Author Topic: Recreational Problems (SM level) (Read 92821 times) Tweet Share

Mao · « **Reply #150 on:** May 18, 2008, 12:54:12 am »

Question by enwiabe
Request made by captain
Solutions posted by Mao coz he's bored
hehe

Quote from: enwiabe on irc

implicitly differentiate:
$\frac{x^2}{y^2}+2x = \sinh^{-1}\left(e^{\tanh(x)}\right)$

$\frac{d}{dx}\left(\frac{x^2}{y^2}+2x\right) = \frac{d}{dx}\left(\sinh^{-1}\left(e^{\tanh(x)}\right)\right)$

looking at LHS:

$\frac{d}{dx}\left(\frac{x^2}{y^2}+2x\right)$

$\implies \frac{d}{dx}\left(\frac{x^2}{y^2}\right)+\frac{d}{dx}\left(2x\right)$

$\implies \frac{2x \cdot y^2 + x^2 \cdot 2y \cdot \frac{dy}{dx}}{y^4} + 2$

looking at RHS:

$\frac{d}{dx}\left(\sinh^{-1}\left(e^{\tanh(x)}\right)\right)$

$\implies \frac{d}{du}\left(\sinh^{-1}(u)\right) \cdot \frac{d}{dx}\left(e^{\tanh(x)}\right)$

$\implies \frac{1}{\sqrt{1+u^2}} \cdot \frac{d}{dx}\left(\tanh(x)\right) \cdot e^{\tanh(x)}$

$\implies \frac{sech^2(x) \cdot e^{\tanh(x)}}{\sqrt{1+\left(e^{\tanh(x)}\right)^2}}$

bringing it all together:

$\frac{2x \cdot y^2 + x^2 \cdot 2y \cdot \frac{dy}{dx}}{y^4} + 2 = \frac{sech^2(x) \cdot e^{\tanh(x)}}{\sqrt{1+\left(e^{\tanh(x)}\right)^2}}$

$2x \cdot y^2 + x^2 \cdot 2y \cdot \frac{dy}{dx} = y^4 \cdot \left(\frac{sech^2(x) \cdot e^{\tanh(x)}}{\sqrt{1+\left(e^{\tanh(x)}\right)^2}}-2\right)$

$\frac{dy}{dx} = \frac{y^4 \cdot \left(\frac{sech^2(x) \cdot e^{\tanh(x)}}{\sqrt{1+\left(e^{\tanh(x)}\right)^2}}-2\right)-2x \cdot y^2}{x^2 \cdot 2y}$

$\frac{dy}{dx} = \frac{y^3 \cdot \left(\frac{sech^2(x) \cdot e^{\tanh(x)}}{\sqrt{1+\left(e^{\tanh(x)}\right)^2}}-2\right)-2x \cdot y}{2x^2}$

$\frac{dy}{dx} = \frac{y^3}{2x^2} \cdot \left(\frac{sech^2(x) \cdot e^{\tanh(x)}}{\sqrt{1+e^{2\tanh(x)}}}-\frac{2 \left(x+y^2\right)}{y^2}\right)$

PS: this is absolute bullcrap, haha. how terrible!

AppleXY · « **Reply #151 on:** May 18, 2008, 12:30:22 pm »

Quote from: Mao

(Image removed from quote.)

woah.

Mao · « **Reply #152 on:** May 18, 2008, 01:50:05 pm »

Quote from: AppleXY on May 18, 2008, 12:30:22 pm

Quote from: Mao
(Image removed from quote.)
woah.

yep... enwiabe gave an ugly ugly ugly question

dcc · « **Reply #153 on:** May 24, 2008, 12:23:32 pm »

dont steal too many ahmad/coblin/neobeo

$\mbox{1. Evaluate: } \int_{-\infty}^{\infty} xe^{-x^2}\mbox{ }dx$
- solved by Neobeo: http://vcenotes.com/forum/index.php/topic,11.msg40792.html#msg40792

$\mbox{2. Find: } \int x\sqrt{x^4 + 1}\mbox{ }dx$
- solved by Mao: http://vcenotes.com/forum/index.php/topic,11.msg40843.html#msg40843
- alternative solution by dcc: http://vcenotes.com/forum/index.php/topic,11.msg40919.html#msg40919

$\mbox{3. Find: } \int \dfrac{x - 3}{4 - 2x - x^2}\mbox{ }dx$
- solved by Mao: http://vcenotes.com/forum/index.php/topic,11.msg40834.html#msg40834

$\mbox{4. Find: } \int \dfrac{x \cdot \cos{2x} - \sin{2x}}{x^3}\mbox{ }dx$
- solved by Mao: http://vcenotes.com/forum/index.php/topic,11.msg40957.html#msg40957

$\mbox{5. Find: } \int \sqrt{x^2 + 4x + 5}\mbox{ }dx$
- solved by ice_blockie: http://vcenotes.com/forum/index.php/topic,11.msg40795.html#msg40795
- alternative solution by dcc: http://vcenotes.com/forum/index.php/topic,11.msg40800.html#msg40800

$\mbox{6. Find the general solution of the differential equation: } \dfrac{dy}{dx} = (x + y)^2$
- solved by Ahmad: http://vcenotes.com/forum/index.php/topic,11.msg40952.html#msg40952

$\mbox{7. Find: } \int \log_{e}(x^2 + 4x)\mbox{ }dx$
- solved by Mao: http://vcenotes.com/forum/index.php/topic,11.msg40824.html#msg40824

$\mbox{8. A runner starting from rest at the origin has an acceleration defined by: } a = \dfrac{4 - e^{x/2}}{4 + e^{x/2}}\mbox{ }m/s^2 \mbox{ where x is the displacement from the origin. Find the runner's speed when they are 4 metres in the positive direction from the origin.}$
- solved by Mao: http://vcenotes.com/forum/index.php/topic,11.msg40884.html#msg40884

$\mbox{9. Find: } \int \dfrac{1}{x^4 + 1}\mbox{ }dx$
- solved by Mao: http://vcenotes.com/forum/index.php/topic,11.msg41572.html#msg41572

$\mbox{10. Find: } \int \sec{x}\mbox{ }dx$
- solved by AppleXY: http://vcenotes.com/forum/index.php/topic,11.msg40810.html#msg40810

edit: some mofo edited my post!

Neobeo · « **Reply #154 on:** May 24, 2008, 03:59:48 pm »

Quote from: dcc on May 24, 2008, 12:23:32 pm

$\mbox{1. Evaluate: } \int_{-\infty}^{\infty} xe^{-x^2}\mbox{ }dx$

$\mbox{Let } f(x)=xe^{-x^2}$
$f(-x)=(-x)e^{-(-x)^2}=-f(x)$
$\Rightarrow f(x) \mbox{ is odd}$

$\therefore \int_{-\infty}^{\infty} xe^{-x^2}\mbox{ }dx= 0$

ice_blockie · « **Reply #155 on:** May 24, 2008, 04:03:46 pm »

Me bags question 5...

Quote from: dcc on May 24, 2008, 12:23:32 pm

$\mbox{5. Find: } \int \sqrt{x^2 + 4x + 5}\mbox{ }dx$

$Let\mbox{ } I = \int \sqrt{x^2 + 4x + 5}\mbox{ }dx$
$Let \mbox{ }v = x + 2, hence \mbox{ }v^2=x^2+4x+4$
$dv=dx$
$Therefore \int \sqrt{x^2 + 4x + 5}\mbox{ }dx = \int \sqrt{v^2+1}\mbox{ }dv$

$Let \mbox{ }sinh(u) = v, cosh(u)du=dv$

$Hence:$
$I = \int \sqrt{sinh^2(u)+1}cosh(u)\mbox{ }du$
$= \int cosh^2 (u) \mbox{ }du$
$= \frac{1}{2}\int cosh(2u)+1\mbox{ }du$
$=\frac{1}{2}[sinh(2u)+u]+c$
$=\frac{1}{2}[sinh(u)cosh(u)+u]+c$
$=\frac{1}{2}sinh(u)\sqrt{sinh^2 (u) +1}+\frac{1}{2}u+c$

$Substituting\mbox{ }sinh(u)=v,$
$I=\frac{1}{2}v\sqrt{v^2+1}+\frac{1}{2}arcsinh(v)+c$

$Substituting\mbox{ } v=x+2,$
$I=\frac{1}{2}(x+2)\sqrt{x^2+4x+5}+\frac{1}{2}arcsinh(x+2)+c$

$hence$
$\int \sqrt{x^2 + 4x + 5}\mbox{ }dx = \frac{1}{2}(x+2)\sqrt{x^2+4x+5}+\frac{1}{2}arcsinh(x+2)+c$

Please don't tell me there was a simpler way...

dcc · « **Reply #156 on:** May 24, 2008, 04:51:58 pm »

For that one, another approach:

$I = \int \sqrt{x^2 + 4x + 5}\mbox{ }dx = \int \sqrt{(x + 2)^2 + 1}\mbox{ }dx$

$\mbox{Let: }x + 2 = \tan{\theta} \implies x = \tan{\theta} - 2 \implies \dfrac{dx}{d\theta} = \sec^2 \theta \implies dx = \sec^2 \theta \mbox{ }d\theta$

$\implies I = \int \sqrt{(\tan \theta)^2 + 1} \cdot \sec^2 \theta\mbox{ }d\theta = \int \sec^3 \theta \mbox{ }d\theta$

Wikipedia informs me that the integral of secant cubed is:

$I = \dfrac{1}{2}\sec \theta \tan \theta + \dfrac{1}{2}log_e{|\sec \theta + \tan \theta|} + c$

Using triangles, it becomes apparent that $\sec \theta = \sqrt{(x + 2)^2 + 1} = \sqrt{x^2 + 4x + 5}$

$\therefore I = \dfrac{1}{2}(x + 2)\sqrt{x^2 + 4x + 5} + \dfrac{1}{2}\log_{e}{|\sqrt{x^2 + 4x + 5} + (x + 2)|} + c$

AppleXY · « **Reply #157 on:** May 24, 2008, 05:51:49 pm »

$\int \sec(x) dx$

$\sec(x) = \frac{1}{\cos(x)} = \frac{cos(x)}{cos^{2}(x)}$

$\mbox{as cos^2(x) is 1-sin^2(x)}$

$\therefore$

$\frac{cos(x)}{1-sin^2(x)}$
$\mbox{let u=sin(x)}$
$\frac{du}{dx} = cos(x)$
$dx = \frac{du}{cos(x)}$
$\mbox{they cancel out}$
$\therefore$
$\frac{1}{1-u^2}$
$\int \frac{1}{1-u^2}$

sub for u and yeah. , convert to partial fractions , intergrate. (soz lack of time)

gfb · « **Reply #158 on:** May 24, 2008, 06:24:54 pm »

Are those recreational problems or homework problems?. They're easy.

Mao · « **Reply #159 on:** May 24, 2008, 08:09:41 pm »

Quote from: gfb on May 24, 2008, 06:24:54 pm

Are those recreational problems or homework problems?. They're easy.

easy?!
are you out of your mind? [or a hidden genious within?

]

Quote from: dcc on May 24, 2008, 12:23:32 pm

$\mbox{7. Find: } \int \log_{e}(x^2 + 4x)\mbox{ }dx$

integration by recognition:

$\begin{align*} \frac{d}{dx}x\cdot log_e (x^2+4x) &= log_e(x^2+4x) + \frac{x\cdot (2x+4)}{x^2+4x}\\ &= log_e(x^2+4x) + \frac{2x+8-4}{x+4}\\ &= log_e(x^2+4x) + 2 - \frac{4}{x+4}\\ \end{aligh*}$

$\begin{align*} \int \log_{e}(x^2 + 4x)\mbox{ }dx &= x\cdot log_e (x^2+4x) - \int 2 \cdot dx + \int \frac{4}{x+4}\cdot dx +c \\ &= x\cdot log_e (x^2+4x) -2x + 4log_e (x+4) + c\\ \end{align*}$

ice_blockie · « **Reply #160 on:** May 24, 2008, 08:50:01 pm »

Quote from: gfb on May 24, 2008, 06:24:54 pm

Are those recreational problems or homework problems?. They're easy.

Prove it.

Mao · « **Reply #161 on:** May 24, 2008, 09:12:59 pm »

Quote from: dcc on May 24, 2008, 12:23:32 pm

$\mbox{3. Find: } \int \dfrac{x - 3}{4 - 2x - x^2}\mbox{ }dx$

$\int \dfrac{x - 3}{4 - 2x - x^2}\mbox{ }dx =- \int \dfrac{3-x}{(x+1)^2-7}\mbox{ }dx$

$\mbox{Let } u = x+1 \implies x=u-1$
also: $du=dx$

$\implies -\int \dfrac{4-u}{u^2-7} \cdot dx$

with partial fractions:

$\begin{align*} \dfrac{4-u}{u^2-7} &= \frac{A}{u-\sqrt{7}} + \frac{B}{u+\sqrt{7}}\\ 4-u &= A(u+\sqrt{7}) + B(u-\sqrt{7})\\ A &= \frac{4-\sqrt{7}}{2 \sqrt{7}} \\ B &= -\frac{4+ \sqrt{7}}{2\sqrt{7}} \\ \end{align*}$

$\implies -\frac{1}{2\sqrt{7}} \int \left(\frac{4-\sqrt{7}}{u-\sqrt{7}} - \frac{4+\sqrt{7}}{u+\sqrt{7}}\right)\cdot dx$

$\implies -\frac{1}{2\sqrt{7}} \cdot \left[ (4-\sqrt{7})log_e | u-\sqrt{7}| - (4+\sqrt{7})log_e |u+\sqrt{7}| \right] + c$

$\implies \frac{(4+\sqrt{7})log_e |x+1+\sqrt{7}|-(4-\sqrt{7})log_e |x+1-\sqrt{7}|}{2\sqrt{7}} + c$

Mao · « **Reply #162 on:** May 24, 2008, 09:45:19 pm »

Quote from: dcc on May 24, 2008, 12:23:32 pm

$\mbox{2. Find: } \int x\sqrt{x^4 + 1}\mbox{ }dx$

$\begin{align*} \mbox{Let } u &= x^2\\ \Rightarrow \frac{du}{dx} &= 2x \\ \end{align*}$

$\implies \frac{1}{2} \int \sqrt{u^2+1} du$

$\begin{align*} \mbox{Let } u &=sinh(v)\\ \Rightarrow \frac{du}{dv} &= cosh(v) \\ \end{align*}$

$\implies \frac{1}{2} \int \sqrt{sinh^2 (v) + 1}\cdot cosh(v) \cdot dv$

$\implies \frac{1}{2} \int cosh^2 (v) \cdot dv$

$\implies \frac{1}{4} \int (cosh(2v) + 1) \cdot dv$

$\implies \frac{1}{4} \cdot \left(\frac{1}{2}sinh(2v) + v\right) + c$

$\implies \frac{sinh(2v)}{8}+ \frac{v}{4}+c$

$\implies \frac{sinh(v)\cdot \sqrt{sinh^2(v)+1}}{4} + \frac{v}{4} + c$

$\implies \frac{u\cdot \sqrt{u^2+1}+sinh^{-1}(u)}{4} + c$

$\implies \frac{x^2 \cdot \sqrt{x^4+1} + sinh^{-1}\left(x^2\right)}{4} + c$

ps: thanks ice_blockie

Mao · « **Reply #163 on:** May 25, 2008, 10:50:01 am »

with our kinematics relationships, we know that:

$a=\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\cdot \frac{dv}{dx}$

hence

$\implies v \cdot \frac{dv}{dx} = \frac{4-e^{x/2}}{4+e^{x/2}}$

$\mbox{Let }u=4+e^{x/2}, \implies e^{x/2}=u-4$

$\frac{du}{dx} = \frac{e^{x/2}}{2} = \frac{u-4}{2}$

$\begin{align*} \implies \int v dv &= \int \frac{8-u}{u} \cdot \frac{dx}{du} \cdot du \\ \frac{v^2}{2} &= \int \frac{8-u}{u} \cdot \frac{2}{u-4} \cdot du \\ \frac{v^2}{2} &= 2\int \frac{8-u}{u(u-3)} \cdot du \\ \end{align*}$

by partial fractions:

$\begin{align*} \frac{8-u}{u(u-4)} &= \frac{A}{u} + \frac{B}{u-4} \\ 8-u &= A(u-4) + B(u) \\ A &= -2 \\ B &= 1 \\ \end{align*}$

$\begin{align*} \implies \frac{v^2}{2} &= 2\int \left(\frac{-2}{u} + \frac{1}{u-4}\right) du\\ \frac{v^2}{2} &= 2\cdot \left[ -2 log_e |u| + log_e |u-4| \right] + c\\ \frac{v^2}{2} &= 2\cdot \left( -2 log_e |4+e^{x/2}| + log_e e^{x/2} \right) + c\\ v^2 &= -4 log_e |4+e^{x/2}| + x + c \mbox{ ... given that when x=0, v=0}\\ c &= 4 log_e 5\\ \frac{v^2}{2} &= -4 log_e |4+e^{x/2}| + x + 4 log_e 5\\ \end{align*}$

therefore, when x=5:

$\begin{align*} \frac{v^2}{2} &= 5 + 4 log_e 5 - 4 log_e (4+e^{5/2})\\ v &= \sqrt{10 + 8 log_e 5 - 8 log_e (4+e^{5/2})}\\ \end{align*}$

and for practicality's sake:

$v\approx 0.777215 ms^{-1}$

which is at last not imaginary.

dcc · « **Reply #164 on:** May 25, 2008, 12:42:25 pm »

I have fixed question 8, I had written the sign wrong.

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Author Topic: Recreational Problems (SM level) (Read 92821 times) Tweet Share

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