Author Topic: Differential Calculuss =/ (Read 3143 times) Tweet Share

Piney.z · « **on:** May 14, 2008, 09:40:31 pm »

Can anyone help with this question?! i am greatly STuck! $:-\$

A curve has a gradient given by $\frac{dy}{dx} = 3x^2 + \frac{1}{1+4x^2}$ and its graph passes through $( \frac{1}{2} , \frac{pi}{8} )$ . Find the equation of the curve.

Thankss~ =]

Mao · « **Reply #1 on:** May 14, 2008, 09:48:17 pm »

antidifferentiate $\frac{dy}{dx}$ , second part is an arctan =)

ed_saifa · « **Reply #2 on:** May 14, 2008, 10:00:48 pm »

$y=x^3 + \frac{arctan2x}{2} + C$
Sub in $(\frac{1}{2},\frac{\pi}{8})$
$y= \frac{1}{8} + \frac{arctan1}{2} + C=\pi$
$\frac{1}{8} + \frac{\pi}{8} + C= \frac{\pi}{8}$
$C= -\frac{1}{8}$
$y= x^3 + \frac{arctan2x}{2} - \frac{1}{8}$

Piney.z · « **Reply #3 on:** May 14, 2008, 10:10:21 pm »

THx mao and Ed!

now it makes sense.....
why dint i think of that.
:buck2:

yep. so the answer would be

$y = x^3 + \frac{1}{2} tan(-1) 2x - \frac{1}{8}$

right?

Piney.z · « **Reply #4 on:** May 14, 2008, 10:11:20 pm »

tan inverse i mean
not tan(-1)

ed_saifa · « **Reply #5 on:** May 14, 2008, 10:12:30 pm »

yeah it should be right..unless i made a careless error

Piney.z · « **Reply #6 on:** May 14, 2008, 10:22:29 pm »

looks right to me.
so i'll say no to tht careless error

Thx

Piney.z · « **Reply #7 on:** May 14, 2008, 10:29:50 pm »

hmm..
Another Q

Find the coordinates of the point where the maximum gradient $f(x) = tan^{-1} \frac{x}{a}$ occurs (a is a constant )

ed_saifa · « **Reply #8 on:** May 14, 2008, 10:39:13 pm »

Wouldn't the maximum gradient occur at $(0,0)$ ? I got this from looking at the graph.In addition, the max gradient sometimes occurs at the non stationary point of inflection.

mark_alec · « **Reply #9 on:** May 14, 2008, 10:45:12 pm »

(0,0) is the point, but it is better if you show it using calculus.

$f'(x) = \frac{a^2}{a^2 + x^2}$

This will have a maximum value when the denominator is minimised, which occurs at $x=0$ .

Piney.z · « **Reply #10 on:** May 15, 2008, 09:27:07 pm »

tyty =]

dont mind me if i'm asking too many q's... =/

Differentiate $sin^{-1} x^2$ and hence find $\int \frac{2x}{\sqrt(1-x^4)} dx$

since it's $x^4$ .. i'm lost lol

dcc · « **Reply #11 on:** May 15, 2008, 09:49:46 pm »

Consider:

$\dfrac{d}{dx} (\sin^{-1} x^2})$

Let u = x^2

$\dfrac{du}{dx} = 2x\mbox{, } \dfrac{d}{du} (\sin^{-1} u) = \dfrac{1}{\sqrt{1 - u^2}} = \dfrac{1}{\sqrt{1 - x^4}}$

By the chain rule:

$\dfrac{d}{dx} (\sin^{-1} x^2}) = 2x \cdot \dfrac{1}{\sqrt{1 - x^4}} = \dfrac{2x}{\sqrt{1 - x^4}}$

:. $\int \dfrac{2x}{\sqrt{1 - x^4}}\mbox{ dx} = \int \dfrac{d}{dx} (\sin^{-1} x^2})\mbox{ dx} = \sin^{-1} x^2 + C$

Piney.z · « **Reply #12 on:** May 15, 2008, 10:07:40 pm »

.. ..

AHH!! i see where i went wrong now. tyty

Piney.z · « **Reply #13 on:** May 15, 2008, 10:30:48 pm »

what about this one?

$cos^{-1} \sqrt{x}$ ... find $\int \dfrac{-1}{\sqrt{x (1 - x)}} dx$

wouldn't differentiating $cos^{-1} \sqrt{x}$ = $\int \dfrac{-1}{2 \sqrt{x(1- x)}}dx$ ?
or did i go somewhere wrong again... $:-\$

Mao · « **Reply #14 on:** May 15, 2008, 10:39:20 pm »

Search the forums now!

We have moved!

Author Topic: Differential Calculuss =/ (Read 3143 times) Tweet Share

Piney.z

Differential Calculuss =/

Mao

Re: Differential Calculuss =/

ed_saifa

Re: Differential Calculuss =/

Piney.z

Re: Differential Calculuss =/

Piney.z

Re: Differential Calculuss =/

ed_saifa

Re: Differential Calculuss =/

Piney.z

Re: Differential Calculuss =/

Piney.z

Re: Differential Calculuss =/

ed_saifa

Re: Differential Calculuss =/

mark_alec

Re: Differential Calculuss =/

Piney.z

Re: Differential Calculuss =/

dcc

Re: Differential Calculuss =/

Piney.z

Re: Differential Calculuss =/

Piney.z

Re: Differential Calculuss =/

Mao

Re: Differential Calculuss =/

Recent Posts

User:
Password: