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TrueTears · « **Reply #330 on:** May 01, 2009, 11:22:52 pm »

Thanks kamil and Mao, I understand it now.

kamil9876 · « **Reply #331 on:** May 01, 2009, 11:26:24 pm »

Also, general rule: You cannot change the domain of a function by algebraically manipulating it:

consider y=x with domain R.
$ y=\frac{x^2}{x}$ has domain R/{0}

by writing $x \neq 0$ we havnt changed the domain of the original function, just said that our new form is only true for $x \neq 0$ and hence our new form is not the complete picture:

This is analogous to ur case. $y \neq x$ does not mean that u magically changed the domain of the function but that ur new form only allow values for which the condition $y \neq x$ holds.

TrueTears · « **Reply #332 on:** May 01, 2009, 11:28:02 pm »

Quote from: kamil9876 on May 01, 2009, 11:26:24 pm

Also, general rule: You cannot change the domain of a function by algebraically manipulating it:

consider y=x with domain R.
$ y=\frac{x^2}{x}$ has domain R/{0}

by writing $x \neq 0$ we havnt changed the domain of the original function, just said that our new form is only true for $x \neq 0$ and hence our new form is not the complete picture:

This is analogous to ur case. $y \neq x$ does not mean that u magically changed the domain of the function but that ur new form only allow values for which the condition $y \neq x$ holds.

True, true. very true.

Thanks for explanation again! :coolsmiley:

TrueTears · « **Reply #333 on:** May 03, 2009, 11:35:22 pm »

And also this question:

The equation of a curve C is $x^3+xy+2y^3 = k$ , where $k$ is a constant

a) find $\frac{dy}{dx}$ $\implies$ $\frac{dy}{dx} = \frac{-3x^2-y}{x+6y^2}$

C does have a tangent parallel to the y axis

b) show that the y coordinate at the point of contact satisfies $216y^6+4y^3+k = 0$

c) Hence show that $k \le \frac{1}{54}$

stomped on part c lol

Mao · « **Reply #334 on:** May 03, 2009, 11:44:21 pm »

the equation in b is a quadratic, it has at least one solution (as per part a), hence the discriminant must be greater or equal to 0

$\implies \Delta \ge 0$

TrueTears · « **Reply #335 on:** May 03, 2009, 11:49:54 pm »

Thanks Mao and kamil (MSN)

TrueTears · « **Reply #336 on:** May 03, 2009, 11:58:24 pm »

$f : [\pi , \frac{3\pi}{2}] \to R , f(x) = sin(x)$

Find the derivative of $f^{-1}(x)$ with respect to $x$

TrueTears · « **Reply #337 on:** May 04, 2009, 12:23:11 am »

Consider $f(x) = tan^{-1}(x) + tan^{-1}(\frac{1}{x})$ , $x \neq 0$

If x > 0 , find $f(x)$

humph · « **Reply #338 on:** May 04, 2009, 02:14:14 am »

Quote from: TrueTears on May 04, 2009, 12:23:11 am

Consider $f(x) = tan^{-1}(x) + tan^{-1}(\frac{1}{x})$ , $x \neq 0$

If x > 0 , find $f(x)$

I remember this from one of my UMEP tests

So for $x \neq 0$ , we have that
$f'(x) = \frac{1}{x^2 + 1} + \frac{-1}{x^2} \frac{1}{(1/x)^2 + 1} = \frac{1}{x^2 + 1} - \frac{1}{1 + x^2} = 0$ .
Thus $f$ is constant. Taking $x=1$ gives
$f(1) = \arctan(1) + \arctan(1/1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$
and so for all $x \neq 0$ ,
$f(x) = \frac{\pi}{2}$ .

Other interesting inverse trigonometric identities:
$ \begin{align*} \arcsin x + \arccos x & = \pi/2 , \qquad \mathrm{arccsc}\: x & = \arcsin(1/x) \\ \mathrm{arccsc}\: x + \mathrm{arcsec}\: x & = \pi/2 , \qquad \mathrm{arcsec}\: x & = \arccos(1/x) \\ \arctan x + \mathrm{arccot}\: x & = \pi/2 , \qquad \mathrm{arccot}\: x & = \arctan(1/x) \\ \end{align*} $

TrueTears · « **Reply #339 on:** May 04, 2009, 08:31:31 am »

Thanks humph!

kamil9876 · « **Reply #340 on:** May 05, 2009, 01:55:05 pm »

To do that without calculus:

since x>0 and arctan has domain $(-\frac{\pi}{2},\frac{\pi}{2})$ , $0<x<\frac{\pi}{2}$

Let $tan\theta=x$
$ tan(\frac{\pi}{2}-\theta)=\frac{1}{x}$ , this can be verified by constructing a right angle triangle with adjacent and opposite sides of $1$ and $x$ respecticely, and an angle of $\theta$

taking inverse tan of both equations and adding them gives QED.

TrueTears · « **Reply #341 on:** May 05, 2009, 04:08:13 pm »

^^^ thanks kamil!

TrueTears · « **Reply #342 on:** May 15, 2009, 04:23:23 pm »

$\int cot^2(x) dx$

I don't think there is a way to integrate it without using $cot(x) = tan(\frac{\pi}{2} - x)$

What do you think?

Mao · « **Reply #343 on:** May 15, 2009, 10:00:22 pm »

$\int \cot^2 (x)\; dx = \int \csc^2 (x) - 1\; dx = -\cot (x) - x + C$

remember that $\frac{d}{dx} \cot (x) = -\csc^2 (x)$

the relationship between cot/csc in many respects resembles tan/sec

TrueTears · « **Reply #344 on:** May 16, 2009, 01:01:20 pm »

Thank you so much Mao

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