Author Topic: Spec '10 - Help forum (Read 9997 times) Tweet Share

Chavi · « **Reply #15 on:** January 04, 2010, 02:41:06 pm »

Could i please get some help with
$\int sin^2(x) cos^3(x) dx$
Thanks

dekoyl · « **Reply #16 on:** January 04, 2010, 02:48:05 pm »

$\int sin^2(x)\left(1-sin^2(x)\right)cos(x) dx$

Let $u = sin(x)$

$du = cos(x) dx$

$\int u^2(1-u^2)du$

Etc.

Juddinator · « **Reply #17 on:** January 24, 2010, 05:21:17 pm »

In Essentials for Spec, an example which is: tan(2x - π) = square root of 3, for x is an element of [-π, π]. Let theta = 2x - π.

The next line states therefore -2π ≤ 2x – π and thus -3 π ≤ 2x – π ≤ π and -3 π ≤ theta ≤ π.

I don't understand how they have gone about this. Is it a worry if I cannot figure this out I could be in trouble for Spec this year? :S

Thanks

superflya · « **Reply #18 on:** January 24, 2010, 05:36:23 pm »

$\Theta =2x-\Pi$

$\Rightarrow$

ur original restrictions are $-\Pi\leq x\leq \Pi$

multiple all by 2 $-2\Pi\leq 2x\leq 2\Pi$

then subtract the $\Pi$ which was in the original restriction

$\Rightarrow$ $-3\Pi\leq 2x-\Pi\leq \Pi$

yea u shood be able to do this

Edit: thanks matt

googoo · « **Reply #19 on:** January 24, 2010, 08:29:55 pm »

Quote from: Chavi on December 15, 2009, 02:44:16 pm

Could i get some help with:
Factorize over R: z^4 + 64
and Factorize over C z^4 + 64

Thanks

By completing squares: z^4 + 64 = z^4 + 16z^2 + 64 - 16z^2
= (z^2 +

^2 - (4z)^2 = (z^2 - 4z +

(z^2 + 4z +

etc.

m@tty · « **Reply #20 on:** January 24, 2010, 08:40:52 pm »

Quote from: superflya on January 24, 2010, 05:36:23 pm

$\Theta =2x-\Pi$

$\Rightarrow$

ur original restrictions are $-\Pi\leq x\leq \Pi$

multiple all by 2 $-\Pi\leq x\leq \Pi$

then subtract the $\Pi$ which was in the original restriction

$\Rightarrow$ $-3\Pi\leq 2x-\Pi\leq \Pi$

yea u shood be ablt to do this

You missed multiplying by two in LaTeX, and yeah, that's how to do it.

mandy · « **Reply #21 on:** January 24, 2010, 10:00:10 pm »

SCALAR PRODUCT OF TWO VECTORS.

Can I get help with this question, please

OAB is an equilateral triangle with side length 1 unit. O is the origin, OA = i and OB = xi + yj.
a. Find the values of x and y, and hence express OB in terms of i and j.

Mao · « **Reply #22 on:** January 24, 2010, 10:20:11 pm »

Angle of AOB - 60 degrees
$\angle AOB = 60^o$

$\implies OA \cdot OB = |OA|\cdot |OB| \cos (60^o)$

$\implies x = \cos(60^o) = \frac{1}{2}$

$|OB| = 1 \implies x^2 + y^2 = 1^2 \implies y = \pm \frac{\sqrt{3}}{2}$

fady_22 · « **Reply #23 on:** January 24, 2010, 10:23:58 pm »

Quote from: Mao on January 24, 2010, 10:20:11 pm

Angle of AOB - 60 degrees
$\angle AOB = 60^o$

$\implies OA \cdot OB = |OA|\cdot |OB| \cos (60^o)$

$\implies x = \cos(60^o) = \frac{1}{2}$

$|OB| = 1 \implies x^2 + y^2 = 1^2 \implies y = \frac{\sqrt{3}}{2}$

Technically, couldn't it be both positive and negative $\frac{\sqrt{3}}{2}$ ?

kamil9876 · « **Reply #24 on:** January 24, 2010, 10:35:46 pm »

Yep you are right, with those conditions as specified in the question you can tell without even doing any working that there will be two answers since you can reflect the triangle in the y-axis and you get another triangle that satisfies those conditions.

mandy · « **Reply #25 on:** January 24, 2010, 10:38:01 pm »

Thank youuuuu, I get it now

mandy · « **Reply #26 on:** January 25, 2010, 12:45:23 pm »

If a = i + 3j - 2k and b = -2i + 4j -8k, resolve a into two components, one parallel to b and the other perpendicular to b.

Help please

fady_22 · « **Reply #27 on:** January 25, 2010, 12:50:43 pm »

Parallel to b, by definition: $\frac{a.b}{b.b}*b$
Perpendicular to b: Subtract the above vector from a.

Hope that helps!

cipherpol · « **Reply #28 on:** January 25, 2010, 01:04:09 pm »

Quote from: mandy on January 25, 2010, 12:45:23 pm

If a = i + 3j - 2k and b = -2i + 4j -8k, resolve a into two components, one parallel to b and the other perpendicular to b.

Help please

Let u be the component parallel to b:

$u=(a.\hat{b})\hat{b}$

$\hat{b}=\frac{1}{2\sqrt{21}}(-2i+4j-8k)$

$a.\hat{b}=\frac{1}{2\sqrt{21}}(i+3j-2k).(-2i+4j-8k)$

$\implies \frac{1}{2\sqrt{21}} (26) = \frac{13\sqrt{21}}{21}$

$u=\frac{13\sqrt{21}}{21}*\frac{1}{2\sqrt{21}}(-2i+4j-8k)$

$\implies u=\frac {13}{42}(-2i+4j-8k)$

Can you tell me if this is the correct answer :S

EDIT: looking over it, fady's way is much faster, use that instead

mandy · « **Reply #29 on:** January 25, 2010, 01:13:54 pm »

When I did it, I got the same answer as you, but its not right

Quote from: cipherpol on January 25, 2010, 01:04:09 pm

Quote from: mandy on January 25, 2010, 12:45:23 pm
If a = i + 3j - 2k and b = -2i + 4j -8k, resolve a into two components, one parallel to b and the other perpendicular to b.

Help please

$\implies u=\frac {13}{42}(-2i+4j-8k)$

The answer to the question is $\implies u=\frac {13}{21}(-i+2j-4k) + \frac {1}{21}(34i+37j+10k)$

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