Author Topic: 3U Maths Question Thread (Read 1506342 times) Tweet Share

Never.Give.Up · « **Reply #3375 on:** April 01, 2018, 09:31:17 pm »

Hey again

sorry for all the questions lately

im a little stuck with finding this indefinite integral
3x(x+1)^1/2 dx by using the substitution u=x+1

i have this part of it... 2 ((x+1)^3)^1/2...but i dont know how to get the first part....

btw how do i use LaTex? especially with integrals??

sorry not real good with this sort of stuff....

thanks heaps

clovvy · « **Reply #3376 on:** April 01, 2018, 09:47:26 pm »

Quote from: Dragomistress on February 12, 2018, 07:34:46 pm

I am actually interested in this more clever method.

Also, how do I do this?

By the way what textbook is that from?

Also Rui, the whole working out you showed is very interesting, some of it you do mention goes a bit beyond.... So are there more 'forbidden techniques' that can potentially be used apart from L'Hopital's Rule and Euler's formula (I tend to use it when I get pissed off with complex numbers)

RuiAce · « **Reply #3377 on:** April 01, 2018, 11:50:00 pm »

Quote from: Never.Give.Up on April 01, 2018, 09:31:17 pm

Hey again sorry for all the questions lately

im a little stuck with finding this indefinite integral
3x(x+1)^1/2 dx by using the substitution u=x+1

i have this part of it... 2 ((x+1)^3)^1/2...but i dont know how to get the first part....

btw how do i use LaTex? especially with integrals?? sorry not real good with this sort of stuff....

thanks heaps

$u=x+1 \implies du = dx$
\begin{align*}\int 3x(x+1)^{1/2}\,dx&= \int 3(u-1) u^{1/2}\,du\\ &= \int 3\left( u^{3/2} - u^{1/2} \right)du\\ &= 3\left(\frac{2u^{5/2}}{5} - \frac{2u^{3/2}}{3} \right)+C\\ &= \frac{6}{5}(x+1)^{5/2} - 2 (x+1)^{3/2} +C \end{align*}
LaTeX guide

RuiAce · « **Reply #3378 on:** April 01, 2018, 11:52:37 pm »

Quote from: clovvy on April 01, 2018, 09:47:26 pm

By the way what textbook is that from?

Also Rui, the whole working out you showed is very interesting, some of it you do mention goes a bit beyond.... So are there more 'forbidden techniques' that can potentially be used apart from L'Hopital's Rule and Euler's formula (I tend to use it when I get pissed off with complex numbers)

You're not allowed to use any of them; they're forbidden for a reason. You will lose marks for using any of those in an exam.

As an added tool to help check your working out though, or just to save time (that's what Euler's formula does), those have been the main two I've used. I've also used the Heaviside cover-up method for partial fractions in integration, which is (as far as I know of) an allowed technique. Note that these are 4U concepts and not relevant to students here.
______________________

That working out is perfectly within the bounds of 3U, and one cannot be marked down for it. The problem is that the level of thinking is not required at a 3U level; the intuition is the problem, not the algebra.

Abdul_k · « **Reply #3379 on:** April 02, 2018, 04:14:33 pm »

Hi, I'm just stuck in proving the k+1 for the induction in part ii)
https://drive.google.com/folderview?id=19PGEtnnVwv3CAGN6MPCxuYg4Nv3uTrSZ

RuiAce · « **Reply #3380 on:** April 02, 2018, 04:33:13 pm »

Quote from: Abdul_k on April 02, 2018, 04:14:33 pm

Hi, I'm just stuck in proving the k+1 for the induction in part ii)
https://drive.google.com/folderview?id=19PGEtnnVwv3CAGN6MPCxuYg4Nv3uTrSZ

$\text{Assume that}\\ \boxed{\frac12 \tan \frac{x}{2} + \frac{1}{2^2} \tan \frac{x}{2^2} + \dots + \frac{1}{2^k} \tan \frac{x}{2^k} = \frac{1}{2^k} \cot \frac{x}{2^k} - \cot x}\\ \text{and recall from part i) that}\\ \cot u = \frac12 \cot \frac{u}{2} - \frac12 \tan \frac{u}{2}$

What we seek to prove.

\[ \frac12 \tan \frac{x}{2} + \frac{1}{2^2} \tan \frac{x}{2^2} + \dots + \frac{1}{2^k} \tan \frac{x}{2^k} + \frac{1}{2^k} \tan \frac{x}{2^{k+1}} = \frac{1}{2^{k+1}} \cot \frac{x}{2^{k+1}} - \cot x \]

\begin{align*}&\quad \frac12 \tan \frac{x}{2} + \frac{1}{2^2} \tan \frac{x}{2^2} + \dots + \frac{1}{2^k} \tan \frac{x}{2^k}+ \frac{1}{2^{k+1}} \tan \frac{x}{2^{k+1}}\\ &= \frac{1}{2^k} \cot \frac{x}{2^k} - \cot x + \frac{1}{2^k} \tan \frac{x}{2^k}\tag{inductive assumption}\\ &= \frac{1}{2^k} \left( \frac12\cot \frac{x}{2^{k+1}} - \frac12\tan \frac{x}{2^{k+1}} \right) -\cot x + \frac{1}{2^{k+1}} \tan \frac{x}{2^{k+1}} \tag{from part i)}\\ &= \frac{1}{2^{k+1}}\cot \frac{x}{2^{k+1}} - \frac{1}{2^{k+1}} \tan \frac{x}{2^{k+1}} -\cot x + \frac{1}{2^{k+1}} \tan \frac{x}{2^{k+1}}\\ &= \frac{1}{2^{k+1}}\cot \frac{x}{2^{k+1}}-\cot x\end{align*}
as required.

Calley123 · « **Reply #3381 on:** April 07, 2018, 09:28:41 pm »

Hey,
I know I have to use the auxiliary rule to solve this but I am not getting the correct answer !

Solve for 0< or = x< or =2pi
cosx+ root 3 *sinx=2

Answer: pi/3

Thanks

Hey again,

How do I solve this equation using general solutions?

sinx+ cosx=1 ?

Thanks

Mod Edit: Post merge

jamonwindeyer · « **Reply #3382 on:** April 08, 2018, 12:14:31 am »

Quote from: Calley123 on April 07, 2018, 09:28:41 pm

Hey,
I know I have to use the auxiliary rule to solve this but I am not getting the correct answer !

Solve for 0< or = x< or =2pi
cosx+ root 3 *sinx=2

Answer: pi/3

Thanks
Hey again,

How do I solve this equation using general solutions?

sinx+ cosx=1 ?

Thanks

Mod Edit: Post merge

Hey! We can try and track down the error on that first one, let's perform the reduction first:

$A=\sqrt{3+1}=2\\\cos{x}+\sqrt{3}\sin{x}\equiv2\cos{\left(x-\alpha\right)}\\cos{x}+\sqrt{3}\sin{x}\equiv2\cos{x}\cos{\alpha}+2\sin{x}\sin{\alpha}\\\therefore2\cos{\alpha}=1\\\cos{\alpha}=\frac{1}{2}\\\alpha=\frac{\pi}{3}\\\therefore\cos{x}+\sqrt{3}\sin{x}\equiv2\cos{\left(x-\frac{\pi}{3}\right)}$

Does that match what you've got? If not, there's your issue, if not then we can figure out what went wrong in the trig equation itself!

As for your second question, you'll again start with the auxillary angle method. You'll get:

$\sin{x}+\cos{x}\equiv\sin{\left(x+\frac{\pi}{4}\right)}$

So the equation you end up solving is:

$\sin{\left(x+\frac{\pi}{4}\right)}=1$

Now we have a formula for the general solution for sine on our reference sheet - Only this time it is $x+\frac{\pi}{4}$, not just $x$. So all you do is:

$x+\frac{\pi}{4}=n\pi+\left(-1\right)^n\sin^{-1}{(1)}$

And solve that for $x$!

beeangkah · « **Reply #3383 on:** April 08, 2018, 11:33:06 am »

Could someone please help with part c?

Calley123 · « **Reply #3384 on:** April 08, 2018, 04:49:14 pm »

Quote from: jamonwindeyer on April 08, 2018, 12:14:31 am

Hey! We can try and track down the error on that first one, let's perform the reduction first:

$A=\sqrt{3+1}=2\\\cos{x}+\sqrt{3}\sin{x}\equiv2\cos{\left(x-\alpha\right)}\\cos{x}+\sqrt{3}\sin{x}\equiv2\cos{x}\cos{\alpha}+2\sin{x}\sin{\alpha}\\\therefore2\cos{\alpha}=1\\\cos{\alpha}=\frac{1}{2}\\\alpha=\frac{\pi}{3}\\\therefore\cos{x}+\sqrt{3}\sin{x}\equiv2\cos{\left(x-\frac{\pi}{3}\right)}$

Does that match what you've got? If not, there's your issue, if not then we can figure out what went wrong in the trig equation itself!

As for your second question, you'll again start with the auxillary angle method. You'll get:

$\sin{x}+\cos{x}\equiv\sin{\left(x+\frac{\pi}{4}\right)}$

So the equation you end up solving is:

$\sin{\left(x+\frac{\pi}{4}\right)}=1$

Now we have a formula for the general solution for sine on our reference sheet - Only this time it is $x+\frac{\pi}{4}$, not just $x$. So all you do is:

$x+\frac{\pi}{4}=n\pi+\left(-1\right)^n\sin^{-1}{(1)}$

And solve that for $x$!

The first worked out well but I am still unsure how there is two solutions for the second question....
If the second solution is 2 pi, then a some must have also been zero. Im not sure how I lost that second solution
Thank you !!!

RuiAce · « **Reply #3385 on:** April 08, 2018, 09:24:46 pm »

Quote from: Calley123 on April 08, 2018, 04:49:14 pm

The first worked out well but I am still unsure how there is two solutions for the second question....
If the second solution is 2 pi, then a some must have also been zero. Im not sure how I lost that second solution
Thank you !!!

Unfortunately it looks like Jamon made an error in his auxiliary angle transform.
$\text{He dropped the coefficient in the front.}\\ \text{We actually have }\sin x + \cos x \equiv\sqrt{2} \sin \left(x + \frac\pi4 \right)$
$\text{So what we're solving is }\boxed{\sin \left(x+\frac\pi4 \right) = \frac{1}{\sqrt2}}\\ \text{This gives }x + \frac\pi4 = \pi n + (-1)^n \frac\pi4$
$\text{Now this is where things get tricky.}\\ \text{Observe how this rearranges to}\\ \boxed{x = \pi n + \frac\pi4 \left((-1)^n - 1 \right)}\\ \text{To decompose this, we must break into cases.}$
$\text{When }n \text{ is even, we may let }n = 2k.\\ \text{This gives }x = 2k\pi + \frac\pi4 (1 - 1)\\ \text{noting that even powers of }(-1)\text{ equal to }1.$
Which simplifies to the second one, i.e. $ x = 2k\pi $. This is of course, the second solution exactly found.
$\text{When }n\text{ is odd, we may let }n = 2k+1.\\ \text{This gives }x = (2k+1)\pi + \frac\pi4 (-1-1)\\ \text{noting that odd powers of }(-1)\text{ equal to }-1.$
Which expands out to $ x = 2k\pi + \pi - \frac\pi2$, i.e. $x = 2k\pi +\frac\pi2$. Note that this one doesn't exactly match the first one, but it's an equivalent form of it. As a challenge exercise, you may like to prove why this is true.

(Comparatively speaking, this one is a bit more involved than most other ones.)

RuiAce · « **Reply #3386 on:** April 08, 2018, 09:28:49 pm »

Quote from: beeangkah on April 08, 2018, 11:33:06 am

Could someone please help with part c?

$\text{Follow the usual process.}$
$\text{When }t = 4, T = 160.\text{So therefore,}\\ \begin{align*}28 + 172e^{-4k}&= 160\\ 172e^{-4k} &= 144 \\ e^{-4k} &= \frac{144}{172}\\ -4k &= \ln \frac{144}{172}\\ -k &= \frac14 \ln \frac{36}{43}.\end{align*}$
$\text{Therefore }\boxed{T = 28 + 173 e^{\frac{t}{4} \ln \frac{36}{43}} }\\ \text{Now you need to find when }T = 0.2 \times 200\text{, i.e. when }T = 40$

Calley123 · « **Reply #3387 on:** April 10, 2018, 03:54:38 pm »

Quote from: RuiAce on April 08, 2018, 09:24:46 pm

Unfortunately it looks like Jamon made an error in his auxiliary angle transform.
$\text{He dropped the coefficient in the front.}\\ \text{We actually have }\sin x + \cos x \equiv\sqrt{2} \sin \left(x + \frac\pi4 \right)$
$\text{So what we're solving is }\boxed{\sin \left(x+\frac\pi4 \right) = \frac{1}{\sqrt2}}\\ \text{This gives }x + \frac\pi4 = \pi n + (-1)^n \frac\pi4$
$\text{Now this is where things get tricky.}\\ \text{Observe how this rearranges to}\\ \boxed{x = \pi n + \frac\pi4 \left((-1)^n - 1 \right)}\\ \text{To decompose this, we must break into cases.}$
$\text{When }n \text{ is even, we may let }n = 2k.\\ \text{This gives }x = 2k\pi + \frac\pi4 (1 - 1)\\ \text{noting that even powers of }(-1)\text{ equal to }1.$
Which simplifies to the second one, i.e. $ x = 2k\pi $. This is of course, the second solution exactly found.
$\text{When }n\text{ is odd, we may let }n = 2k+1.\\ \text{This gives }x = (2k+1)\pi + \frac\pi4 (-1-1)\\ \text{noting that odd powers of }(-1)\text{ equal to }-1.$
Which expands out to $ x = 2k\pi + \pi - \frac\pi2$, i.e. $x = 2k\pi +\frac\pi2$. Note that this one doesn't exactly match the first one, but it's an equivalent form of it. As a challenge exercise, you may like to prove why this is true.

(Comparatively speaking, this one is a bit more involved than most other ones.)

Thank you so much!

KT Nyunt · « **Reply #3388 on:** April 12, 2018, 10:12:17 am »

Hi, could someone please show me the working for this? The answer is 110km/h.

Thanks !

beeangkah · « **Reply #3389 on:** April 12, 2018, 03:32:02 pm »

Could someone help with 19&20 please x

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Author Topic: 3U Maths Question Thread (Read 1506342 times) Tweet Share

Never.Give.Up

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RuiAce

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RuiAce

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Abdul_k

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