Author Topic: VCE Specialist 3/4 Question Thread! (Read 2551380 times) Tweet Share

Bestie · « **Reply #3165 on:** May 31, 2014, 07:00:53 pm »

oh... thank you but the ans says f(x) = pi/2?

Phy124 · « **Reply #3166 on:** May 31, 2014, 07:04:48 pm »

Yes $f(x) = \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}, x>0$

I'm not sure whether the latter part is assumed knowledge in specialist if not maybe just do a google search for something along the lines of "Prove arctan(x) + arccot(x) = pi/2" or wait for someone else to reply with more rigorous answers

Bestie · « **Reply #3167 on:** May 31, 2014, 07:09:13 pm »

yup can you please explain how you got:
tan^-1(x) + cot^-1(x) = pi/2 ?

Phy124 · « **Reply #3168 on:** May 31, 2014, 07:23:43 pm »

Screw it I shall do it an easier way...

Note that the functions in ii and iii are constant for all $x$ as

$\frac{d}{dx}\left(\(\tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right)\right) = \frac{1}{x^2+1} - \frac{1}{x^2+1} = 0$

So you can substitute any value into f(x) in the given domain and this will yield your answer.

e.g. for ii $x>0$ so sub in $x = 1$ for simplicity $\Rightarrow f(1) = \tan^{-1}(1) + \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$ and then you can do a similar thing for the next part.

My apologies for any confusion I may have caused

Alternatively if you form a right angled triangle on a erm "non" unit circle (

)with non-hypotenuse lengths of 1 and x you can show that $\tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \text{when x} > \text{0 and =} -\frac{\pi}{2} \text{when x} < \text{0 due to the quadrant}$

Bestie · « **Reply #3169 on:** May 31, 2014, 09:57:10 pm »

thank you.. it's ok... it wasn't very confusing i got it...

can you please help me with this question as well?

1/[(2x + 1)^2 + 9] - whats the antiderivative of that?

thank you

Phy124 · « **Reply #3170 on:** May 31, 2014, 10:13:12 pm »

Bestie · « **Reply #3171 on:** June 01, 2014, 11:46:21 am »

yes thank you so much.

sorry but what about this one question 27a)

Stevensmay · « **Reply #3172 on:** June 01, 2014, 12:08:18 pm »

Quote from: Bestie on June 01, 2014, 11:46:21 am

yes thank you so much.

sorry but what about this one question 27a)

I'm not very confident of this answer.

$U_n = \int_0^\frac{\pi}{4} \tan^n(x) \,dx$ and $U_{n-2} = \int_0^\frac{\pi}{4} \tan^{n-2}(x) \,dx$

Adding these together gives us $U_n + U_{n-2} = \int_0^\frac{\pi}{4} \tan^n(x) + \tan^{n-2}(x) \,dx$

Probably wrong, someone please correct me.

kinslayer · « **Reply #3173 on:** June 01, 2014, 04:50:59 pm »

You are on the right track, but you need to go further:

$\begin{alignedat}1 U_n + U_{n-2} & = \int_0^\frac{\pi}{4}\left( \tan^n(x) + \tan^{n-2}(x)\right)\mbox{d}x \\ &= \int_0^\frac{\pi}{4} \tan^n{(x)}\left( 1 + \cot^2(x)\right ) \, \mbox{d}x \\ &= \int_0^\frac{\pi}{4} \tan^n{(x)}\csc^2(x) \, \mbox{d}x \end{alignedat}$

$\mbox{Now let } u = \tan(x) \Rightarrow \frac{\mbox{d}u}{\mbox{d}x} = \sec^2(x) \Rightarrow \mbox{d}x = \cos^2(x)\, \mbox{d}u$

$\begin{alignedat}1 \mbox{So } U_n + U_{n-2} & = \int_0^1 u^{n-2} \,\mbox{d} u \\ &= \left[ \frac{u^{n-1}}{n-1}\right]_0^1 \\ &= \frac{1}{n-1} \end{alignedat}$

Finally,

$\begin{alignedat}1 U_0 &= \int_0^\frac{\pi}{4}\mbox{d}x = \frac{\pi}{4} \\ U_0 + U_2 &= 1 \\ U_2 + U_4 &= \frac{1}{3} \\ U_4 + U_6 &= \frac{1}{5} \\ \Rightarrow U_6 &= \frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4} \\ &= \frac{13}{15} - \frac{\pi}{4} \end{alignedat}$

lzxnl · « **Reply #3174 on:** June 01, 2014, 06:10:41 pm »

Quote from: kinslayer on June 01, 2014, 04:50:59 pm

You are on the right track, but you need to go further:

$\begin{alignedat}1 U_n + U_{n-2} & = \int_0^\frac{\pi}{4}\left( \tan^n(x) + \tan^{n-2}(x)\right)\mbox{d}x \\ &= \int_0^\frac{\pi}{4} \tan^n{(x)}\left( 1 + \cot^2(x)\right ) \, \mbox{d}x \\ &= \int_0^\frac{\pi}{4} \tan^n{(x)}\csc^2(x) \, \mbox{d}x \end{alignedat}$

$\mbox{Now let } u = \tan(x) \Rightarrow \frac{\mbox{d}u}{\mbox{d}x} = \sec^2(x) \Rightarrow \mbox{d}x = \cos^2(x)\, \mbox{d}u$

$\begin{alignedat}1 \mbox{So } U_n + U_{n-2} & = \int_0^1 u^{n-2} \,\mbox{d} u \\ &= \left[ \frac{u^{n-1}}{n-1}\right]_0^1 \\ &= \frac{1}{n-1} \end{alignedat}$

Finally,

$\begin{alignedat}1 U_0 &= \int_0^\frac{\pi}{4}\mbox{d}x = \frac{\pi}{4} \\ U_0 + U_2 &= 1 \\ U_2 + U_4 &= \frac{1}{3} \\ U_4 + U_6 &= \frac{1}{5} \\ \Rightarrow U_6 &= \frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4} \\ &= \frac{13}{15} - \frac{\pi}{4} \end{alignedat}$

Erm...you had a cosec squared there, how does your substitution work? It has dx = cosine squared * du, not cosec

Oh wait you just typoed. You should have tan^(n-2) x (1 + tan^2 x)
Instead of using cot
Funnily enough your answer is correct. You've made two mistakes in your working and fortuitously reached the right answer xD
(second one being going from tan^n x * csc^2 x to u^(n-2); where the n-2 come from!?)

hyunah · « **Reply #3175 on:** June 01, 2014, 06:26:20 pm »

just a quick question what is the derivative of cos^-1(-x/3) is it just -1/(9-x^2)^(1/2)

thanks

b^3 · « **Reply #3176 on:** June 01, 2014, 06:30:59 pm »

Quote from: lzxnl on June 01, 2014, 06:10:41 pm

Erm...you had a cosec squared there, how does your substitution work? It has dx = cosine squared * du, not cosec

Oh wait you just typoed. You should have tan^(n-2) x (1 + tan^2 x)
Instead of using cot
Funnily enough your answer is correct. You've made two mistakes in your working and fortuitously reached the right answer xD
(second one being going from tan^n x * csc^2 x to u^(n-2); where the n-2 come from!?)

It looks ok to me?
You don't have to take $\tan^{n-2}$ out, he's taken $\tan^{n}$ out and worked with that.

As for his substitution, it checks out, I think you've flipped the $u$ and $x$ .
$\begin{alignedat}{1}u & =\tan\left(x\right) \\ \frac{du}{dx} & =\sec^{2}\left(x\right) \\ du & =\sec^{2}\left(x\right)dx \\ du & =\frac{1}{\cos^{2}\left(x\right)}dx \\ dx & =\cos^{2}\left(x\right)du \end{alignedat}$
Then this feeds back into to give another $\cot^{2}$ , combining the $\cos^2{x}$ term with the $\csc^2{x}$ term.
The $n-2$ comes from getting the $\cot^{2}(x)$ from the substitution.
$\begin{alignedat}{1}\int\tan^{n}\left(x\right)\csc^{2}\left(x\right)dx & =\int\tan^{n}\left(x\right)\frac{1}{\sin^{2}\left(x\right)}\cos^{2}\left(x\right)du\:\left(\text{technically shouldn't show these lines but will for explanation}\right) \\ & =\int\tan^{n}\left(x\right)\cot^{2}\left(x\right)du \\ & =\int\tan^{n}\left(x\right)\frac{1}{\tan^{2}\left(x\right)}du \\ & =\int\tan^{n-2}\left(x\right)du \\ & =\int u^{n-2}du \end{alignedat}$

kinslayer · « **Reply #3177 on:** June 01, 2014, 06:31:07 pm »

Quote from: lzxnl on June 01, 2014, 06:10:41 pm

Erm...you had a cosec squared there, how does your substitution work? It has dx = cosine squared * du, not cosec

Oh wait you just typoed. You should have tan^(n-2) x (1 + tan^2 x)
Instead of using cot
Funnily enough your answer is correct. You've made two mistakes in your working and fortuitously reached the right answer xD
(second one being going from tan^n x * csc^2 x to u^(n-2); where the n-2 come from!?)

I don't think I got anything wrong, I just left out some working since I got lazy typing it up. For the first line,

$\tan^n( x) + \tan^{n-2}(x) = \tan^n(x) ( 1 + \tan^{-2}(x)) = \tan^n(x) (1 + \cot^2(x))$

and for the substitution you get $\mbox{d}x = \frac{\mbox{d}u}{\sec^2(x)} = \cos^2(x)\, \mbox{d}u$

and the integrand becomes $u^n \frac{\cos^2(x)}{\sin^2(x)} = u^n \cot^2 (x) = u^{n-2}$

lzxnl · « **Reply #3178 on:** June 01, 2014, 11:01:12 pm »

OK. Perhaps I misread the dx and dus

Even so, I stand by my original opinion that taking out a tan^(n-2) x is simpler as then your bracket actually has 1 + tan^2 x, but whatever. Your method works and that's what matters

Bestie · « **Reply #3179 on:** June 02, 2014, 10:51:59 pm »

yup thank you everyone

i really appreciate you helping this little lost lamb

but i have yet another question....

what do you guys get when you guys differentiate sin^-1(a/x)?, where a>0 and is a contant.
in the end i got -a/[squareroot(x^2-a^2)] but i have a feelng im missing something and when i do it on a cas it gives me mod x in the denominator too???

thanks again!

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