Author Topic: Dekoyl's Questions (Read 23458 times) Tweet Share

dekoyl · « **Reply #15 on:** March 20, 2009, 09:54:22 pm »

Quote from: /0 on March 20, 2009, 09:47:01 pm

I think this is what it is asking: $m\mathbf{a}+n\mathbf{b}=\mathbf{i}$

Ah okay then - Thanks /0 :]

Thank you too TT.
v
v

TrueTears · « **Reply #16 on:** March 20, 2009, 09:57:43 pm »

yeap /0 is correct, the UNIT vector in the direction of x axis is 'i' and UNIT vector in the direction of the y axis is 'j' likewise if it was 3D the unit vector in the z axis would be 'k'

dekoyl · « **Reply #17 on:** March 21, 2009, 10:11:32 am »

/sigh. Slightly stuck on this one =\

Let a and b be two vectors such that $|a|=2, |b|=1$ and the angle between a and b is $60^{o}$ . Let $\phi$ be the angle between vectors $2ta + 7b$ and $a + tb$ . If $90^{o}\leq \phi \leq 180^{o}$ , find all the possible values of real number $t.$

I got $a.b=1$ then I'm not sure what to do. I suck.

dekoyl · « **Reply #18 on:** March 21, 2009, 05:32:48 pm »

Sorry nevermind about the question. Just a bit tired.

dekoyl · « **Reply #19 on:** April 07, 2009, 05:35:48 pm »

Easy one..
$\int_0^{\frac{\pi}{4}}cos^4(x).dx$

No matter what I do I keep getting
$\frac{3}{32}\pi + \frac{5}{16}$

The answer is $\frac{3}{32}\pi + \frac{1}{4}$

I don't have my CAS here to check =\ Is the book wrong?

dekoyl · « **Reply #20 on:** April 07, 2009, 08:35:24 pm »

$\int7cos^7(t).dt$

I got: $-sin^7(t) + \frac{21}{5}sin^5(t) - 7sin^3(t) + 7sin(t) + c$ but the book differs.

Am I wrong or am I wrong?

TrueTears · « **Reply #21 on:** April 07, 2009, 08:43:13 pm »

Quote from: dekoyl on April 07, 2009, 05:35:48 pm

Easy one..
$\int_0^{\frac{\pi}{4}}cos^4(x).dx$

No matter what I do I keep getting
$\frac{3}{32}\pi + \frac{5}{16}$

The answer is $\frac{3}{32}\pi + \frac{1}{4}$

I don't have my CAS here to check =\ Is the book wrong?

book is right, i get $\frac{3}{32}\pi + \frac{1}{4}$ as well

Quote from: dekoyl on April 07, 2009, 08:35:24 pm

$\int7cos^7(t).dt$

I got: $-sin^7(t) + \frac{21}{5}sin^5(t) - 7sin^3(t) + 7sin(t) + c$ but the book differs.

Am I wrong or am I wrong?

That is correct, sometimes the book has a more simplified answer, but I don't think that's necessary. I think i had a similar kind of problem: http://vcenotes.com/forum/index.php/topic,9192.60.html

Mao · « **Reply #22 on:** April 07, 2009, 09:39:21 pm »

dekoyl · « **Reply #23 on:** April 07, 2009, 09:44:05 pm »

Quote from: Mao on April 07, 2009, 09:39:21 pm

$\int_0^{\frac{\pi}{4}} \cos^4 (x) \; dx = \int_0^{\frac{\pi}{4}} \cos^2 (x) (1 - \sin^2 (x))\; dx = \int_0^{\frac{\pi}{4}} \cos^2 (x) - \frac{1}{4}\sin^2 (2x)\; dx = \int_0^{\frac{\pi}{4}} \frac{\cos(2x) + 1}{2} - \frac{1}{8}(1-\cos(4x))\; dx$

$\implies \left[ \frac{1}{4}\sin(2x) + \frac{3}{8}x + \frac{1}{32}\sin (4x)\right]_0^{\frac{\pi}{4}} = \frac{1}{4} + \frac{3\pi}{32}$

=X Thanks Mao haha.
I $\mbox{facepalmed}^n$ when I realised I did $sin^2(2x) = \frac{1}{2}(1-cos(2x))$

dekoyl · « **Reply #24 on:** April 27, 2009, 11:00:41 pm »

Okay I think I'm losing it. I'm having trouble with stupid things tonight.

Using the knowledge of $\int{xsin(x)}.dx = sin(x)-cos(x)+c$ , hence find $\int_{0}^{\pi}(x-\pi)sin(x).dx$

Answer: $-\pi$

Thanks =_=

Mao · « **Reply #25 on:** April 27, 2009, 11:45:23 pm »

$\int_0^{\pi} (x-\pi)\sin (x)\; dx = \int_0^{\pi} x\sin x\; dx - \int_0^{\pi} \pi \sin (x)\; dx = [\sin (x) - x\cos(x)]_0^{\pi} + \pi [\cos (x)]_0^\pi = \pi - 2\pi = -\pi$

it should be: $\int x\sin(x)\; dx = \sin(x) - x\cos (x) + C$

dekoyl · « **Reply #26 on:** April 27, 2009, 11:52:36 pm »

Bah. That little $-xcos(x)$ when copying out my own work screwed me over.
Thanks Mao @_@

kamil9876 · « **Reply #27 on:** April 29, 2009, 08:05:31 pm »

1.) If this identity is true for negative numbers $n$ too(which i cbf verifying) then let n=-2. and hence u will get $cos^{-4}$ on the far right which can be turned into $sec^{4}$ which is what u want. You will get $cos^{-2}$ on the left which is easy (derivative of tan).

dekoyl · « **Reply #28 on:** April 29, 2009, 10:09:32 pm »

^Hmm not sure if that works. Probably due to my own mistake, but it didn't work for me.

Anyway another one.. Because I haven't officially covered inverse circular functions [in class], please bear with me.
$\int_{\frac{-1}{2}}^{\frac{1}{2}}cos^{-1}(2x).dx$

Thanks.

Mao · « **Reply #29 on:** April 29, 2009, 10:33:20 pm »

This one is tricky, but it goes like this:

performing linear transformations on the variable x, $\int_{-1/2}^{1/2} \cos^{-1} (2x)\; dx = \int_{0}^{1} \cos^{-1} (2(x-1/2))\; dx$

This area is the same as taking an integral in the y axis from 0 to pi

$I = \int_{0}^{\pi} \frac{1}{2}\cos (y) + \frac{1}{2},\; dx$

Search the forums now!

We have moved!

Author Topic: Dekoyl's Questions (Read 23458 times) Tweet Share

dekoyl

Re: Dekoyl's Questions

TrueTears

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

TrueTears

Re: Dekoyl's Questions

Mao

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

Mao

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

kamil9876

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

Mao

Re: Dekoyl's Questions

Recent Posts

User:
Password: