Author Topic: Recreational Problems (SM level) (Read 77132 times) Tweet Share

brightsky · « **Reply #255 on:** March 30, 2011, 07:59:53 pm »

I've seen a similar question before...

$\int^{\frac{\pi}{2}}_{0} \frac{\sin^{2011}(x)}{\sin^{2011}(x) + \cos^{2011}(x)} dx$

$= \int^{\frac{\pi}{2}}_{0} \frac{\sin^{2011}\left(\frac{\pi}{2} - x\right)}{\sin^{2011}\left(\frac{\pi}{2} - x\right) + \cos^{2011}\left(\frac{\pi}{2} - x\right)} dx$

$= \int^{\frac{\pi}{2}}_{0} \frac{\cos^{2011}(x)}{\cos^{2011}(x) + \sin^{2011}(x)} dx$

$= I$

Hence:

$2I = \int^{\frac{\pi}{2}}_{0} \frac{\cos^{2011}(x)}{\cos^{2011}(x) + \sin^{2011}(x)} dx + \int^{\frac{\pi}{2}}_{0} \frac{\sin^{2011}(x)}{\sin^{2011}(x) + \cos^{2011}(x)} dx = \int^{\frac{\pi}{2}}_{0} 1 dx$

$2I = \frac{\pi}{2}$

$I = \frac{\pi}{4}$

enpassant · « **Reply #256 on:** March 30, 2011, 08:45:40 pm »

well done

ohexploitable · « **Reply #257 on:** August 20, 2011, 10:49:57 pm »

An interesting question I encountered while studying for NSW trial exams.

A has mass 2m, B has mass m, the whole thing is rotated at constant angular velocity w.

Find the tension in OA and OB.

pi · « **Reply #258 on:** November 29, 2011, 05:16:10 pm »

A few from class tests and SACs, all are CAS-free

Q1
$Evaluate \ in \ surd \ form \ (all \ angles \ in \ degrees): \frac{\cos(157.5) + \cos(97.5) + \cos(352.5)}{\sin(172.5) + \sin(202.5) - \cos(262.5)}$

Q2
$Find \ \frac{d^2y}{dx^2} \ if \ y=\tan(\sin^{-1}(\frac{x}{3}))$

Q3
$Evaluate \ \int{\frac{x-2}{x^2-6x+11}}dx$

Q4
$Evaluate \ \int{\sqrt{\frac{x+1}{1-x}}}}dx$

Q5
$Show \ that \ \frac{cis(\theta)-1}{cis(\theta)+1} = i\tan(\frac{\theta}{2})$

Q6
$Show \ that \ \sin^{-1}(\frac{4}{5}) + \sin^{-1}(\frac{5}{13}) + \sin^{-1}(\frac{16}{65}) = \frac{\pi}{2}$

Q7
$Show \ that \ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi$

Q8
$A \ cissoids \ is \ a \ curve \ described \ by \ the \ equation \ y^2(2a-x)=x^3, \ x \ge 0 \ where \ a>0.$
$The \ cissoid \ is \ rotated \ about \ its \ asymptote \ to \ form \ a \ volume \ of \ solid \ revolution.$
$Find \ this \ exact \ volume \ given \ that \ \ I_{p,q} = \int^{\frac{\pi}{2}}_0{\sin^p x\cos^q x}dx = (\frac{p-1}{p+q})I_{p-2,q}$

Q9
$Find \ the \ solutions \ to \ the \ differential \ equation \ \frac{d^2y}{dx^2} + 9y = 0$

Q10
$Find \ which \ is \ larger: \sin(\cos x) \ or \ \cos(\sin x)$

Enjoy

edit: no idea what is with the random font sizes/error messages, just bear with them please

brightsky · « **Reply #259 on:** December 02, 2011, 03:46:35 pm »

2) draw a right angle triangle with hypotenuse 3 and side lengths x and $\sqrt{9 - x^2}$ .
mark in $\arcsin\left(\frac{x}{3}\right)$ . hence $\tan\left(\arcsin\left(\frac{x}{3}\right)\right) = \frac{x}{\sqrt{9-x^2}} = y$ .

$\frac{dy}{dx} = \frac{\sqrt{9-x^2} +\frac{x^2}{\sqrt{9-x^2}}}{9-x^2} = \frac{9}{(9-x^2)^{\frac{3}{2}}}$

$\frac{d^2 y}{d x^2} = \frac{27x}{(9-x^2)^{\frac{5}{2}}}$

brightsky · « **Reply #260 on:** December 02, 2011, 04:19:38 pm »

3) Let $\int \frac{x-2}{x^2 - 6x + 11} dx = \int \frac{x-2}{(x-3)^2 + 2} = I$

Let $u = x - 3$ , so $du = dx$

$I= \int \frac{u+1}{u^2 + 2} du$

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

For the first part, let $v = u^2 + 2$ , so $dv = 2u$ :

So $\int \frac{u}{u^2 + 2} du = \int \frac{1}{2v} dv = \ln |2v| = \ln |2u^2 + 4| = \ln |2(x-3)^2 + 4| + C_{1}$

For the second part, let $u = \sqrt{2} \tan(t)$ , so $du = \sqrt{2} \sec^2(t) dt$ :

So $\int \frac{1}{u^2 + 2} du = \int \frac{\sqrt{2} \sec^2(t)}{2 \tan^2(t) + 2} dt = \int \frac{\sqrt{2} \sec^2(t)}{2 \sec^2(t)}dt = \int \frac{\sqrt{2}}{2} dt = \frac{\sqrt{2}}{2} t = \frac{\sqrt{2}}{2} \arctan(\frac{u}{\sqrt{2}}) =\frac{\sqrt{2}}{2} \arctan(\frac{x-3}{\sqrt{2}}) + C_2$

So $I = \ln |2(x-3)^2 + 4|+ \frac{\sqrt{2}}{2} \arctan(\frac{x-3}{\sqrt{2}}) + C$

brightsky · « **Reply #261 on:** December 02, 2011, 04:55:13 pm »

9) auxiliary equation is $r^2 + 9 = 0$

$(r-3i)(r+3i) = 0$

so the roots are $r = 3i, -3i$

hence general solution is: $y = c_1 \cos(3x) + c_2 \sin(3x)$

brightsky · « **Reply #262 on:** December 02, 2011, 05:32:11 pm »

7) swear i've seen this one before...

draw triangle ABC. let $\angle A = \arctan(1)$ and $\angle B = \arctan(2)$ .

draw line CD perpendicular to AB. Let CD = h. thus AD = h since angle A is arctan(1). also, DB = h/2 since angle B = arctan(2). by pythagoras' theorem then, AC = sqrt(h^2 + h^2) = sqrt(2) h. similarly, BC = sqrt(h^2 + (h/2)^2) = sqrt(5)h/2.

using the cosine rule, AB^2 = AC^2 + AB^2 - 2*AB*AC*2cos(angle C)

9h^2/4 = 2h^2 + 5h^2/4 - sqrt(10) h^2 cos(angle C)

9h^2 = 8h^2 + 5h^2 - 4sqrt(10) h^2 cos (angle C)

4sqrt(10) h^2 cos(angle C) = 4h^2

cos(angle C) = 1/sqrt(10)

draw a triangle relating to the above equation. it is clear that the unknown side of the triangle (opposite to angle C) is sqrt(10 - 1^2) = 3. hence tan(angle C) = 3/1 = 3, so arctan(3) = angle C.

since angle A + angle B + angle C = pi

arctan(1) + angle(2) + angle(3) = pi as required.

pi · « **Reply #263 on:** December 02, 2011, 08:47:52 pm »

Very nice brightsky! I did q9 in a much more tedious way, but same result! Question 3 is a b*tch, nicely done (the other ones of its kind are q6, 8, 10 imo).

Quote from: brightsky on December 02, 2011, 05:32:11 pm

7) swear i've seen this one before...

Probably, it's pretty famous

dc302 · « **Reply #264 on:** December 02, 2011, 10:34:08 pm »

10) what does larger mean?

dc302 · « **Reply #265 on:** December 02, 2011, 10:38:10 pm »

4) seen this one many times, multiply top and bottom by sqrt(1+x) and it follows from that that the answer should be something like arcsinx - 1/2 ln|1-x^2| + C

edit: changed a sign

TrueTears · « **Reply #266 on:** December 02, 2011, 10:38:16 pm »

think he means whether it is true in general that sin(cosx)<cos(sinx) or cos(sinx)<sin(cosx)

you can prove it using calculus or just use the identity sin(x) = cos(pi/2 - x) then some manipulations, from memory i think cos(sinx) is always greater

brightsky · « **Reply #267 on:** December 03, 2011, 04:48:21 pm »

5) gonna use x in place of theta for efficiency of expression

$i \tan(x)$

$= \frac{e^{\frac{1}{2}ix} - e^{-\frac{1}{2}ix}}{e^{\frac{1}{2}ix} + e^{-\frac{1}{2}ix}}$

$= \frac{e^{\frac{1}{2}ix}(e^{\frac{1}{2}ix} - e^{-\frac{1}{2}ix})}{e^{\frac{1}{2}ix}(e^{\frac{1}{2}ix} + e^{-\frac{1}{2}ix})}$

$= \frac{e^{ix} - 1}{e^{ix} + 1}$

$= \frac{cis(x) - 1}{cis(x) + 1}$ as required.

dc302 · « **Reply #268 on:** December 03, 2011, 05:31:43 pm »

Quote from: brightsky on December 03, 2011, 04:48:21 pm

5) gonna use x in place of theta for efficiency of expression

$i \tan(x)$

$= \frac{e^{\frac{1}{2}ix} - e^{-\frac{1}{2}ix}}{e^{\frac{1}{2}ix} + e^{-\frac{1}{2}ix}}$

$= \frac{e^{\frac{1}{2}ix}(e^{\frac{1}{2}ix} - e^{-\frac{1}{2}ix})}{e^{\frac{1}{2}ix}(e^{\frac{1}{2}ix} + e^{-\frac{1}{2}ix})}$

$= \frac{e^{ix} - 1}{e^{ix} + 1}$

$= \frac{cis(x) - 1}{cis(x) + 1}$ as required.

Unfortunately, you can't use the complex exponential in spesh.

dc302 · « **Reply #269 on:** December 03, 2011, 05:34:16 pm »

Quote from: brightsky on December 02, 2011, 04:55:13 pm

9) auxiliary equation is $r^2 + 9 = 0$

$(r-3i)(r+3i) = 0$

so the roots are $r = 3i, -3i$

hence general solution is: $y = c_1 \cos(3x) + c_2 \sin(3x)$

Lol again you can't use that in spesh, you need to separate the variables manually.

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Author Topic: Recreational Problems (SM level) (Read 77132 times) Tweet Share

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