Ng and Pang’s Breakthrough Calculus

September 26, 2020October 24, 2022

202009260344 Exercise 2.5.1

Show that $x=e^{2\theta}\sin\theta$ satisfies the equation $x''-4x'+5x=0$ .

Solution.

(bottom-up)

Let $x'$ denote $\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}\theta}}$ , $x''$ denote $\displaystyle{\frac{\mathrm{d}^2x}{\mathrm{d}\theta^2}}$ .

$\begin{aligned} x' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \sin\theta \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( \sin\theta \big) + \sin\theta \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = e^{2\theta} \cos\theta + 2e^{2\theta}\sin\theta \\ & = e^{2\theta}(\cos\theta + 2\sin\theta ) \\ \end{aligned}$

$\begin{aligned} x'' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( x' \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} (\cos\theta +2\sin\theta ) + (\cos\theta + 2\sin\theta )\cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = (e^{2\theta})(-\sin\theta + 2\cos\theta ) + (\cos\theta + 2\sin\theta )(2e^{2\theta }) \\ & = e^{2\theta}(3\sin\theta + 4 \cos\theta ) \\ \end{aligned}$

To know whether or not $x=e^{2\theta}\sin\theta$ is a solution, I simply do substitution in the equation $x''-4x'+5x=0$ .

$\begin{aligned} \textrm{LHS} & = x''-4x'+5x \\ & = e^{2\theta}(3\sin\theta +4\cos\theta )-4\big[ e^{2\theta}(\cos\theta +2\sin\theta )\big] + 5\big(e^{2\theta}\sin\theta \big) \\ & = \dots \\ & = 0 \\ & = \textrm{RHS} \\ \end{aligned}$

Revision.

(top-down)

We are given a second-order linear homogeneous ordinary differential equation (ODE):

$x''-4x'+5x = 0$

with some independent variable $\theta$ and some dependent variable $x(\theta )=e^{2\theta }\sin\theta$ , the coefficients of $x''$ , $x'$ , and $x$ being constants $1$ , $-4$ , and $5$ .

Let the primed $[ *]'$ be the function derived wrt. to $\theta$ . Following the routine procedures,

$\begin{aligned} x & = e^{r\theta} \\ x' & = re^{r\theta }\\ x'' & = r\cdot ( e^{r\theta })' + r' \cdot ( e^{r\theta }) \\ & = r^2e^{r\theta} \end{aligned}$

rewrite it,

$\begin{aligned} x''-4x'+5x & = 0 \\ r^2e^{r\theta} - 4re^{r\theta} + 5e^{r\theta} & = 0 \\ \dots \textrm{\quad excepting\quad} & x=e^{r\theta}=0\textrm{\quad \dots} \\ r^2 -4r+5 & = 0 \\ \end{aligned}$

and we shall obtain the auxiliary equation (aka. the characteristic equation) on the very last line.

$r=\displaystyle{\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(5)}}{2(1)}}=2\pm \textrm{i}$ .

Note.

If the roots of the auxiliary equation $ar^2+br+c=0$ are the complex numbers $r_1=\alpha +\textrm{i}\beta$ and $r_2=\alpha -\textrm{i}\beta$ , the general solution of $ay''+by'+cy=0$ is

$y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)$

_{(please refer to J. Stewart’s Calculus, Second-Order Linear Differential Equations)}

Now that $r_1=2+\textrm{i}$ and $r_2=2-\textrm{i}$ and $c_1$ , $c_2$ are arbitrary constants, to this $2^{\textrm{nd}}$ -order ODE, the general solution is

$x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta )$ .

Examination.

It is a good practice, no matter how much time one would allow oneself, to countercheck one’s solution…

Here it goes,

$\begin{aligned} x & =e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (c_2e^{2\theta })(\sin\theta ) + (c_1e^{2\theta })(\cos\theta ) \\ \end{aligned}$ .

$\begin{aligned} x' & = e^{2\theta}(-c_1\sin\theta +c_2\cos\theta ) + 2e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) + (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \\ \end{aligned}$

$\begin{aligned} x''& = \big[ (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) \big]' + \big[ (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \big]' \\ & = \big[ (2c_2e^{2\theta}-c_1e^{2\theta })(\cos\theta )+(4c_2e^{2\theta }-2c_1e^{2\theta })(\sin\theta ) \big] \\ & \qquad\quad + \big[ (2c_1e^{2\theta }+c_2e^{2\theta })(-\sin\theta )+(4c_1e^{2\theta }+2c_2e^{2\theta })(\cos\theta ) \big] \\ & = (-4c_1e^{2\theta }+3c_2e^{2\theta })(\sin\theta ) + (3c_1e^{2\theta }+4c_2e^{2\theta })(\cos\theta ) \\ \end{aligned}$

Then,

$\begin{aligned} \textrm{LHS} & = x'' - 4x' + 5x \\ & = \big( (-4c_1e^{2\theta }+3c_2e^{2\theta })-4(2c_2e^{2\theta}-c_1e^{2\theta})+5c_2e^{2\theta}\big)(\sin\theta ) \\ & \qquad\quad + \big( (3c_1e^{2\theta}+4c_2e^{2\theta}) - 4(2c_1e^{2\theta}+c_2e^{2\theta})+5c_1e^{2\theta} \big)(\cos\theta ) \\ & = 0 \\ & = \textrm{RHS} \end{aligned}$

In conclusion, $x=e^{2\theta}\sin\theta$ is a particular solution, the general solution being $x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta )$ .

September 25, 2020October 24, 2022

202009250041 Exercise 5.8.1

Find out if any the asymptote(s) of the curve given by:

$y=\displaystyle{\frac{x^2-6x+3}{x-3}}$ .

Attempts.

$\begin{aligned} & & x & -3 & & \\\cline{3-5} x & -3 \Big) & x^2 & -6x & +3 & \\ & & x^2 & -3x & & \\ \cline{3-5} & & & -3x & +3 \\ & & & -3x & +9 \\\cline{4-5} & & & & -6 \end{aligned}$

The curve is given by

$\begin{aligned} y & =\displaystyle{\frac{x^2-6x+3}{x-3}} \\ & = x-3 - \displaystyle{\frac{6}{x-3}} \end{aligned}$ .

Ans. Of the given curve, $x=3$ is a vertical asymptote and $y=x-3$ an oblique asymptote.

Retrospectively, for $f(x)=\displaystyle{\frac{x^2-6x+3}{x-3}}$ , maybe one could also have checked the conditions as follow:

i. $\displaystyle{\lim_{x\to 0}}f(x)$ ;

ii. $\displaystyle{\lim_{x\to 3^{+}}}f(x)$ ;

iii. $\displaystyle{\lim_{x\to 3^{-}}}f(x)$ ;

iv. $\displaystyle{\lim_{x\to +\infty}}f(x)$ ; and

iv. $\displaystyle{\lim_{x\to -\infty}}f(x)$ .

i.

$\begin{aligned} \lim_{x\to 0}f(x) & = \lim_{x\to 0} \frac{x^2-6x+3}{x-3} \\ & = \frac{(0)^2-6(0)+3}{(0)-3} \\ & = -1 \end{aligned}$

iii.

$\begin{aligned} \lim_{x\to 3^{-}}f(x) & = \lim_{x\to 3^{-}} \frac{x^2-6x+3}{x-3} \\ & = \lim_{x\to 3^{-}} \bigg( x-3-\frac{6}{x-3} \bigg) \\ & = 3^{-} -3-\frac{6}{3^{-}-3} \\ & = (3^{-}-3) + \bigg( \frac{6}{3-3^{-}} \bigg) \\ & \\ \hline & \\ & \because 3^{-} < 3 \\ & \therefore 3^{-}-3<0\qquad \textrm{and}\qquad 3-3^{-}>0 \\ & \dots \textrm{excuse me for writing with no more ado,}\dots \\ & \\ \hline & \\ & = 0^{-} + \frac{6}{0^{+}} \\ & = +\infty \\ \end{aligned}$

ii., iv., and iv. are not checked.

Just work with it like whom I was taught:

First, find the slope of tangent to the curve,

$\begin{aligned} f(x) & = \frac{x^2-6x+3}{x-3} \\ f'(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x) \big) \\ & = \frac{(x-3)(2x-6)-(x^2-6x+3)(1)}{(x-3)^2} \\ & = \frac{(2x^2-12x+18)-(x^2-6x+3)}{(x-3)^2} \\ & = \frac{x^2-6x+15}{x^2-6x+9} \\ & = 1 + \frac{6}{x^2-6x+9} \end{aligned}$

The purpose is to find out any crests or troughs, because at the tip or bottom of a plotted curve, the slope of tangent will be zero, i.e., $f'(x)=0$ .

Secondly, find the change of slope of tangent to the curve,

$\begin{aligned} f''(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big[ f'(x) \big] \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( 1+\frac{6}{x^2-6x+9} \bigg) \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \big( 1 \big) + \frac{\mathrm{d}}{\mathrm{d}x} \bigg( \frac{6}{x^2-6x+9} \bigg) \\ & = 0 + \frac{(x^2-6x+9)(0)+(6)(2x-6)}{(x^2-6x+9)^2} \\ & = \frac{12x-36}{(x^2-6x+9)^2} \\ \end{aligned}$

The purpose is to find out if there were any inflexion points (i.e., $f''(x)=0$ ) upon which curvature changes sign.

It begins, assuming $x\neq 3$ , I will set $f(x)=0$ :

$\begin{aligned} f(x) & = 0 \\ \Leftrightarrow \qquad\qquad x^2-6x+3 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(3)}}{2(1)} \\ & = 3 \pm \sqrt{6} \\ \end{aligned}$

The $x$ -intercepts are thus two, $3+\sqrt{6}$ and $3-\sqrt{6}$ .

Then I needs set $f'(x)=0$ with the same assumption $x\neq 3$ .

$\begin{aligned} f'(x) & = 0 \\ \Leftrightarrow \qquad \qquad 1+\frac{6}{x^2-6x+9} & = 0 \\ & \\ \textrm{\dots \quad by assumption }x\neq 3 & \enspace \textrm{jump legitimately to the next step\quad \dots} \\ & \\ x^2-6x+9+6 & = 0 \\ x^2 -6x +15 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(15)}}{2(1)} \\ & = 3 \pm \sqrt{6}\, \textrm{i} \end{aligned}$

As curvature here permits of no nonreal points, there are neither convex point(s) nor concave point(s).

Besides, there are none any one point of inflexion because if on one hand $f''(x)=0$ and on the other $x\neq 3$ be assumed,

$\begin{aligned} \frac{12x-36}{(x^2-6x+9)^2} & = 0 \\ \Rightarrow 12x-36 & = 0 \\ \Rightarrow x & =3\qquad\qquad\qquad \bot \\ \end{aligned}$

it would have resulted in contradiction.

In conclusion, this exercise was overdone.

July 26, 2019January 19, 2023

201907260633 Exercise 3.1.1

Evaluate the indefinite integral:

$\displaystyle{\int\frac{x}{1-2x}}\,\mathrm{d}x$

Solution.

(bad example)

By change of variable, set $u=1-2x$ ,

then $x=\displaystyle{\frac{1-u}{2}}$ .

$\begin{aligned} \displaystyle{\frac{\mathrm{d}u}{\mathrm{d}x}} & =\displaystyle{\frac{\mathrm{d}(1-2x)}{\mathrm{d}x}}=-2 \\ \mathrm{d}x & = -\frac{1}{2}\,\mathrm{d}u \end{aligned}$ .

$\begin{aligned} & \quad \int \frac{x}{1-2x}\,\mathrm{d}x\\ & = \int \frac{\frac{1-u}{2}}{u} \bigg( -\frac{1}{2} \,\mathrm{d}u \bigg) \\ & = -\frac{1}{4} \int \bigg( \frac{1}{u} -1 \bigg) \,\mathrm{d}u \\ & = -\frac{1}{4} \big( \ln |u| - u + C \big) \\ & = -\frac{1}{4}\ln |1-2x| +\frac{1}{4} (1-2x) -\frac{1}{4}C \\ & = -\frac{1}{4}\ln |1-2x| -\frac{x}{2} + \frac{1}{4}(1-C) \\ & = -\frac{1}{4}\ln |1-2x| -\frac{x}{2} + C' \end{aligned}$

(better example)

Express $x$ in terms of $1-2x\enspace (=u)$ ,

$\begin{aligned} 1-2x & = u \\ x & = \frac{1}{2} (1-u) = \frac{1}{2} [1-(1-2x)] \end{aligned}$

Then

$\begin{aligned} & \quad \int \frac{x}{1-2x} \,\mathrm{d}x\\ & = \frac{1}{2} \int \frac{1-(1-2x)}{1-2x} \,\mathrm{d}x \\ & = \frac{1}{2} \Bigg\{ \int \frac{\mathrm{d}x}{1-2x} - \int \mathrm{d}x \Bigg\} \\ & = \frac{1}{2} \Bigg\{ \bigg(\frac{\ln |1-2x|}{-2} +C' \bigg) - \Big( x+C'' \Big) \Bigg\} \\ & = -\frac{1}{4}\ln |1-2x| - \frac{x}{2} + C \end{aligned}$