Calculus – Physicspupil

August 21, 2024August 22, 2024

202408211136 Exercise 14.C.10

A stone is thrown vertically upwards so that its height, $s$ metres, above the ground, after $t$ seconds is given by $s=40t-5t^2$ . Find

(a) its velocity after $2\,\mathrm{s}$ ;
(b) its height above the ground when it is momentarily at rest;
(c) its initial velocity;
(d) its velocity when it is $15\,\mathrm{m}$ above the ground, giving your answer correct to the nearest $\mathrm{m\,s^{-1}}$ .

_{Extracted from K. S. Teh & C. Y. Loh. (2007). New Syllabus Additional Mathematics (8e)}

Roughwork.

Initially (i.e., at $t=0$ ), the stone is

$s(t=0)=40(0)-5(0)^2=0\,\mathrm{m}$

above the ground, i.e., at ground level $s=0$ .

$\begin{aligned} & s = 40t-5t^2 = t(40-5t) \\ \Rightarrow & \begin{cases} s =0 \Leftrightarrow t = \{0\}\cup\{ 8\} \\ s > 0 \Leftrightarrow t\in (0,8) \\ s < 0 \Leftrightarrow t<0\enspace\textrm{(rej.)}\enspace\textrm{\scriptsize{OR}}\enspace t>8 \\ \end{cases} \\ \end{aligned}$

Take upward positive. The (vertical) displacement of the stone is

$\begin{aligned} \mathbf{s}(t) & =s(t)\,\hat{\mathbf{j}} \\ & = (40t-5t^2)\,\hat{\mathbf{j}} \\ \end{aligned}$

its (vertical) instantaneous velocity

$\begin{aligned} \mathbf{v}(t) & = v(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}s(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40t-5t^2)'\,\hat{\mathbf{j}} \\ & = (40-10t)\,\hat{\mathbf{j}} \\ \end{aligned}$

and its (vertical) instantaneous acceleration

$\begin{aligned} \mathbf{a}(t) & = a(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}v(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40-10t)'\,\hat{\mathbf{j}} \\ \mathbf{a} & = -10\,\hat{\mathbf{j}} \\ \end{aligned}$

so are the answers to

(a) $\mathbf{v}(2)$
(b) $t'\in\{ t:v(t)=0\}\enspace\rightsquigarrow\enspace s(t')$
(c) $\mathbf{v}(0)$
(d) $t'\in\{ t:\mathbf{s}(t)=+15\,\hat{\mathbf{j}}\} \enspace\rightsquigarrow\enspace \mathbf{v}(t')$

This problem is not to be attempted.

August 20, 2024August 20, 2024

202408201533 Example 13.8

Let $\displaystyle{x=\frac{y-2}{y+2}}$ .

(a) Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $y$ .
(b) Make $y$ the subject of the given function. Hence, find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $x$ .
(c) Are the answers obtained in (a) and (b) different? Explain your assertion.

_{Extracted from S. W. Li et al. (2002). New Progress in Additional Mathematics Book 3}

Roughwork.

$\begin{aligned} x & = \frac{y-2}{y+2} \\ x & = \frac{y+2-4}{y+2} \\ \textrm{Eq. (1):}\quad x=f(y) & = 1-\frac{4}{y+2} \\ \frac{4}{y+2} & = 1-x \\ \frac{4}{1-x} & = y+2 \\ \textrm{Eq. (2):}\quad y=g(x) & = \frac{4}{1-x} - 2 \\ \end{aligned}$

domain: what can go into a function
codomain: what may possibly come out of a function
range: what actually comes out of a function

_{Kevin Sookocheff on Range, Domain, and Codomain (2018)}

Write, for function $f$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ -2\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

and for function $g$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ 1\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

We see $f$ discontinuous at $-2$ ; and $g$ , at $1$ ; whereas non-differentiable.

(to be continued)

February 15, 2024February 15, 2024

202402151329 Solution to 1985-CE-AMATH-II-5

If the equation

$x^2+y^2+kx-(2+k)y=0$

represents a circle with radius $\sqrt{5}$ ,

(a) find the value(s) of $k$ ;
(b) find the equation(s) of the circle(s).

As if sitting an exam in additional mathematics, it certainly not being showing off, we should perhaps involve ourselves with some sort of calculus.

Roughwork.

As always, have in mind some pictures. Hence, we draw:

and also

after the stage got set, we write

$\begin{aligned} 0 & = x^2+y^2+kx-(2+k)y \\ 0 & = 2x\,\mathrm{d}x+2y\,\mathrm{d}y + k\,\mathrm{d}x - (2+k)\,\mathrm{d}y \\ y' & = \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x+k}{k-2y+2} \\ \end{aligned}$

and furthermore,

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ & \Rightarrow 2x+k=0 \\ & \Rightarrow x = -\frac{k}{2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \infty \\ & \Rightarrow k-2y+2 = 0 \\ & \Rightarrow y = \frac{k}{2}+1 \\ \end{aligned}$

In the case of $x=-\frac{k}{2}$ :

$\begin{aligned} 0 & = \bigg( -\frac{k}{2}\bigg)^2+y^2+k\bigg( -\frac{k}{2}\bigg) -(2+k)y \\ 0 & = y^2 - (2+k)y - \frac{k^2}{4} \\ \Delta & = \big( -(2+k)\big)^2 - 4(1)\bigg( -\frac{k^2}{4}\bigg) \\ & = 2(k^2+2k+2)\qquad (> 0) \\ y_1,y_3 & = \frac{-\big( -(2+k)\big) \pm \sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = y_3-y_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

and the case of $y=\frac{k}{2}+1$ :

$\begin{aligned} 0 & = x^2 + \bigg( \frac{k}{2}+1\bigg)^2 + kx - (2+k)\bigg( \frac{k}{2}+1\bigg) \\ 0 & = x^2+kx - \bigg(\frac{k}{2}+1\bigg)^2\\ \Delta & = (k)^2 - 4(1)\bigg( -\bigg(\frac{k}{2}+1\bigg)^2\bigg) \\ & = 2(k^2+2k+2) \qquad (> 0) \\ x_1,x_3 & = \frac{-(k)\pm\sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = x_3 - x_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

Targeting the centre $C(a,b)$ :

$\begin{aligned} a_{i=1,2} & = \frac{x_3 - x_1}{2} = -\frac{k}{2}\bigg|_{k=2,-4} = -1,2 \\ b_{i=1,2} & = \frac{y_3-y_1}{2} = \frac{-\big( -(2+k)\big)}{2}\bigg|_{k=2,-4} = 2,-1 \\ (a,b) & = (-1,2)\cup (2,-1) \\ \end{aligned}$

Of the equation of circle, the standard form is

$\left\{ \begin{aligned} (x+1)^2 + (y-2)^2 & = (\sqrt{5})^2 \\ (x-2)^2 + (y+1)^2 & = (\sqrt{5})^2 \\ \end{aligned}\right\}$

and the general form

$\left\{ \begin{aligned} x^2+y^2+2x-4y & = 0 \\ x^2+y^2-4x+2y & = 0 \\ \end{aligned}\right\}$

This problem is not to be attempted.

January 31, 2024January 31, 2024

202401311548 Pastime Exercise 008

Let there be a cone $R\,\mathrm{(unit)}$ in radius of its base and $h\,\mathrm{(unit)}$ in height.

Find the volume $V\,\mathrm{(cubic\, unit)}$ of the cone, using what are so-called i. the disc (/washer) method and ii. the (cylindrical) shell method, if the names my memory serves me right.

Answer. $\displaystyle{V=\frac{1}{3}\pi R^2h}$ .

Roughwork.

Visualize the cone.

i. By disc method,

$\begin{aligned} \frac{R}{h} & = \frac{r}{h-z} = \tan\theta \\ r & = R\bigg( 1-\frac{z}{h}\bigg) \\ V & = \pi R^2\int_{0}^{h}\bigg( 1-\frac{z}{h}\bigg)^2\,\mathrm{d}z \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ \textrm{let }& z'=1-\frac{z}{h} \\ \frac{\mathrm{d}z'}{\mathrm{d}z} & = -\frac{1}{h} \\ \mathrm{d}z & = -h\,\mathrm{d}z' \\ z=0 & \Leftrightarrow z'=1 \\ z=h & \Leftrightarrow z'= 0 \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ V& = \pi R^2\int_{1}^{0} z'^2(-h\,\mathrm{d}z') \\ & = -\pi R^2h \bigg[ \frac{z'^3}{3}\bigg]\bigg|_{1}^{0} \\ & = -\pi R^2h \bigg(\frac{(0)^3}{3} - \frac{(1)^3}{3}\bigg) \\ & = \frac{\pi R^2h}{3}\\ \end{aligned}$

ii. By shell method,

we are obtaining the volume of a solid of revolution about the $z$ -axis by integrating the slices (/cross sections) ranging between $\theta\in [0,2\pi ]$ .

It is left the reader as an exercise.

This problem is not to be attempted.

October 19, 2023October 19, 2023

202310191146 Exercise 7.9.26

Evaluate the integral

$\displaystyle{\int x\sin x\cos x\,\mathrm{d}x}$ .

_{Extracted from James Stewart. (2012). Calculus (8e)}

Roughwork.

(abortive attempt)

$\begin{aligned} & \quad\enspace \int x\sin x\cos x\,\mathrm{d}x \\ & = \int x\sin x\cos x \bigg(\frac{\mathrm{d}(\sin x)}{\cos x}\bigg) \\ & = \int x\sin x\,\mathrm{d}(\sin x) \\ & = (x\sin x)(\sin x) - \int (\sin x)\bigg(\frac{\mathrm{d}(x\sin x)}{\mathrm{d}x}\bigg)\,\mathrm{d}x + C\quad\textrm{for some constant} \\ & = x\sin^2x - \int \sin x\,\mathrm{d}(x\sin x) \\ & = x\sin^2x - \bigg\{ (\sin x)(x\sin x) - \int (x\sin x)\bigg(\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)\bigg) \,\mathrm{d}x \bigg\} \\ & = x\sin^2x - \bigg\{ x\sin^2x - \int x\sin x\cos x\,\mathrm{d}x \bigg\} \\ & = \int x\sin x\cos x\,\mathrm{d}x \\ \end{aligned}$

As a last resort, I need help from the following

Fundamental Theorem of Calculus. Let $f$ be a continuous real-valued function defined on a closed interval $[a,b]$ , Let $F$ be the function defined, for all $x$ in $[a,b]$ , by

$\displaystyle{F(x)=\int_{a}^{x}f(t)\,\mathrm{d}t}$

Then $F$ is uniformly continuous on $[a,b]$ and differentiable on the open interval $(a,b)$ , and

$F'(x)=f(x)$

for all $x$ in $(a,b)$ so $F$ is an antiderivative of $f$ .

_{Cited from Wikipedia on Fundamental theorem of calculus}

Let $\displaystyle{\frac{\mathrm{d}F(x)}{\mathrm{d}x}=x\sin x\cos x}$ .

But for using foresight, one should not have written

$\begin{aligned} F(x) & :=\frac{x}{2}\sin^2x+G(x) \\ \frac{\mathrm{d}F(x)}{\mathrm{d}x} & = x\sin x\cos x + \frac{\sin^2x}{2} + G'(x) \\ G'(x) & := -\frac{\sin^2x}{2} \\ G(x) & = \int G'(x)\,\mathrm{d}x \\ & = -\frac{1}{2}\int\sin^2x\,\mathrm{d}x \\ & = -\frac{1}{2}\int \bigg( \frac{1-\cos 2x}{2}\bigg) \,\mathrm{d}x\\ & = -\frac{1}{4}\bigg\{ \int \mathrm{d}x - \int \cos 2x\,\mathrm{d}x \bigg\} \\ & = -\frac{1}{4}\bigg\{ x - \frac{\sin 2x}{2} \bigg\} \\ F(x) & = \frac{x}{2}\sin^2x - \frac{1}{4}\bigg( x - \frac{\sin 2x}{2} \bigg) \\ & = \frac{x\sin^2x}{2} + \frac{\sin 2x}{8} - \frac{x}{4} \\ \end{aligned}$

with some constant $C$ .

The case is closed.

October 19, 2023

202310191053 Exercise 7.9.5

Evaluate the integral

$\displaystyle{\int \frac{\mathrm{d}t}{2t^2+3t+1}}$ .

_{Extracted from James Stewart. (2012). Calculus (8e)}

Roughwork.

$\begin{aligned} &\quad\enspace \int\frac{1}{2t^2+3t+1}\,\mathrm{d}t \\ & = \int\frac{1}{(2t+1)(t+1)}\,\mathrm{d}t \\ & = \int\bigg\{ \frac{A}{2t+1} + \frac{B}{t+1}\frac{}{}\bigg\}\,\mathrm{d}t \\ & = \int\frac{A(t+1)+B(2t+1)}{(2t+1)(t+1)}\,\mathrm{d}t\\ & = \int\frac{(A+2B)t+(A+B)}{(2t+1)(t+1)}\,\mathrm{d}t\\ & = \int\bigg\{\frac{2}{2t+1}-\frac{1}{t+1}\bigg\}\,\mathrm{d}t \\ & = \int\frac{2}{2t+1}\,\mathrm{d}t - \int\frac{1}{t+1}\,\mathrm{d}t \\ & = \int\frac{1}{2t+1}\,\mathrm{d}(2t+1) - \int\frac{1}{t+1}\,\mathrm{d}(t+1) \\ & = \ln |2t+1| - \ln |t+1| +C\quad\textrm{for some constant}\\ &= \ln \bigg| \frac{2t+1}{t+1} \bigg| +C \\ \end{aligned}$

This problem is not to be attempted.

October 18, 2023November 26, 2023

202310181051 Exercise 7.9.21

Evaluate the integral

$\displaystyle{\int \frac{\mathrm{d}x}{\sqrt{x^2-4x}}}$ .

_{Extracted from James Stewart. (2012). Calculus (8e)}

Roughwork.

There are as many ways to evaluate definite/indefinite integrals as there are to evaluate derivatives/differentials, for example, by i. basic integration formulae, ii. substitution by rational polynomials, trigonometric functions, or hyperbolic functions, iii. integration by parts, iv. reduction formulae, iv. simplification by partial fractions, v. parity argument, and vi. parametrization.

Make use of none above but the following

Residue Theorem. Suppose that the function $f$ is analytic within and on a positively oriented simple closed contour $C$ except for a finite number of isolated singular points $\{ z_j, j=1,2,\dots ,N\}$ interior to $C$ , then

$\displaystyle{\int_{C}f(z)\,\mathrm{d}z=2\pi i\sum_{j=1}^{N}\mathrm{Res}_{z=z_j}f(z)}$ .

Residue Formula. If $f(z)$ has a pole of order $k$ at $z=z_0$ , then

$\displaystyle{\mathrm{Res}\, (f,z_0)=\frac{1}{(k-1)!}\frac{\mathrm{d}^{k-1}}{\mathrm{d}z^{k-1}}\Big( (z-z_0)^kf(z)\Big)\bigg|_{z=z_0}}$

Write for some real number $x$

$\begin{aligned} & \quad\enspace \int \frac{\mathrm{d}x}{\sqrt{x^2-4x}} \\ & \hookrightarrow \textrm{let }R(x)=\frac{1}{\sqrt{x^2-4x}} \\ \end{aligned}$

and also for some complex number $z=x+iy$ ,

$\begin{aligned} & \quad\enspace \int \frac{\mathrm{d}z}{\sqrt{z^2-4z}} \\ & \hookrightarrow \textrm{let }R(z)=\frac{1}{\sqrt{z^2-4z}} \\ \end{aligned}$

Then there are two singular points (/singularities) $(0,0)$ and $(4,0)$ on the Argand plane, because, for instance

$\displaystyle{R\big( (0,0)\big)=\frac{1}{\sqrt{(0)^2-4(0)}}}$

is undefined.

From the denominator of $R(z)$ we see there

$\sqrt{z(z-4)}$

$2$ poles of order $\frac{1}{2}$ (*fractional order seems disputable, but no worries as you read further along), by the fact that

$\begin{aligned} & \lim_{z\to 0} (z)^{(\frac{1}{2})}\frac{1}{\sqrt{z^2-4z}} = -\frac{i}{2} \neq 0 \\ & \lim_{z\to 4} (z-4)^{(\frac{1}{2})}\frac{1}{\sqrt{z^2-4z}} = \frac{1}{2} \neq 0 \\ \end{aligned}$

and their residues being:

$z(x,y)=\{ (0,0)\} \cup \{ (4,0)\}$ .

We introduce the below

Lemma 1. (positive fractional derivatives)

$\displaystyle{\frac{\mathrm{d}^nx^m}{\mathrm{d}x^n}=\frac{\Gamma (m+1)}{\Gamma (m-n+1)}x^{m-n}}$

In particular, if letting $m=1$ and $n=1/2$ , then

$\displaystyle{\frac{\mathrm{d}^{1/2}x}{\mathrm{d}x^{1/2}}=\frac{2\sqrt{x}}{\sqrt{\pi}}}$ .

Lemma 2. (negative fractional derivatives)

$\displaystyle{D^{-\nu}x^{\mu}=\frac{\Gamma (\mu +1)}{\Gamma (\mu +\nu +1)}x^{\mu +\nu}}$

where $\nu >0$ , $\mu >-1$ , and $x>0$ . In particular, if $\nu =\frac{1}{2}$ , then

$\begin{aligned} D^{-1/2}x^0 & = \frac{\Gamma (1)}{\Gamma (3/2)}x^{1/2}=2\sqrt{\frac{x}{\pi}} \\ D^{-1/2}x^1 & = \frac{\Gamma (2)}{\Gamma (5/2)}x^{3/2}=\frac{4}{3}\sqrt{\frac{x^3}{\pi}} \\ D^{-1/2}x^2 & = \frac{\Gamma (3)}{\Gamma (7/2)}x^{5/2}=\frac{16}{15}\sqrt{\frac{x^5}{\pi}} \\ \end{aligned}$

where the Legendre’s Gamma function is given by

$\Gamma (n+1)=n!$ .

For brevity we state without proof the definition of factorial of fractions:

$\displaystyle{N!=\prod (N)}$

where $N\in\mathbb{Q}$ and Gauss product Pi is

$\displaystyle{\prod (x) = \int_{0}^{\infty}t^xe^{-t}\,\mathrm{d}t}$

_{Cited from Kimeu, Joseph M., “Fractional Calculus: Definitions and Applications” (2009). Masters Theses & Specialist Projects. Paper 115.}

Therefore

$\begin{aligned} &\quad\enspace \int R(x)\,\mathrm{d}x \\ & = \pi i\bigg\{ \mathrm{Res}\, [R(z),0] + \mathrm{Res}\, [R(z),4] \bigg\} \\ \end{aligned}$

$\begin{aligned} \hookrightarrow\quad\mathrm{Res}(R(z),0) & = \frac{1}{(-1/2)!}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}(z-0)^{1/2}R(z)\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{\sqrt{z}}{\sqrt{z^2-4z}}\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{1}{\sqrt{z-4}}\bigg|_{z=0} \\ & \stackrel{\dagger}{=}\frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z'^3}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z'^{-1/2}}\big( z'\big) \\ & = \frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z'^3}}\bigg(\frac{4}{3}\sqrt{\frac{z'^3}{\pi}}\bigg) \\ & = \frac{4\sqrt{2}i}{3\pi} \\ & \\ \textrm{\dag} :\qquad\qquad & \\ \quad \textrm{let }z' & = \frac{1}{\sqrt{z-4}} \\ \textrm{then }\frac{\mathrm{d}z'}{\mathrm{d}z} & = -\bigg(\frac{1}{2}\bigg)(z-4)^{-3/2}(1) \\ & = -\frac{z'^3}{2} \\ \mathrm{d}z & = -\frac{2}{z'^3}\,\mathrm{d}z' \\ \end{aligned}$

$\begin{aligned} \hookrightarrow\quad\mathrm{Res}(R(z),4) & = \frac{1}{(-1/2)!}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}(z-4)^{1/2}R(z)\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{\sqrt{z-4}}{\sqrt{z^2-4z}}\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{1}{\sqrt{z}}\bigg|_{z=0} \\ & \stackrel{\textrm{\ddag}}{=}\frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z''^3}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z''^{-1/2}}\big( z''\big) \\ & = \frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z''^3}}\bigg(\frac{4}{3}\sqrt{\frac{z''^3}{\pi}}\bigg) \\ & = \frac{4\sqrt{2}i}{3\pi} \\ & \\ \textrm{\ddag} :\qquad\qquad & \\ \quad \textrm{let }z'' & = \frac{1}{\sqrt{z-4}} \\ \textrm{then }\frac{\mathrm{d}z''}{\mathrm{d}z} & = -\bigg(\frac{1}{2}\bigg)(z-4)^{-3/2}(1) \\ & = -\frac{z''^3}{2} \\ \mathrm{d}z & = -\frac{2}{z''^3}\,\mathrm{d}z'' \\ \end{aligned}$

$\begin{aligned} &\quad\enspace \int R(x)\,\mathrm{d}x \\ & = \pi i\bigg\{ \mathrm{Res}\, [R(z),0] + \mathrm{Res}\, [R(z),4] \bigg\} \\ & = \pi i \bigg\{\frac{4\sqrt{2}i}{3\pi}+\frac{4\sqrt{2}i}{3\pi}\bigg\} \\ & = -\frac{8\sqrt{2}}{3} \\ \end{aligned}$

Of course this $\textrm{\scriptsize{MUST}}$ be a joke; how can an indefinite integral give a definite value, let alone be independent of any variable?

Countercheck.

The correct answer is

$2\ln (\sqrt{x}+\sqrt{x-4})+C\enspace\textrm{for some constant}$ .

(to be continued)

October 17, 2023October 17, 2023

202310171238 Exercise 18.6.8

A particle moves along a straight line so that its distance ( $s$ ) from a fixed point $O$ on the line after $t\,\mathrm{s}$ is given by

$s=t^3-12t^2+45t$ .

Find the distances from $O$ when the particle is momentarily at rest and the accelerations at these times.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI units}

Roughwork.

We give the direction to the right a positive sign.

By substituting $t=0$ for time, we found the original position of the particle to be at a distance:

$\begin{aligned} s(t=0) & = |\mathbf{s}(0)| \\ & = |(0)^3-12(0)^2+45(0)\,\hat{\mathbf{i}}| \\ & = 0\,\mathrm{unit} \\ \end{aligned}$

from point $O$ , which shall be called the origin hence.

The instantaneous velocity being

$\begin{aligned} \mathbf{v}(t)& =\lim_{\Delta t\to 0}\frac{\Delta \mathbf{s}}{\Delta t} \\ & = \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}t} \\ & = \frac{\mathrm{d}}{\mathrm{d}t}(t^3-12t^2+45t)\,\hat{\mathbf{i}} \\ & = (3)t^{(3)-1} - 12(2)t^{(2)-1} + 45(1)t^{(1)-1} \\ & = 3t^2 -24t + 45\,\hat{\mathbf{i}} \\ \end{aligned}$

the average velocity during time interval of $t\in [0,t']$ being

$\begin{aligned} v_{\textrm{avg}}(t') & = \frac{s(t')}{t'} \\ & = \frac{(t')^3-12(t')^2+45(t')}{t'} \\ & = t'^2-12t'+45 \\ \end{aligned}$

in magnitude; the instantaneous acceleration being

$\begin{aligned} a(t) & = \bigg| \frac{\mathrm{d}^2}{\mathrm{d}t^2}(s(t))\,\hat{\mathbf{i}}\bigg| \\ & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}}{\mathrm{d}t}(t^3-12t^2+45t) \bigg) \\ & = \frac{\mathrm{d}}{\mathrm{d}t}(3t^2-24t+45) \\ & = 6t-24\\ \end{aligned}$

in magnitude; the momentum being

$\begin{aligned} p & =mv\\ & = m(3t^2 -24t + 45) \\ \end{aligned}$

in magnitude; and the displacement

$\begin{aligned} s(t) & = t^3-12t^2+45t \\ & = t(t-3)(t-15) \\ \end{aligned}$

in magnitude; such that the particle is at the origin $O$ when

$t=\{ 0\}\cup\{ 3\}\cup \{ 15\}$ .

The particle is momentarily at rest when the direction of velocity changes signs, or the instant speed is $v=\lVert \mathbf{v}\rVert = 0$ :

$\begin{aligned} v(t) & = 3t^2-24t+45 \\ & = 3(t-5)(t-3) \\ \end{aligned}$

such as $t=\{ 3\}\cup\{ 5\}$ that the distances from $O$ are respectively

$\begin{aligned} s(3) & =(3)^3-12(3)^2+45(3) = 0 \\ s(5) & =(5)^3-12(5)^2+45(5) = 50 \\ \end{aligned}$

and the accelerations at these times

$\begin{aligned} \mathbf{a}(3) & =6(3)-24 = 6\,(-)\hat{\mathbf{i}} \\ \mathbf{a}(5) & =6(5)-24 = 6\,(+)\hat{\mathbf{i}} \\ \end{aligned}$ .

This problem is not to be attempted.

October 13, 2023October 16, 2023

202310130920 Exercise 14.1.3

If $\cos A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$ where $A$ and $B$ are both acute angles, find, without using tables or a calculator,

(a) $\sin (A+B)$ ,
(b) $\cos (A-B)$ ,
(c) $\tan (A+B)$ .

_{Extracted from J. F. Talbert & H. H. Heng. (1995). Additional Mathematics Pure and Applied (6e).}

Roughwork.

(a)

Not yet to approach a solution, I wish to get a feel of the problem from three perspectives— i. algebraic, ii. geometric, and iii. differential:

(i. algebraic)

$\begin{aligned} \cos A & =\frac{4}{5} \\ \cos B & = \frac{12}{13} \\ & \\ \cos^2A & = \frac{16}{25} \\ \cos^2B & = \frac{144}{169} \\ & \\ \sin^2A & = 1-\cos^2A\\ & = \frac{9}{25} \\ \sin^2B & = 1-\cos^2B\\ & = \frac{25}{169} \\ & \\ \sin A & = \sqrt{\frac{9}{25}} \\ & = \frac{3}{5} \\ \sin B & = \sqrt{\frac{25}{169}} \\ & = \frac{5}{13} \\ \end{aligned}$

(ii. geometric)

(iii. differential)

Let $f(x)=\sin x$ . Then, recall that

$\displaystyle{f(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }$

because of Taylor series/expansion of a function $f(x)$ at some point $a$ :

$\displaystyle{f(x)\big|_{x=a} = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\bigg|_{x=a}}$

such that in our situation, you know, we have $a=0$ (the Maclaurin series):

$\displaystyle{f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n}$

Now that we have enough, here begins the

Solution.

(i. algebraic)

$\begin{aligned} \sin (A+B) & = \sin A\cos B + \cos A\sin B \\ & = \bigg(\frac{3}{5}{\bigg)\bigg(\frac{12}{13}\bigg) + \bigg(\frac{4}{5}\bigg)\bigg(\frac{5}{13}\bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(ii. geometric)

$\begin{aligned} \frac{h}{14} & = \sin (90^\circ -B) \\ h & = 14\cos B \\ & \\ \sin (A+B) & = \frac{h}{15} \\ & = \frac{14\cos B}{15} \\ & = \frac{14}{15}\bigg( \frac{12}{13} \bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(iii. differential)

That angles $A$ and $B$ are acute angles, i.e.,

$0<B<A<90^\circ$

does not imply $A+B$ is also an acute angle, but that

$0<A+B<180^\circ$ .

such that

$\begin{aligned} 0 & <\sin (A+B)<1 \\ -1 & <\cos (A+B)<1 \\ 0 & <\sin B<\sin A<\cos A<\cos B < 1\\ \end{aligned}$

Letting $f(x)=\sin x$ ,

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A+B] \\ & = \int_{0}^{A+B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A+B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A+B} \\ & = 1-\cos (A+B) \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A] \\ & = \int_{0}^{A}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A} \\ & = 1-\cos A \\ & = 1-\frac{4}{5} \\ & = \frac{1}{5} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,B] \\ & = \int_{0}^{B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{B} \\ & = 1-\cos B \\ & = 1-\frac{12}{13} \\ & = \frac{1}{13} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [B,A] \\ & = \frac{1}{5}-\frac{1}{13} \\ & = \frac{8}{65} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [A,A+B] \\ & = 1-\cos (A+B) - \frac{1}{5} \\ & = \frac{4}{5}-\cos (A+B) \\ \end{aligned}$

then letting $g(x)=\cos x$ , and noting in a right-angled triangle the cosine of an angle admits of no negative values, so can it further be said

$\begin{aligned} & \quad\enspace 0<g(x)<1 \\ &\Rightarrow 0<g(A+B)<1 \\ &\Rightarrow 0<A+B<90^\circ \\ \end{aligned}$

I lost my way…

Let $w(u(t),v(t))=u(t)-v(t)$ be the difference in degrees between time-varying angles $u$ and $v$ where assumed is $u(t)>v(t)$ without loss of generality.

$\begin{aligned} f(u,v,w) & \stackrel{\textrm{def}}{=} \sin (u+v) \\ & = \sin (2v+w ) \\ & = \sin (2u-w ) \\ & \\ g(u,v,w) &\stackrel{\textrm{def}}{=} \cos (u+v) \\ & = \cos (2v+w ) \\ & = \cos (2u-w ) \\ \end{aligned}$

Taking partial derivatives wrt $u$ , $v$ , $w$ , and $t$ :

$\begin{aligned} \partial_uf(u,v,w) & = 2\cos (2u-w) \\ \partial_vf(u,v,w) & = 2\cos (2v+w) \\ \partial_wf(u,v,w) & = 0 \\ \partial_tf(u,v,w) & = (u'+v')\cos (u+v)\\ \end{aligned}$

$\begin{aligned} \mathrm{d}f(u,v,w) & = \partial_uf\,\mathrm{d}u + \partial_vf\,\mathrm{d}v + \partial_wf\,\mathrm{d}w \\ & = 2\cos (2u-w )\,\mathrm{d}u + 2\cos (2v+w )\,\mathrm{d}v + 0 \\ \dots\,\mathrm{d}u & = \mathrm{d}v+\mathrm{d}w\,\dots \\ & = 4\cos (u+v)\,\mathrm{d}v + 2\cos (u+v)\,\mathrm{d}w \\ & = 2\big(\cos (u+v)\big)(2\,\mathrm{d}v+\mathrm{d}w) \\ \end{aligned}$

I lost my faith…

$f(x,y)\stackrel{\textrm{(1)}}{=}\sin (x+y)$

then

$f(x'=x+y,0)\stackrel{\textrm{(2)}}{=} f(x,y)\stackrel{\textrm{(3)}}{=} f(0,x+y=y')$ .

By equality $\textrm{(1)}$ it states that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values of two arguments $(\texttt{var1},\texttt{var2})=(x,y)$ ;

by equality $\textrm{(2)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(x+y,0)$ , i.e., by one significant argument $\texttt{var1}=x+y$ , and another trivial argument $\texttt{var2}=0$ ; and

by equality $\textrm{(3)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(0,x+y)$ , i.e., by one trivial argument $\texttt{var1}=0$ , and another significant argument $\texttt{var2}=x+y$ .

Hence, assuming a partial variation

$\begin{aligned} &\quad\enspace \sin (\texttt{var1}+\texttt{var2}) \\ & = k_1(\texttt{var1},\texttt{var2})\sin (\texttt{var1}) + k_2(\texttt{var1},\texttt{var2})\sin (\texttt{var2}) \\ \end{aligned}$

we WTS the following

$\begin{aligned} k_1(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var2}) \\ k_2(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var1}) \\ \end{aligned}$

LHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}\sin (\texttt{var1}+\texttt{var2}) \\ & = \big[\cos (\texttt{var1}+\texttt{var2})\big] (\texttt{var1}'+\texttt{var2}') \\ \end{aligned}$

RHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}(k_1\sin (\texttt{var1})+k_2\sin (\texttt{var2})) \\ & = \big(k_1'\cos (\texttt{var1})\big) \texttt{var1}' + \big(k_2'\cos (\texttt{var2})\big) \texttt{var2}' \\ \end{aligned}$

and LHS=RHS implies:

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}k_1 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var1})} \\ \frac{\mathrm{d}}{\mathrm{d}t}k_2 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var2})} \\ \dots & \dots \dots \dots \\ k_1' & = \bigg( \frac{\cos (\texttt{var2})}{\cos (\texttt{var1})}\bigg) k_2' \\ \end{aligned}$

(to be continued)

January 16, 2023January 16, 2023

202301160933 Exercise 3.4.B (Q29)

A ball is thrown vertically upward from the ground. Its displacement at time $t$ is given by $20t-5t^2$ . Find the maximum distance of the ball from the ground and the time this occurs.