \pmb b$ lie *in* the plane.
|
||||
(The vectors $\pmb a$, $\pmb b$, and $\pmb c$ do not necessarily lie *in* the plane;
|
||||
rather they take you from $O$ **to** the plane.)
|
||||
|
||||
The normal to the plane, $\pmb n$, is then parallel to
|
||||
|
||||
$$(\pmb c - \pmb a)\times(\pmb c - \pmb b)$$
|
||||
|
||||
and so the equation of the plane is
|
||||
|
||||
$$(\pmb r - \pmb a)\cdot((\pmb c - \pmb a)\times(\pmb c - \pmb b)) = 0$$
|
||||
|
||||
|
||||
#### The Angle Between Two Planes
|
||||
|
||||
... is the same as the angle between their normal vectors
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
[^d_associative]: The grouping of elements in in an operation do not matter (e.g. scalar addition: $a+(b+c) = (a+b)+c$)
|
||||
[^d_commutative]: The order of elements in an operation do not matter (e.g. scalar addition: $a+b = b+a$)
|
||||
[^d_distributive]: Easiest to explain with examples. Scalar multiplication is said to be distributive because $(a+b)c = ac + bc$
|
||||
Reference in New Issue
Block a user