Author Topic: Recreational Problems (SM level) (Read 76996 times) Tweet Share

pi · « **Reply #270 on:** December 04, 2011, 12:18:57 pm »

^^^Agree with dc302 there, we would have got 0 for those questions with that working

pi · « **Reply #271 on:** August 29, 2012, 09:18:20 pm »

I left this one in the Spesh Q Thread, but it remained unsolved. Have a go

Quote from: LovesPhysics on August 14, 2012, 08:16:43 pm

A cissoid $y^2(2a-x)=x^3$ (where $x \ge 0$ and $a>0$ ) is rotated about it's asymptote to form a volume of revolution. Find this volume in terms of $a$ . You don't really need a CAS until the very last part to get the answer.

There's a few more unanswered questions throughout this thread too, get on it spesh kiddos!

Jenny_2108 · « **Reply #272 on:** August 29, 2012, 10:38:06 pm »

Quote from: LovesPhysics on August 29, 2012, 09:18:20 pm

I left this one in the Spesh Q Thread, but it remained unsolved. Have a go

Quote from: LovesPhysics on August 14, 2012, 08:16:43 pm
A cissoid $y^2(2a-x)=x^3$ (where $x \ge 0$ and $a>0$ ) is rotated about it's asymptote to form a volume of revolution. Find this volume in terms of $a$ . You don't really need a CAS until the very last part to get the answer.

There's a few more unanswered questions throughout this thread too, get on it spesh kiddos!

$y^2=\frac{x^3}{2a-x}$

$2ydy=\frac{6ax^2-2x^3}{(2a-x)^2}$

$dy =\frac{6ax^2-2x^3}{2y(2a-x)^2}$

$dy=\frac{(3a-x) x^\frac{1}{2}}{(2a-x)^\frac{3}{2}} (1)$

So the asymptote is x=2a
Rotate: radius (2a-x)
height dy

$volume = 2\pi \int_{0}^{2a}(2a-x)^2dy\,dx (2)$

Sub (1) into (2)

$volume = 2\pi \int_{0}^{2a}(3a-x)x^\frac{1}{2} (2a-x)^\frac{1}{2}\,dx$

P/S: OMGGGGGGGGG!!!!!!!!! CANT BELIEVE IT! THE FIRST TIME I TYPE A MATHS ANS IN LATEX

It takes me like 30 mins though

TrueTears · « **Reply #273 on:** August 29, 2012, 10:44:38 pm »

Quote from: Jenny_2108 on August 29, 2012, 10:38:06 pm

Quote from: LovesPhysics on August 29, 2012, 09:18:20 pm
I left this one in the Spesh Q Thread, but it remained unsolved. Have a go

Quote from: LovesPhysics on August 14, 2012, 08:16:43 pm
A cissoid $y^2(2a-x)=x^3$ (where $x \ge 0$ and $a>0$ ) is rotated about it's asymptote to form a volume of revolution. Find this volume in terms of $a$ . You don't really need a CAS until the very last part to get the answer.

There's a few more unanswered questions throughout this thread too, get on it spesh kiddos!

$y^2=\frac{x^3}{2a-x}$

$2ydy=\frac{6ax^2-2x^3}{(2a-x)^2}$

$dy =\frac{6ax^2-2x^3}{2y(2a-x)^2}$

$dy=\frac{(3a-x) x^\frac{1}{2}}{(2a-x)^\frac{3}{2}} (1)$

So the asymptote is x=2a
Rotate: radius (2a-x)
height dy

$volume = 2\pi \int_{0}^{2a}(2a-x)^2dy\,dx (2)$

Sub (1) into (2)

$volume = 2\pi \int_{0}^{2a}(3a-x)x^\frac{1}{2} (2a-x)^\frac{1}{2}\,dx$

P/S: OMGGGGGGGGG!!!!!!!!! CANT BELIEVE IT! THE FIRST TIME I TYPE A MATHS ANS IN LATEX
It takes me like 30 mins though

Well done, my young pandawan.

brightsky · « **Reply #274 on:** December 10, 2013, 08:01:49 pm »

Not sure if all of Ahmad's original questions have been answered yet. It seems 14-18 have yet to be solved. I shall start with some of the more straightforward ones.

14.

$\sqrt{x+3-4\sqrt{x-1}} + \sqrt{x+8-6\sqrt{x-1}}=1$

$\sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}-3)^2} = 1$

There are three cases to consider:

Case 1: $x < 5$ . Then we have:

$2-\sqrt{x-1} + 3-\sqrt{x-1} = 1$

$5 - 2\sqrt{x-1} = 1$

$x = 5$ , which is outside the domain.

So no real solutions for $x$ .

Case 2: $5 \le x \le 10$ . Then we have:

$\sqrt{x-1} -2 + 3-\sqrt{x-1} = 1$ , which is true for all $x$ in the domain.

So $5 \le x \le 10$ .

Case 3: $x > 10$ . Then we have:

$\sqrt{x-1}-2+\sqrt{x-1}-3=1$

$2\sqrt{x-1}-5 = 1$

$x = 10$ , which is outside the domain.

So no real solutions for $x$ .

So it appears that the real roots for the equation are $5 \le x \le 10$ .

16.

Let $P(x) = 6x^4 - 25x^3 + 12x^2 + 25x + 6$

We find that $P(2) = P(3) = 0$ .

So $(x-2)(x-3)$ is a factor. By inspection, then, $P(x) = (x-2)(x-3)(6x^2 + 5x + 1) = (x-2)(x-3)(2x+1)(3x+1)$ .

It is clear then that the solutions to $P(x) = 0$ are $x = -\frac{1}{2}, -\frac{1}{3}, 2, 3$ .

17.

Let $P(x) = 6x^4 - 24x^3 + 12x^2 + 24x + 6$

Now:

$P(x) = 6(x^4 - 4 x^3 + 2 x^2 + 4x + 1)$

$= 6(x^4 - 4x^3 - 2x^2 + 4x^2 + 4x + 1)$

$= 6(x^4 - 2x^2(2x+1)+ (2x+1)^2)$

$= 6(x^2 - (2x+1))^2$

$=6(x^2 - 2x - 1)^2$

$=6((x-1)^2 - (\sqrt{2})^2)^2$

$=6((x-1-\sqrt{2})(x-1+\sqrt{2}))^2$

So the real solutions to $P(x) = 0$ are $x = 1 + \sqrt{2}, 1 - \sqrt{2}$ .

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Author Topic: Recreational Problems (SM level) (Read 76996 times) Tweet Share

pi

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