Author Topic: dcc help me thread (Read 31393 times) Tweet Share

droodles · « **Reply #150 on:** July 10, 2008, 05:07:59 pm »

hmm essential methods 3 4 is quite a prick book, so im using a calculator

/0 · « **Reply #151 on:** July 10, 2008, 06:57:49 pm »

Quote from: droodles on July 10, 2008, 04:46:43 pm

(Image removed from quote.)

Is that the entire question? If they don't give you a hint (i.e. a derivative to solve) then you'll need to use a calculator.

Mao · « **Reply #152 on:** July 10, 2008, 07:13:20 pm »

Quote from: droodles on July 10, 2008, 04:46:43 pm

(Image removed from quote.)

its a graphics calculator question =P

Code: [Select]

fnInt(e^(-x/10)*sin(2x),x,0,3.14159)

... i told you that on tuesday...

droodles · « **Reply #153 on:** July 10, 2008, 08:00:23 pm »

dont lie

you did.

droodles · « **Reply #154 on:** July 21, 2008, 09:25:22 pm »

POINT A IS (0,6) AND POINT B IS (5,5)

SOME INSIGHT PLEASE

dcc · « **Reply #155 on:** July 21, 2008, 09:31:11 pm »

A is when x = 0:

$0 = 6y - y^2 = y(6 - y) \implies A = (0,\ 6)$

$B = (5,\ 5)$ (setting the two equations equal to each other)

$P = \int_0^5 y\ dy + \int_5^6 6y - y^2\ dy = \left[ \dfrac{y^2}{2} \right]_0^5 + \left[ 3y^2 - \dfrac{y^3}{3} \right]_5^6$

$P = \dfrac{25}{2} + \left(108 - 72\right) - \left(75 - \dfrac{125}{3}\right) = \dfrac{91}{6}\ \mbox{sq units}$

Ahmad · « **Reply #156 on:** July 21, 2008, 11:59:28 pm »

droodles, the idea is to integrate along the y-axis. It may feel unnatural, and if you truly wish you may switch all the y's and x's in the question (without changing any areas), and consequently the graph will be flipped about the line y = x, now the question should seem familiar.

droodles · « **Reply #157 on:** July 23, 2008, 09:38:56 pm »

a) is a given

but I have no idea about the rest

Mao · « **Reply #158 on:** July 23, 2008, 09:50:09 pm »

$\frac{d}{dx}(e^{-3x}\sin 2x) = -3\cdot e^{-3x}\sin 2x + 2\cdot e^{-3x}\cos 2x$

$\implies \int (-3\cdot e^{-3x}\sin 2x + 2\cdot e^{-3x}\cos 2x)\; dx = e^{-3x}\sin 2x + C_1$

$\implies -3\int e^{-3x}\sin 2x\; dx + 2\int e^{-3x}\cos 2x\; dx = e^{-3x}\sin 2x + C_1$ as required
(we'll call this [1])

$\frac{d}{dx}(e^{-3}\cos 2x) = -3\cdot e^{-3x}\cos 2x - 2\cdot e^{-3x}\sin 2x$

$\implies -3\int e^{-3x}\cos 2x\; dx - 2\int e^{-3x}\sin 2x\; dx = e^{-3x}\cos 2x + C_2$ as required also
(we'll call this [2])

c)
let $I_1=\int e^{-3x}\sin 2x\; dx$ and $I_2=\int e^{-3x}\cos 2x\; dx$

[1]: $e^{-3x}\sin(2x)+C_1=-3I_1+2I_2$
[2]: $e^{-3x}\cos(2x)+C_2=-3I_2-2I_1$

3*[1]+2*[2]

$3e^{-3x}\sin(2x)+2e^{-3x}\cos(2x)+C = -13I_1$

$\implies I_1=\int e^{-3x}\sin 2x\; dx=-\frac{1}{13}(3e^{-3x}\sin(2x)+2e^{-3x}\cos(2x))+C$

done =]

costargh · « **Reply #159 on:** July 23, 2008, 09:54:43 pm »

Whos better at math; dcc or mao?

bigtick · « **Reply #160 on:** July 23, 2008, 09:56:32 pm »

Mao · « **Reply #161 on:** July 23, 2008, 10:01:19 pm »

Quote from: costargh on July 23, 2008, 09:54:43 pm

Whos better at math; dcc or mao?

why would it matter?
I say dcc though.

costargh · « **Reply #162 on:** July 23, 2008, 10:03:38 pm »

Wondering who VN dux will be, I know you're smart with maths + science, but dcc is my dark horse cause i dont know much about him

dcc · « **Reply #163 on:** July 23, 2008, 10:16:39 pm »

Quote from: costargh on July 23, 2008, 10:03:38 pm

Wondering who VN dux will be, I know you're smart with maths + science, but dcc is my dark horse cause i dont know much about him

maybe I like it that way

ell · « **Reply #164 on:** August 25, 2008, 08:28:56 pm »

Quote from: droodles on August 25, 2008, 08:18:50 pm

since that asshat dcc is away in QLD, i need someone to fill in for her.

A fair coin is tossed 50 times. If X is the number of heads observed, ﬁnd:

$Pr(X\le10)$

Code: [Select]
binomcdf(50,0.5,10)won't work

Why not? Answer should be approx. 0.00001193, or 1.193066584 x 10^-5

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