Author Topic: pi's Specialist Maths Questions (Read 23684 times) Tweet Share

pi · « **Reply #60 on:** July 05, 2011, 05:19:49 pm »

Thanks brightsky!

My crazy teacher will probably let me use them, he hates 'Australian standard' maths anyway

mmonn1906 · « **Reply #61 on:** July 05, 2011, 05:55:49 pm »

I am just scared I won't get a high score..

Andiio · « **Reply #62 on:** July 05, 2011, 06:23:25 pm »

Quote from: Markkiieee on July 05, 2011, 05:55:49 pm

I am just scared I won't get a high score..

Instead of thinking about the technicalities, focus on getting your basic foundations strong and sturdy. Then year 11/12 will be an absolute breeze.

pi · « **Reply #63 on:** July 15, 2011, 10:11:03 pm »

Just a couple of questions:

1.
This is from MathsQuest (

), p. 264
Evaluate: $\int{\sec^2(x)\cos^n(x)}dx$
I got upto $\int{\cos^{n-2}(x)}dx$ , which is only one step from the question, and am now stuck

2. (two parts)
(i)
This is more of a confusion. We have to find an antiderivative using $\int{\frac{-1}{\sqrt{a^2-x^2}}}dx$ .
I know, that by the formula sheet, that one antiderivative would be $\cos^{-1}(\frac{x}{a})$

But if we do $\int{\frac{-1}{\sqrt{a^2-x^2}}}dx = -\int{\frac{1}{\sqrt{a^2-x^2}}}dx = -\sin^{-1}(\frac{x}{a})$ . Isn't this also true?

Bit confused at what to write now if this pops up in the exam...

(ii)
AND, by extension, doesn't $\frac{d}{dx}(-\sin^{-1}(\frac{x}{a})) = \frac{-1}{\sqrt{a^2-x^2}}$ and $\frac{d}{dx}(\cos^{-1}(\frac{x}{a})) = \frac{-1}{\sqrt{a^2-x^2}}$ mean that $-\sin^{-1}(\frac{x}{a}) = \cos^{-1}(\frac{x}{a})$ ?

^I know that is completely NOT TRUE, as $\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}$ is an established identity

Please VN, what have a done wrong? Because I am definitely wrong here

Thanks!

EDIT: fixed up 2 (ii) question, sooo stupid

HenryP · « **Reply #64 on:** July 15, 2011, 10:22:48 pm »

For question 1, its actually a typo. I spent half an hour trying to figure it out as well and realised it was a typo after reading the worked solutions. Cos is meant to be Tan. Then its simple enough to anti diff.
This is why I hate mathquest. Hope this helped.

moekamo · « **Reply #65 on:** July 15, 2011, 10:24:36 pm »

2.
i) you can do either, since arcsin(x) + arccos(x) = pi/2, you will just get a different + C depending on which one you choose

ii) your first line is wrong, $\frac{d}{dx} -\arcsin (x/a) \ne -\int \frac{1}{\sqrt{x^2-a^2}} dx$ , should be $\frac{d}{dx} -\arcsin (x/a) = -\frac{1}{\sqrt{x^2-a^2}}$ , then $\frac{d}{dx} \arccos (x/a) = -\frac{1}{\sqrt{x^2-a^2}}$

Now you get $\frac{d}{dx}( -\arcsin (x/a)) = \frac{d}{dx} \arccos (x/a) \implies -\arcsin(x/a) = \arccos(x)+C$ then you will get that identity you said

pi · « **Reply #66 on:** July 15, 2011, 10:32:08 pm »

Quote from: HenryP on July 15, 2011, 10:22:48 pm

For question 1, its actually a typo. I spent half an hour trying to figure it out as well and realised it was a typo after reading the worked solutions. Cos is meant to be Tan. Then its simple enough to anti diff.
This is why I hate mathquest. Hope this helped.

Thanks! I was like WTF!?!?! Combed the internet for a while, couldn't find any way to integrate that.

That makes it much easier, thanks!

Quote from: moekamo on July 15, 2011, 10:24:36 pm

2.
i) you can do either, since arcsin(x) + arccos(x) = pi/2, you will just get a different + C depending on which one you choose

OK, that's handy

Quote from: moekamo on July 15, 2011, 10:24:36 pm

2.
ii) your first line is wrong, $\frac{d}{dx} -\arcsin (x/a) \ne -\int \frac{1}{\sqrt{x^2-a^2}} dx$ , should be $\frac{d}{dx} -\arcsin (x/a) = -\frac{1}{\sqrt{x^2-a^2}}$ , then $\frac{d}{dx} \arccos (x/a) = -\frac{1}{\sqrt{x^2-a^2}}$

Now you get $\frac{d}{dx}( -\arcsin (x/a)) = \frac{d}{dx} \arccos (x/a) \implies -\arcsin(x/a) = \arccos(x)+C$ then you will get that identity you said

Oh, that was stupid.
What I meant was $\frac{d}{dx}(-\sin^{-1}(\frac{x}{a})) = \frac{-1}{\sqrt{a^2-x^2}}$ and $\frac{d}{dx}(\cos^{-1}(\frac{x}{a})) = \frac{-1}{\sqrt{a^2-x^2}}dx$
So the 'relationship' I found would probably only work for one certain value of x (if even that). So its imperative that I include '+c's here to get to the identity.

Just as a clarification, in your last line, did you just put all the 'constants' to one side? As there would be one from the inverse sine one too?

luken93 · « **Reply #67 on:** July 15, 2011, 10:33:09 pm »

1) You may wanna check the solutions manual, they refer to sec^2(x)tan^n(x) rather than cos^n(x)...

2) And yeah, I've also wondered this, will be interested in the solution...

EDIT: Beaten.

moekamo · « **Reply #68 on:** July 15, 2011, 10:40:18 pm »

no, the relationship you got, -arcsin(x)=arccos(x) (or arcsin(x) + arccos(x)=0) never works, since arcsin(x) + arccos(x) = pi/2 works for any x between -1 and 1 (i.e. any x in the domain of each of these functions), so yes, the constant, + C must be included

and yea i chucked the constants on one side since im a bad-ass

tony3272 · « **Reply #69 on:** July 15, 2011, 10:41:03 pm »

Quote from: Rohitpi on July 15, 2011, 10:11:03 pm

2. (two parts)
(i)
This is more of a confusion. We have to find an antiderivative using $\int{\frac{-1}{\sqrt{a^2-x^2}}}dx$ .
I know, that by the formula sheet, that one antiderivative would be $\cos^{-1}(\frac{x}{a})$

But if we do $\int{\frac{-1}{\sqrt{a^2-x^2}}}dx = -\int{\frac{1}{\sqrt{a^2-x^2}}}dx = -\sin^{-1}(\frac{x}{a})$ . Isn't this also true?

Bit confused at what to write now if this pops up in the exam...

When i asked my teacher about this he said that by convention, if the negative is inside the integral you do cosine, and if it's outside you do sine. I doubt VCAA would take marks off for that tho.

pi · « **Reply #70 on:** July 15, 2011, 10:41:59 pm »

Quote from: moekamo on July 15, 2011, 10:40:18 pm

no, the relationship you got, -arcsin(x)=arccos(x) (or arcsin(x) + arccos(x)=0) never works, since arcsin(x) + arccos(x) = pi/2 works for any x between -1 and 1 (i.e. any x in the domain of each of these functions), so yes, the constant, + C must be included

and yea i chucked the constants on one side since im a bad-ass

Ok, I get it now! Thanks a lot!

EDIT: Thanks for the further clarification tony on 2.(i) and luke on 1.

luken93 · « **Reply #71 on:** July 15, 2011, 10:47:14 pm »

Quote from: Rohitpi on July 15, 2011, 10:41:59 pm

Quote from: moekamo on July 15, 2011, 10:40:18 pm
no, the relationship you got, -arcsin(x)=arccos(x) (or arcsin(x) + arccos(x)=0) never works, since arcsin(x) + arccos(x) = pi/2 works for any x between -1 and 1 (i.e. any x in the domain of each of these functions), so yes, the constant, + C must be included

and yea i chucked the constants on one side since im a bad-ass

Ok, I get it now! Thanks a lot!

EDIT: Thanks for the further clarification tony on 2.(i) and luke on 1.

hahaha, big help I am!

pi · « **Reply #72 on:** July 18, 2011, 08:41:37 pm »

OK, more questions/queries!

1. More of a query
If we have $\int{\frac{3x+10}{x^2 + 2x}}dx$ , and the answer is $5\ln{|x|} - 2\ln{|x+2|} + c$ , do we need to include the line: x is element of R\{0, -2} in the answer? The textbook seems to, and I wonder if its necessary.

2. More of a query
When do we have to have the modulus in the log when antidifferentiating functions? MQ seems to randomly remove it from some answers (eg. $\int{\frac{x-1}{x+5}}dx = x - 6\ln{(x+5)} + c$ ). Bit confused here...

3. Help/query
Is there a proof that is fairly easy to understand for the Gaussian integral $\int^{\infty}_{-\infty} e^{-x^2}dx = \sqrt{\pi}$ ? Even a link would be very helpful and much appreciated.

Thanks

(btw, to those who though MQ was easy, try Ex 5E, q4. part e, its a decent question

)

TrueTears · « **Reply #73 on:** July 18, 2011, 08:46:15 pm »

3. You won't understand it yet, but the integral is easily computed by converting to polar coordinates and using a double integral. If you want to see the working out: http://en.wikipedia.org/wiki/Gaussian_integral#Careful_proof

i think this thread would be funner with the presence of some nt, combs ; )

pi · « **Reply #74 on:** July 18, 2011, 09:00:18 pm »

Quote from: TrueTears on July 18, 2011, 08:46:15 pm

3. You won't understand it yet, but the integral is easily computed by converting to polar coordinates and using a double integral. If you want to see the working out: http://en.wikipedia.org/wiki/Gaussian_integral#Careful_proof

Hmmm, thought so. Maybe I'll have to pursue this later in life.

Quote from: TrueTears on July 18, 2011, 08:46:15 pm

i think this thread would be funner with the presence of some nt, combs ; )

Noooooooooooooooooooooooo! Bad yr11 experience (damn poker!)

Search the forums now!

We have moved!

Author Topic: pi's Specialist Maths Questions (Read 23684 times) Tweet Share

pi

Re: pi's Specialist Maths Questions

mmonn1906

Re: pi's Specialist Maths Questions

Andiio

Re: pi's Specialist Maths Questions

pi

Re: pi's Specialist Maths Questions

HenryP

Re: pi's Specialist Maths Questions

moekamo

Re: pi's Specialist Maths Questions

pi

Re: pi's Specialist Maths Questions

luken93

Re: pi's Specialist Maths Questions

moekamo

Re: pi's Specialist Maths Questions

tony3272

Re: pi's Specialist Maths Questions

pi

Re: pi's Specialist Maths Questions

luken93

Re: pi's Specialist Maths Questions

pi

Re: pi's Specialist Maths Questions

TrueTears

Re: pi's Specialist Maths Questions

pi

Re: pi's Specialist Maths Questions

Recent Posts

User:
Password: