Author Topic: BEC'S methods questions (Read 107505 times) Tweet Share

Glockmeister · « **Reply #300 on:** May 05, 2008, 08:57:34 pm »

Btw: Just you let you know, you don't need to know sinh for methods. You don't even need to for Spech ('tis a First-Year Uni maths thing)

Mao · « **Reply #301 on:** May 05, 2008, 09:02:22 pm »

Quote from: coblin on May 05, 2008, 08:08:35 pm

Mao, it is very easy because you end up with $32\sinh^5x = 32\sinh x(\sinh^2x)^2 = 32\sinh x(\cosh^2x -1)^2$ . A substitution from here for an easy kill.

yeah, i forgot that trig stuff applies

silly me xD

bec · « **Reply #302 on:** May 08, 2008, 08:14:07 am »

Find the gradient of the curve f(x) = |sin(2x)| at x = $\frac {2 \pi}{3}$
How do I do this? I think i need to find a general equation but i'm not sure how...

AppleXY · « **Reply #303 on:** May 08, 2008, 08:42:41 am »

as the graph of an absolute function has sharp end-points which are non-differentiable it's difficult to find a general derivative function.

Thus, use a tangent at x = $\frac{2\pi}{3}$

take values close to $\frac{2\pi}{3}$ say, $\frac{2\pi}{3} - \frac{\pi}{20}$ and $\frac{2\pi}{3} + \frac{\pi}{20}$

sub in into f

(remember with the abs)

:. m = $\frac{f[x_2] -f[x_1]}{x_2-x_1}$

:. m $\approx$ 1

Neobeo · « **Reply #304 on:** May 08, 2008, 09:00:29 am »

At $x=\frac{2\pi}{3}$ , sin(2x) is negative, so the effective function at that point is $g(x)=-\sin(2x)$ .

Differentiating it normally:
$g'(x)=-2\cos(2x)$
$g'(\frac{2\pi}{3})=-2\cos\frac{4\pi}{3}=1$

Note that this works at $x=\frac{2\pi}{3}$ because it is negative, it is not a general derivative that will hold for all values of x.

bec · « **Reply #305 on:** May 08, 2008, 05:12:57 pm »

thank you!

AppleXY · « **Reply #306 on:** May 08, 2008, 05:18:42 pm »

Neobeo's way is much more efficient, however

Collin Li · « **Reply #307 on:** May 08, 2008, 07:12:52 pm »

In general, when finding the derivative of an absolute function, you should convert the absolute function into a hybrid function, and differentiate each part of the hybrid individually. In this case, you are working with the $-\sin 2x$ part of the hybrid.

bec · « **Reply #308 on:** May 17, 2008, 07:36:39 pm »

can i work out questions like this without a calculator?

Let g(x)=e^f(x). If g'(x)=-2xe^-x², find a rule for f.

or

Let g(x)= cos(f(x)). If g'(x)=3e^2xsin(f(x)), find a rule for f.

<mod action: fixed superscript tags>

dcc · « **Reply #309 on:** May 17, 2008, 07:49:07 pm »

Consider:

$u = f(x) \therefore \dfrac{du}{dx} = f'(x)$ (please ignore my mixture of notation)

$g = e^{u} \therefore \dfrac{dg}{du} = e^{u}$

THEREFORE:

$\dfrac{dg}{dx} = \dfrac{dg}{du} \cdot \dfrac{du}{dx} = e^{f(x)} \cdot f'(x)$

Comparing this to our original statement:

$e^{f(x)} \cdot f'(x) = -2xe^{-x^2} \implies f(x) = -x^2$

dcc · « **Reply #310 on:** May 17, 2008, 07:56:42 pm »

Similiarly for the second one:

$g(x) = \cos f(x) \implies g'(x) = f'(x) \cdot -\sin f(x)$ (thanks bec, i've been doing too much integration recently

)

$\implies f'(x) \cdot -\sin f(x) = 3e^{2x} \sin f(x)$

$\implies f'(x) = -3e^{2x} \implies f(x) = \int -3e^{2x} dx = -\dfrac{3}{2}e^{2x} + c$

$f(x) = -\dfrac{3}{2}e^{2x} + c$ , c being the constant of integration.

bec · « **Reply #311 on:** May 17, 2008, 09:59:40 pm »

thanks, i understand it now. one thing though, the second answer you gave wasn't right - you forgot the negative in the first line: g'x should = -f'x sin (f(x))

thanks!

Mao · « **Reply #312 on:** May 17, 2008, 10:33:05 pm »

Quote from: gonzo on May 17, 2008, 09:08:32 pm

Quote from: Glockmeister on May 05, 2008, 08:57:34 pm
Btw: Just you let you know, you don't need to know sinh for methods. You don't even need to for Spech ('tis a First-Year Uni maths thing)

Tell my spesh teacher that. His maxim is "This is not examinable but......"

that's not necessarily a bad thing though

knowing more advanced things can, in many occasions, improve your understanding.

bec · « **Reply #313 on:** May 18, 2008, 12:05:20 pm »

i keep getting this wrong:

Find the integer value of p, if $\int_5^p (x^{3}-3x^{2})dx = 30$

is my method wrong? i'm trying to solve F(5)-F(p)=30, where $F(x)= \frac {x^{4}}{4}-x^{3}$

I'm getting two solutions on the calc, regardless of whether i put in
solve( $\int_5^p (x^{3}-3x^{2})dx = 30$ ,p) or the above eqn that i worked out by hand...

edit: integrals fixed

Mao · « **Reply #314 on:** May 18, 2008, 01:32:03 pm »

quick thing first:

second fundamental theorem: $\int_a^b f(x) dx = F(b)-F(a)$
it appears you have got the two front-to-back.

$\int_5^p (x^3-3x^2)dx = 30$

$\left[ \frac{x^4}{4} - x^3 \right]_5^p = 30$

$\left( \frac{p^4}{4}-p^3\right) - \left(\frac{5^4}{4} - 5^3\right) = 30$

$\frac{p^4}{4} -p^3 - \frac{245}{4}=0$

$p^4 - 4p^3 - 245 = 0$

using calculator:

$p\approx \{-3.235,5.485\}$

there are naturally two solutions:

ps windows snipping tool sucks... it only has one colour highlighter on there! =S

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Author Topic: BEC'S methods questions (Read 107505 times) Tweet Share

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