Author Topic: TyErd's questions (Read 43013 times) Tweet Share

brightsky · « **Reply #420 on:** July 03, 2010, 06:52:40 pm »

$3\sin(x) = 4\cos(x)$

$\frac{3\sin(x)}{4\cos(x)} = 1$

$\frac{3}{4} \tan(x) = 1$

$\tan(x) = \frac{4}{3}$

$x = \arctan\left(\frac{4}{3}\right)$

Sub back to the original equation to find y coordinate.

Lol, I don't know if this method is legitimate or not...:S

TyErd · « **Reply #421 on:** July 03, 2010, 06:56:23 pm »

hmm never learnt arctan

TyErd · « **Reply #422 on:** July 03, 2010, 06:57:39 pm »

another problem while I quickly go have dinner

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

is this even in the methods course?

superflya · « **Reply #423 on:** July 03, 2010, 06:58:38 pm »

Quote from: TyErd on July 03, 2010, 06:56:23 pm

hmm never learnt arctan

inverse tan.

brightsky · « **Reply #424 on:** July 03, 2010, 07:36:48 pm »

First find the indefinite integral $\int \frac{1 - \cos(2x)}{1 +\cos(2x)} dx$

$= \int \frac{1 - \frac{e^{2ix} + e^{-2ix}}{2}}{1 + \frac{e^{2ix} + e^{-2ix}}{2}} dx$

$= \int \frac{\frac{2 - e^{2ix} - e^{-2ix}}{2}}{\frac{2 + e^{2ix} + e^{-2ix}}{2}} dx$

$= \int \frac{\frac{e^{2ix} - 2 + e^{-2ix}}{4i^2}}{\frac{2 + e^{2ix} + e^{-2ix}}{4}} dx$

$= \int \frac{\sin^2(x)}{\cos^2(x)} dx$

$= \int \tan^2(x) dx$

But $\tan^2(x) = \sec^2(x) - 1$

So the original integrand becomes:

$\int \sec^2(x) - 1 dx$

$= \int \sec^2(x) dx - \int 1 dx$

$= \int \sec^2(x) dx - x + C$

$= \tan(x) - x + C$

EDIT: Latex error.

Using this:

$\int^{\frac{\pi}{4}}_{0} \frac{1 - \cos(x)}{1 +\cos(x)} dx$

$= \tan\left(\frac{\pi}{4}\right) - \frac{\pi}{4} - \tan(0) + 0$

$= 1 - \frac{\pi}{4}$

moekamo · « **Reply #425 on:** July 03, 2010, 07:38:37 pm »

Quote from: TyErd on July 03, 2010, 06:57:39 pm

another problem while I quickly go have dinner

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

is this even in the methods course?

a bit of simplifying using double angle formulas: since $\cos 2x = 2\cos ^2 x - 1 = 1 - 2 \sin ^2 x$ we can do this:

$\frac{1-\cos 2x}{1 + \cos 2x} = \frac{1-1+2\sin ^2 x}{1+2\cos ^2 x - 1} = \frac{\sin ^2 x}{\cos ^2 x} = \tan ^2 x = \sec ^2 x - 1$

therefore $\int_0^{\frac{\pi}{4}} \frac{1-\cos 2x}{1 + \cos 2x} dx = \int_0^{\frac{\pi}{4}} \sec ^2 x - 1 dx = \left [ \tan x -x \right ]_0^{\frac{\pi}{4}} = 1-\frac{\pi}{4}$

since you have to use double angle formulas and stuff, its not going to appear on a methods exam, but if you can follow it and understand then thats better for you

TyErd · « **Reply #426 on:** July 03, 2010, 08:01:46 pm »

so it definitely wont appear in the exam?

TyErd · « **Reply #427 on:** July 04, 2010, 06:13:08 pm »

The straight line $y=2x+1$ cuts the curve $y=e^x + 1 at (0,1)$ . At this point the acute angle between the line and the curve is $\theta$ . What is $tan \theta$

I'm confused with this question because they don't intersect. Question is from essentials pg 500.

brightsky · « **Reply #428 on:** July 04, 2010, 06:15:55 pm »

Quote from: TyErd on July 04, 2010, 06:13:08 pm

The straight line $y=2x+1$ cuts the curve $y=e^x + 1 at (0,1)$ . At this point the acute angle between the line and the curve is $\theta$ . What is $tan \theta$

I'm confused with this question because they don't intersect. Question is from essentials pg 500.

LOL. :p

TyErd · « **Reply #429 on:** July 04, 2010, 06:20:30 pm »

Also $\int_0^{36} \frac{dx}{2x + 9} = log_e k$ , then $k$ is ?

TyErd · « **Reply #430 on:** July 04, 2010, 06:25:46 pm »

and this one too $\int_a^b sin(2x) dx = 0$ . What does a and b equal?

m@tty · « **Reply #431 on:** July 04, 2010, 06:27:22 pm »

$\int \frac{dx}{2x+9}=\frac{1}{2}\log_e(2x+9)+C$

$\int_0^{36}\frac{dx}{2x+9}=\frac{1}{2}\left(\log_e{81}-\log_e{9}\right)=\frac{1}{2}\log_e9$

Therefore $k=\sqrt{9}=3$

m@tty · « **Reply #432 on:** July 04, 2010, 06:30:55 pm »

Quote from: TyErd on July 04, 2010, 06:25:46 pm

and this one too $\int_a^b sin(2x) dx = 0$ . What does a and b equal?

For this there are infinite solutions.

It will equal zero for all $a=b$ .

But I assume that you are meant to use symmetry to find a situation such as $a=0$ and $b=\pi$ . Where the net result is zero.

crappy · « **Reply #433 on:** July 04, 2010, 06:35:06 pm »

Quote from: m@tty on July 04, 2010, 06:27:22 pm

$\int \frac{dx}{2x+9}=\frac{1}{2}\log_e(2x+9)+C$

$\int_0^{36}\frac{dx}{2x+9}=\frac{1}{2}\left(\log_e{81}-\log_e{9}\right)=\frac{1}{2}\log_e9$

Therefore $k=\frac{9}{2}$

Shouldn't k = 3?

m@tty · « **Reply #434 on:** July 04, 2010, 06:36:53 pm »

Ah, yes. How stupid of me.

Search the forums now!

We have moved!

Author Topic: TyErd's questions (Read 43013 times) Tweet Share

brightsky

Re: TyErd's questions

TyErd

Re: TyErd's questions

TyErd

Re: TyErd's questions

superflya

Re: TyErd's questions

brightsky

Re: TyErd's questions

moekamo

Re: TyErd's questions

TyErd

Re: TyErd's questions

TyErd

Re: TyErd's questions

brightsky

Re: TyErd's questions

TyErd

Re: TyErd's questions

TyErd

Re: TyErd's questions

m@tty

Re: TyErd's questions

m@tty

Re: TyErd's questions

crappy

Re: TyErd's questions

m@tty

Re: TyErd's questions

Recent Posts

User:
Password: