From 7a0a6938d3292b37a0f25b604a4e751118682996 Mon Sep 17 00:00:00 2001 From: Andrew Lorimer Date: Mon, 22 Apr 2019 18:12:30 +1000 Subject: [PATCH] [spec] tidy up calculus notes --- spec/calculus.md | 75 +++++++++++++++++++++--------------------------- 1 file changed, 33 insertions(+), 42 deletions(-) diff --git a/spec/calculus.md b/spec/calculus.md index ddda405..9025d37 100644 --- a/spec/calculus.md +++ b/spec/calculus.md @@ -1,12 +1,3 @@ ---- -geometry: margin=2cm -columns: 2 -graphics: yes -tables: yes -author: Andrew Lorimer -classoption: twocolumn ---- - # Differential calculus ## Limits @@ -63,14 +54,14 @@ Can also be used with functions, where $h=\delta x$. $$f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={dy \over dx}$$ -$$m_{\operatorname{tangent}}=\lim_{h \rightarrow 0}f^\prime(x)$$ +$$m_{\tan}=\lim_{h \rightarrow 0}f^\prime(x)$$ -$$m_{\operatorname{chord PQ}}=f^\prime(x)$$ +$$m_{\vec{PQ}}=f^\prime(x)$$ first principles derivative: -$${m_{\operatorname{tangent at P}} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$ +$${m_{\text{tangent at }P} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$ ## Gradient at a point @@ -125,15 +116,11 @@ ${dy \over du} = 7u^6$ $${dy \over dx} = u{dv \over dx} + v{du \over dx}$$ -Surds can be left on denomintaors. - ## Quotient rule for $y={u \over v}$ $${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$$ -If $f(x)={u(x) \over v(x)}$, then $f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}$ - -If $y={u(x) \over v(x)}$, then derivative ${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$ +$$f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}$$ ## Logarithms @@ -166,7 +153,7 @@ $${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$ ## Derivative rules -| $f(x)$ | $f^\prime(x)$ |xs +| $f(x)$ | $f^\prime(x)$ | | ------ | ------------- | | $\sin x$ | $\cos x$ | | $\sin ax$ | $a\cos ax$ | @@ -184,23 +171,21 @@ $${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$ | $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ | | $\tan^{-1} x$ | $1 \over {1 + x^2}$ | - +## Reciprocal derivatives -Reciprocal derivatives: - -$${{dy \over dx} \over 1} = dx \over dy$$ +$${1 \over {dy \over dx}} = {dx \over dy}$$ ## Differentiating $x=f(y)$ -Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}}$. +Find $dx \over dy$. Then ${dx \over dy} = {1 \over {dy \over dx}} \implies {dy \over dx} = {1 \over {dx \over dy}}$. $${dy \over dx} = {1 \over {dx \over dy}}$$ ## Second derivative -$$f(x) \implies f^\prime (x) \implies f^{\prime\prime}(x)$$ +$$f(x) \longrightarrow f^\prime (x) \longrightarrow f^{\prime\prime}(x)$$ -$$\therefore y \implies {dy \over dx} \implies {d({dy \over dx}) \over dx} \implies {d^2 y \over dx^2}$$ +$$\therefore y \longrightarrow {dy \over dx} \longrightarrow {d({dy \over dx}) \over dx} \longrightarrow {d^2 y \over dx^2}$$ Order of polynomial $n$th derivative decrements each time the derivative is taken @@ -209,16 +194,16 @@ Order of polynomial $n$th derivative decrements each time the derivative is take *Stationary point* - point of zero gradient (i.e. $f^\prime(x)=0$) *Point of inflection* - point of maximum $|$gradient$|$ (i.e. $f^{\prime\prime} = 0$) -- if $f^\prime (a) = 0$ and $f^{\prime\prime}(a) > 0$, then point $(a, f(a))$ is a local min (curve is concave up) -- if $f^\prime (a) = 0$ and $f^{\prime\prime} (a) < 0$, then point $(a, f(a))$ is local max (curve is concave down) -- if $f^{\prime\prime}(a) = 0$, then point $(a, f(a))$ is a point of inflection -- - if also $f^\prime(a)=0$, then it is a stationary point of inflection +* if $f^\prime (a) = 0$ and $f^{\prime\prime}(a) > 0$, then point $(a, f(a))$ is a local min (curve is concave up) +* if $f^\prime (a) = 0$ and $f^{\prime\prime} (a) < 0$, then point $(a, f(a))$ is local max (curve is concave down) +* if $f^{\prime\prime}(a) = 0$, then point $(a, f(a))$ is a point of inflection + + if also $f^\prime(a)=0$, then it is a stationary point of inflection ![](graphics/second-derivatives.png) ## Implicit Differentiation -On CAS: Action $\rightarrow$ Calculation $\rightarrow$ `impDiff(y^2+ax=5, x, y)`. Returns $y^\prime= \dots$. +**On CAS:** Action $\rightarrow$ Calculation $\rightarrow$ `impDiff(y^2+ax=5, x, y)`. Returns $y^\prime= \dots$. Used for differentiating circles etc. @@ -226,19 +211,15 @@ If $p$ and $q$ are expressions in $x$ and $y$ such that $p=q$, for all $x$ nd $y $${dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}$$ -## Antidifferentiation - -$$y={x^{n+1} \over n+1} + c$$ - ## Integration -$$\int f(x) dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)$$ +$$\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)$$ + +$$\int x^n \cdot dx = {x^{n+1} \over n+1} + c$$ - area enclosed by curves - $+c$ should be shown on each step without $\int$ -$$\int x^n = {x^{n+1} \over n+1} + c$$ - ### Integral laws $\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$ @@ -255,18 +236,28 @@ $\int k f(x) dx = k \int f(x) dx$ | $e^k$ | $e^kx + c$ | | $\sin kx$ | $-{1 \over k} \cos (kx) + c$ | | $\cos kx$ | ${1 \over k} \sin (kx) + c$ | +| $\sec^2 kx$ | ${1 \over k} \tan(kx) + c$ | +| $1 \over \sqrt{a^2-x^2}$ | $\sin^{-1} {x \over a} + c \>\vert\> a>0$ | +| $-1 \over \sqrt{a^2-x^2}$ | $\cos^{-1} {x \over a} + c \>\vert\> a>0$ | +| $a \over {a^2-x^2}$ | $\tan^{-1} {x \over a} + c$ | | ${f^\prime (x)} \over {f(x)}$ | $\log_e f(x) + c$ | | $g^\prime(x)\cdot f^\prime(g(x)$ | $f(g(x))$ (chain rule)| | $f(x) \cdot g(x)$ | $\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx$ | +Note $\sin^{-1} {x \over a} + \cos^{-1} {x \over a}$ is constant for all $x \in (-a, a)$. + ### Definite integrals -$$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)_{}$$ +$$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)$$ + +- Signed area enclosed by: $\> y=f(x), \quad y=0, \quad x=a, \quad x=b$. +- *Integrand* is $f$. +- $F(x)$ may be any integral, i.e. $c$ is inconsequential ## Applications of antidifferentiation - $x$-intercepts of $y=f(x)$ identify $x$-coordinates of stationary points on $y=F(x)$ -- the nature of any stationary point of $y=F(x)$ is determined by the way the sign of the graph of $y=f(x)$ changes about its $x$-intercepts +- nature of stationary points is determined by sign of $y=f(x)$ on either side of its $x$-intercepts - if $f(x)$ is a polynomial of degree $n$, then $F(x)$ has degree $n+1$ To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function. @@ -275,13 +266,13 @@ To find stationary points of a function, substitute $x$ value of given point int ### Related rates -$${da \over db} \quad \text{change in } a \text{ with respect to } b$$ +$${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$ #### Gradient at a point on parametric curve -$${dy \over dx} = {{dy \over dt} \over {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$ +$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$ -$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$ +$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \div {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$ ## Rational functions -- 2.43.2