Author Topic: Bawse's SM Question Thread (Read 2606 times) Tweet Share

bawse · « **Reply #15 on:** March 27, 2011, 11:53:03 am »

Quote from: enpassant on March 27, 2011, 07:12:56 am

8, 165 deg; 8, -75 deg

Do you mind showing me how you got that?

Thank you

luken93 · « **Reply #16 on:** March 27, 2011, 11:58:08 am »

If it is a cubic equation, then an Argand diagram will have 3 lines that kinda look like a hippy sign - lines with equal angles in between.

Therefore:
45,
45 + 120,
45 - 120
= 45, 165, - 75

The magnitude always stays the same, hence enppassant's solutions...

kamil9876 · « **Reply #17 on:** March 27, 2011, 12:17:10 pm »

Of course that is assuming that by "cubic equation" we mean an equation of the form $x^3-a=0$ . If it was just a general degree 3 polynomial then we cannot say that the angle between them is equal, or that they even have the same modulus.

bawse · « **Reply #18 on:** March 27, 2011, 12:20:20 pm »

Thanks luken.

bawse · « **Reply #19 on:** March 27, 2011, 09:35:17 pm »

Have I done this correctly?

Quote

Jeff and Henry are picturing their position as if theu were on an Argand diagram. Jeff's standing at point G on level ground 8 units from the origin on a 45 degree angle. Henry is at point F on level ground with the coordinates $(6\sqrt{3}, -6i)$

Let c represent point G and d represent point H

$\frac{c}{d} = \frac{8+8i}{6\sqrt{3}-6i} * \frac{6\sqrt{3}+6i}{6\sqrt{3}+6i}<br />$

$= \frac{48\sqrt{3} +48i + 48\sqrt{3}i +48 (-1)}{108+36\sqrt3i -36\sqrt3i - 36(-1)}$

$=\frac{(48\sqrt{3}-48)+(48+48\sqrt{3})i}{(108+36)+(36\sqrt3-36\sqrt3)i}$

$=\frac{(48\sqrt{3}-48)+(48+48\sqrt{3})i}{144}$

$Cartesian form = (\frac{48\sqrt{3}-48}{144}, \frac{48+48\sqrt{3}}{144})$

luken93 · « **Reply #20 on:** March 27, 2011, 09:58:34 pm »

c is wrong...

8 units from the origin on a 45 degree angle = $8cis(\frac{\pi}{4})$

For ease of division, you can also convert d into polar form:
$\sqrt{(6\sqrt{3})^2 + (-6)^2} \times cis(arc tan\frac{-6}{6\sqrt{3}})$

$= 12 cis(\frac{-\pi}{3})$

Hence,

$\frac{c}{d} = \frac{8}{12}cis(\frac{\pi}{4} - \frac{-\pi}{3})$

$\frac{c}{d} = \frac{2}{3}cis(\frac{3\pi}{12} - \frac{-4\pi}{12})$

$\frac{c}{d} = \frac{2}{3}cis(\frac{7\pi}{12})$

You can then get the co-ords from here:

$x = r cos(\theta)$

$x = \frac{2}{3}cos(\frac{7\pi}{12})$

$x = \frac{\sqrt{2} - \sqrt{6}}{6}$

$y = r sin(\theta)$

$y = \frac{2}{3}sin(\frac{7\pi}{12})$

$y = \frac{\sqrt{2} + \sqrt{6}}{6}$

Hence co-ords are:

$\frac{\sqrt{2} - \sqrt{6}}{6}, \frac{\sqrt{2} + \sqrt{6}}{6}i$

bawse · « **Reply #21 on:** March 27, 2011, 10:02:16 pm »

Ah ok, I totally misread it

Thanks again luken

bawse · « **Reply #22 on:** June 11, 2011, 10:50:54 pm »

$f'(x)=x\sqrt[]{x^2+1}$ and $f(0)=1$ , then $f(x)=$

Anyone mind helping? Thanks.

xZero · « **Reply #23 on:** June 11, 2011, 11:16:05 pm »

use substitution, let u=x^2. see if you can work it out from there

bawse · « **Reply #24 on:** June 11, 2011, 11:45:37 pm »

Lol, I'm an idiot. I just realised it's

$\int x(\sqrt[]{x^2+1})\,dx$

$u=x^2+1$
$du=2xdx$

$\frac{1}{2}$ $\int (\sqrt[]{u}) \,du$

integral of $\sqrt[]{u}$ is $\frac{2u^(\frac{3}{2})}{3}$

$=\frac{u^\frac{3}{2}}{3}$

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

xZero · « **Reply #25 on:** June 12, 2011, 12:44:09 am »

or you can let u=x^2+1 ahahha, nice work!

might wanna fix your latex tho, the integral isn't displaying properly

bawse · « **Reply #26 on:** June 12, 2011, 12:54:08 am »

Lol, thanks.

I fixed the latex, was using the incorrect command.

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