Author Topic: over9000 questions thread (Read 18025 times) Tweet Share

/0 · « **Reply #90 on:** April 04, 2009, 10:54:57 pm »

3)

If T = {circle K including all its inside points} then the intersection $T \cap S$ must be only the points inside circle K which are part of the line S, i.e. the chord through QO.

$S = \{(x,y):y=2x\}$

The centre of circle K is the midpoint of DM, since DM is a diameter.

$\therefore centre = (\frac{\frac{-9}{5}+4}{2},0) = (\frac{11}{10},0)$

Distance from centre to M = $\frac{29}{10}$

$\therefore T = \{u:|u-\frac{11}{10}| \leq \frac{29}{10}\}$

I CANT BRAIN TODAY SO THIS IS LIKELY TO BE WONG

Mao · « **Reply #91 on:** April 04, 2009, 11:34:21 pm »

As requested:

T is the circle. The centre of the circle is at a point $C(x,0)$

$CM = CP\implies (4-x)^2 = (x + 1)^2 + 2^2\implies 16 - 8x = 2x + 5 \implies x = \frac{11}{10}$

which makes /0 correct

kamil9876 · « **Reply #92 on:** April 04, 2009, 11:40:16 pm »

Quote from: Over9000 on April 04, 2009, 02:07:46 pm

(circle K just passes through the points P (-1,2i), Q(-1,-2i), M(4,0) and D $(\frac{-9}{5},0)$ on the graph above)

Lol how come i missed that earlier? I guess it was a bit strange that no information was given on K, making it completely random and hence the wide possibility of answers

edit: btw, just got my century ;p

Over9000 · « **Reply #93 on:** April 04, 2009, 11:45:04 pm »

Quote from: kamil9876 on April 04, 2009, 11:40:16 pm

Quote from: Over9000 on April 04, 2009, 02:07:46 pm

(circle K just passes through the points P (-1,2i), Q(-1,-2i), M(4,0) and D $(\frac{-9}{5},0)$ on the graph above)

Lol how come i missed that earlier? I guess it was a bit strange that no information was given on K, making it completely random and hence the wide possibility of answers

edit: btw, just got my century ;p

you didnt miss it, I forgot to give that info, I just put it up now

kamil9876 · « **Reply #94 on:** April 04, 2009, 11:50:27 pm »

lol yea true. Btw /0 fact is true but it doesn't answer the question. You are right about the centre of the circle but that is not what the question is asking for. Btw, T does not equal K. Once the equation of the circle is known. U must find the two intersections of S and K. Once u know these two points you must now restrict the domain of S so that you get the line segment that we're after(the chord) by selecting an appriopriate T.

Edit: but yes, T can equal to a disk that has the same equation as circle K. SO no point of finding the intercepts.

TrueTears · « **Reply #95 on:** April 04, 2009, 11:53:25 pm »

lol soz kamil forgot that info earlier too

(congratz on ur 100 man)

Over9000 · « **Reply #96 on:** April 04, 2009, 11:56:00 pm »

Yeh, thanks for the help mao and kamil

Mao · « **Reply #97 on:** April 05, 2009, 01:40:56 am »

Quote from: kamil9876 on April 04, 2009, 11:50:27 pm

lol yea true. Btw /0 fact is true but it doesn't answer the question. You are right about the centre of the circle but that is not what the question is asking for. Btw, T does not equal K. Once the equation of the circle is known. U must find the two intersections of S and K. Once u know these two points you must now restrict the domain of S so that you get the line segment that we're after(the chord) by selecting an appriopriate T.

Edit: but yes, T can equal to a disk that has the same equation as circle K. SO no point of finding the intercepts.

I thought the question only asked for a possible set of T, not the intersection...?

confused

kamil9876 · « **Reply #98 on:** April 05, 2009, 11:14:22 am »

it says "T is a subset of Complex plane".

Now, it says that the intersection of T and S is the chord. So you must find a T such that T intersection S is the chord. The easiest way is to find the equation of K (which was almost done when u guys found the centre) and then change that circle into a disk, and that disk will be T
ie: x^2 + y^2=1 is circle, however x^2 + y^2 <=1 is a disk with same circumference.

Mao · « **Reply #99 on:** April 05, 2009, 11:24:23 am »

Quote from: kamil9876 on April 05, 2009, 11:14:22 am

it says "T is a subset of Complex plane".

Now, it says that the intersection of T and S is the chord. So you must find a T such that T intersection S is the chord. The easiest way is to find the equation of K (which was almost done when u guys found the centre) and then change that circle into a disk, and that disk will be T
ie: x^2 + y^2=1 is circle, however x^2 + y^2 <=1 is a disk with same circumference.

I thought describing T as a loci would be enough, as /0 has done, it is a subset of complex plane and its intersection with S is the chord. A circle in the complex plane is fairly easily described, $|z-z_0| = r$ , where the centre is at $z_0$

kamil9876 · « **Reply #100 on:** April 05, 2009, 11:31:02 am »

Quote from: Mao on April 05, 2009, 11:24:23 am

Quote from: kamil9876 on April 05, 2009, 11:14:22 am
it says "T is a subset of Complex plane".

Now, it says that the intersection of T and S is the chord. So you must find a T such that T intersection S is the chord. The easiest way is to find the equation of K (which was almost done when u guys found the centre) and then change that circle into a disk, and that disk will be T
ie: x^2 + y^2=1 is circle, however x^2 + y^2 <=1 is a disk with same circumference.

I thought describing T as a loci would be enough, as /0 has done, it is a subset of complex plane and its intersection with S is the chord. A circle in the complex plane is fairly easily described, $|z-z_0| = r$ , where the centre is at $z_0$

Oh yes he did that, just scrolled up. Sorry guys didn't see his last line of working.

Over9000 · « **Reply #101 on:** April 05, 2009, 06:41:07 pm »

Given that $3\overrightarrow{OX}$ + $4\overrightarrow{OY}$ = $7\overrightarrow{OZ}$ , then $\frac{XZ}{ZY}$ equals: ??
This is the diagram that came with the question http://img10.imageshack.us/img10/69/fuckfknfucjkfuckfuck.jpg

kamil9876 · « **Reply #102 on:** April 05, 2009, 08:25:14 pm »

$\vec{XZ}=\vec{OZ}-\vec{OX}$
$=\frac{3\vec{OX} + 4\vec{OY}}{7} - \frac{7\vec{OX}}{7}$
$=\frac{-4\vec{OX} + 4\vec{OY}}{7}$
$=\frac{-4}{7} (\vec{OY}-\vec{OX})$
$=\frac{-4}{7} \vec{YX}$

See if you can do the rest

Over9000 · « **Reply #103 on:** April 05, 2009, 09:30:20 pm »

Yep, thanks alot kamil

Over9000 · « **Reply #104 on:** July 05, 2009, 07:04:51 pm »

If the thermostat in an electric heater failed, the rate of increase of temperature, $\frac{d\theta}{dt}$ , would be $0.01\theta$ degrees per minute where \theta is in degrees Kelvin and t is in minutes. If a heater was switched on at a room temperature of 300K and the thermostat did not function, what would the temperature of the heater be after 10 minutes.
I did this:
$\frac{d\theta}{dt} = 0.01\theta$
$\frac{dt}{d\theta} = \frac{100}{\theta}$
$t= 100log_e(\theta)$
$10 = 100log_e(\theta)$
$0.1 = log_e(\theta)$
$\theta = e^{0.1}$
$\theta = 1.01$
$\therefore 301.1K$
But the answer is 331.55, what did I do wrong, I tried many different ways and got many different answers but I dont see whats wrong with this way.

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