Author Topic: TrueTears question thread (Read 54180 times) Tweet Share

kamil9876 · « **Reply #585 on:** October 23, 2009, 06:28:32 pm »

Quote from: TrueTears on October 23, 2009, 06:19:38 pm

Quote from: TrueTears on October 23, 2009, 05:58:42 pm
Okay, so the range of $x(t)$ for $t \ge 0$ is $[-2,2]$ which is right domain for cartesian, however range of $y(t)$ is $[0, \infty)$ however the cartesian is not defined for $[0, \infty)$ does that mean we take the highest possible range that is defined for the cartesian? ie, $[0, 30]$
So... what range do I take?

it all depends on what they defiend as t. e.g if t can go over 10seconds, say t=11, then the cos^{-1} expression is not complete. Domain of t must be provided.

TrueTears · « **Reply #586 on:** October 23, 2009, 08:22:57 pm »

Yeah thanks guys, the question doesn't specify a domain... quite a stupid Q heh

TrueTears · « **Reply #587 on:** October 24, 2009, 11:26:58 pm »

If $3\int_0^k \frac{1}{x+2} - \frac{1}{x-2} dx = 6$ and $0<k<2$ , find the exact value of $k$ .

Okay so I did:

$[\log_e|\frac{x+2}{x-2}|]_0^k = 2$

$e^2 = |\frac{k+2}{k-2}|$

How do I know which half to pick?

I did this (not sure if reasoning is right):

Case 1: if $\frac{k+2}{k-2} > 0$

then $k+2 >0$ and $k-2 > 0$ or $k+2<0$ and $k-2<0$

so $k>-2$ and $k>2$ or $k<-2$ and $k<2$

which means overall we have $k>2$ or $k<-2$

However the domain they gave for k is $0<k<2$ so if can't be the positive half of the modulus ie $\frac{k+2}{k-2} > 0$ is false.

Case 2: $\frac{k+2}{k-2} < 0$

Then we must have $k+2 <0$ and $k-2>0$ or $k+2 >0$ and $k-2<0$

so $k<-2$ and $k>2$ or $k>-2$ and $k<2$

Which means overall we have $-2<k<2$ . However the domain they give $0<k<2$ is just a subset of this which means the right equation to solve is:

$e^2 = -\left(\frac{k+2}{k-2}\right)$

Is that how you go about deciding which half of the modulus to use or is it complete wrong?

kamil9876 · « **Reply #588 on:** October 24, 2009, 11:35:50 pm »

Yes, that is completely correct reasoning

.

Although a bit convoluted as I would just do this by noticing that all I need to decide is whether $\frac{k+2}{k-2}$ is positive or negative over the domain given. We know it is positive if numerator and denominator are the same sign, and negative when they are opposite signs.

we have $0<k<2$

$\implies -2<k-2<0 (1)$ (subtracting 2 from both sides)
and $2<k+2<4 (2)$ (adding two to both sides).

(1) implies the denominator must be negatve, (2) implies the numerator is positive. Hence opposite signs and the result follows.

TrueTears · « **Reply #589 on:** October 25, 2009, 02:19:09 am »

Thanks kamil.

Also what is the convention used for naming polygons, say you a parallelogram ABCD, is the 4 vertices named in a clockwise way? Or anti clockwise?

TrueTears · « **Reply #590 on:** October 25, 2009, 03:51:00 am »

And also for any complex number $z$ , then $|z^x| = |z|^x$ for $x \in R$ .

Is that true? If so, how do you prove it?

humph · « **Reply #591 on:** October 25, 2009, 05:39:53 am »

Quote from: TrueTears on October 25, 2009, 03:51:00 am

And also for any complex number $z$ , then $|z^x| = |z|^x$ for $x \in R$ .

Is that true? If so, how do you prove it?

Not sure how to prove it using spec techniques. Using the complex-valued exponential and logarithm functions, we have that for $z \in \mathbb{C}$ , $a \in \mathbb{R}$ ,
$z^a = (\exp(\log z))^a = (\exp(\log|z| + i \arg z))^a = (\exp(\log|z|)^a (\exp(i \arg z))^a = |z|^a \exp(i a \arg z)$ ,
and so
$|z^a| = ||z|^a \exp(i a \arg z)| = |z|^a |\exp(i a \arg z)| = |z|^a$ ,
as
$|\exp(i a \arg z)| = |\cos(a \arg z) + i \sin(a \arg z)| = \sqrt{\cos^2(a \arg z) + \sin^2(a \arg z)} = 1$ .

TrueTears · « **Reply #592 on:** October 25, 2009, 03:07:29 pm »

Thanks humph so it is true for $x \in \mathbb{R}$ what about $x \in \mathbb{C}$ ?

Mao · « **Reply #593 on:** October 25, 2009, 11:08:37 pm »

Quote from: TrueTears on October 25, 2009, 03:07:29 pm

Thanks humph so it is true for $x \in \mathbb{R}$ what about $x \in \mathbb{C}$ ?

raising a number to a complex power?

$a^{z} = a^{x+yi} = a^x \cdot a^{yi} = a^x \cdot e^{\log_e (a) y i}$

As you can see, it is just another complex number.

Extending $a$ to a complex plane:

$a^{z} = \left(|a|e^{i\arg a}\right)^{z} = |a|^z \cdot e^{i \arg a \cdot z} = |a|^x \cdot e^{y\log_e |a| i} \cdot \left( e^{i x \arg a} \cdot e^{-y\arg a}\right) = |a|^x \cdot e^{-y\arg a} \cdot \left( e^{(y\log_e |a| + x \theta ) i} \right)$

$|a^z| = |a|^x e^{-y\arg a}$

So no, it does not extend over the complex plane.

TrueTears · « **Reply #594 on:** October 31, 2009, 08:04:05 pm »

Just a few technicalities.

What's the definition of a line, line segment and ray?

Line: Both ends extend out to infinity.

Ray: Has a fixed starting point, other end goes to infinity.

Line segment: Has 2 fixed ends and thus a fixed length. (Does that mean a point would also be a line segment since you can consider it as 0 length lol)

Is that right..?

bem9 · « **Reply #595 on:** October 31, 2009, 08:33:32 pm »

but a point doesnt have '2 fixed ends' so its cant be a line segment

TrueTears · « **Reply #596 on:** October 31, 2009, 09:01:33 pm »

Quote from: bem9 on October 31, 2009, 08:33:32 pm

but a point doesnt have '2 fixed ends' so its cant be a line segment

What if you define the starting to be 'A'. The ending point is 'B' (even though it's the same point).

Thus the point can be called AB with a length of 0.

Just like a random line segment can be called AB with a length of 5.

Or is the definition of the length of a line segment with length $x$ such that $x \in R^+$ only? In which case a point would not be a line segment.

Over9000 · « **Reply #597 on:** October 31, 2009, 09:17:17 pm »

Quote from: TrueTears on October 31, 2009, 09:01:33 pm

Quote from: bem9 on October 31, 2009, 08:33:32 pm
but a point doesnt have '2 fixed ends' so its cant be a line segment
What if you define the starting to be 'A'. The ending point is 'B' (even though it's the same point).

Thus the point can be called AB with a length of 0.

Just like a random line segment can be called AB with a length of 5.

Or is the definition of the length of a line segment with length $x$ such that $x \in R^+$ only? In which case a point would not be a line segment.

By doing that, wouldn't point a = point b since their exactly the same point.

TrueTears · « **Reply #598 on:** October 31, 2009, 09:17:45 pm »

Quote from: Over9000 on October 31, 2009, 09:17:17 pm

Quote from: TrueTears on October 31, 2009, 09:01:33 pm
Quote from: bem9 on October 31, 2009, 08:33:32 pm
but a point doesnt have '2 fixed ends' so its cant be a line segment
What if you define the starting to be 'A'. The ending point is 'B' (even though it's the same point).

Thus the point can be called AB with a length of 0.

Just like a random line segment can be called AB with a length of 5.

Or is the definition of the length of a line segment with length $x$ such that $x \in R^+$ only? In which case a point would not be a line segment.
By doing that, wouldn't point a = point b since their exactly the same point.

No I'm saying a point is also a line segment does not imply that any point equals any other point.

However I don't know the rigorous definition of a line segment yet (nor a line or a ray), just that according to my current understanding, they can be proved to be the same.

TrueTears · « **Reply #599 on:** October 31, 2009, 09:37:15 pm »

None of the choices are right? WDF

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