Author Topic: 4U Maths Question Thread (Read 946115 times) Tweet Share

RuiAce · « **Reply #930 on:** March 03, 2017, 06:30:35 pm »

Quote from: Wales on March 03, 2017, 04:39:10 pm

I've just started the Graphs topic and this question seems completely foreign to me. Could anyone please guide me through it? I've had a look at the solutions but nothing makes too much sense.

Cheers, Wales

$\text{In this case, }f(x)\text{ is quite similar to }x^2-1\\ \text{and probably the exact same. So we can derive the actual equations of the rest.}$
__________________________________________
$\text{But we will act as though we don't know this}\\ \text{These are very important skills in the graphing topic and should've been taught.}$
$\text{For example, }y=xf(x)\text{, we are }\textbf{MULTIPLYING}\text{ the }y\text{-coordinates of }y=f(x)\\ \text{to the }y\text{-coordinates of }y=x\\ \text{Clearly, the }x\text{-intercepts of the original graph will be preserved. since }f(x)=0\text{ so }xf(x)=0x=0\\ \text{The }x\text{-intercept on }y=x \text{ (at }x=0)\text{ will be introduced}\\ \text{and of course, if you multiply two negatives you get a positive}\\ \text{If you multiply a positive with a negative, you get a negative}\\ \text{These are some of the things that should run in your head}\\ \text{as soon as you see the need to multiply}$
$\text{Adding is easy. Simply add the }y\text{-coordinates by just looking at the graphs}$
$\text{Note that these questions are NOT meant to be easy}\\ \text{unless you can stare at }y=x. y=x+1, y=x^2\\ \text{for reference sake.}$

hinakamishiro · « **Reply #931 on:** March 04, 2017, 02:47:09 pm »

Hey guys could I please have some help with this conics question? $:-\$

jakesilove · « **Reply #932 on:** March 04, 2017, 05:03:27 pm »

Quote from: hinakamishiro on March 04, 2017, 02:47:09 pm

Hey guys could I please have some help with this conics question? $:-\$

Definitely a bloody tough question. I'm no good at drawing diagrams, so I'm gonna just say some diagram-related stuff and you can draw it out if it doesn't make conceptual sense.

First, let's let the tangent at P meet x=a and x=-a at Q, R respectively. QR can hit the y=-axis at C.

The tangent PR has equation

$\frac{x\sin(\theta)}{a}-\frac{y\cos(\theta)}{b}=1$

So, Q has coordinates

$(a, \frac{b}{\cos(\theta)}(\sin(\theta)-1))$

and R has coordinates

$(-a, -\frac{b}{\cos(\theta)}(\sin(\theta)+1))$

We can easily find the gradient of QS and RS, and we multiply them together to find whether they are at right angles to each other. Using our knowledge of eccentricity,

$m_{QS} \times m_{RS}=\frac{b(\cos(\theta)-1)}{a\sin(\theta)(1-e)} \times \frac{-b(\cos(\theta)+1)}{-a\sin(\theta)(1+e)}$
$=\frac{b^2}{a^2(1-e^2)} \times \frac{\cos^2(\theta)-1}{sin^2(\theta)}$

Now, recalling that

$b^2=a^2(e^2-1)$

by definition, we find that the gradient of QS* the gradient of RS =-1. Thus, they are at right angles to each other. We only need to replace e with -e to prove the same relation for QS' and RS' (Potentially a good idea to write this out in full in an exam, but that's up to you).

As QR subtends a right angle at S and S', QSRS' are concyclic. By circle geometry rules, QR must be the diameter of the circle. As the y'axis must bisect the chord SS', the centre of the circle is the point C.

Now, we use our point.

$P(1,\frac{2}{3}\sqrt{2})$

on the hyperbola stated, then QR has equation

$\frac{x}{9}-\frac{2\sqrt{2}y}{3}=1$

Clearly, the point C (where the line meets the y axis) will be

$C(0, -\frac{3}{2\sqrt{2}})$

As

$b^2=a^2(e^2-1)$
$e^2=\frac{10}{9}$

So we can find coordinates S

$S(\sqrt{10}, 0)$
$CS^2=\frac{89}{8}$

So, the circle will have equation

$x^2+(y+\frac{3}{2\sqrt{2}})^2=\frac{89}{8}$

Subject to error here or there (I hate conics), but the method is correct!

cutiepie30 · « **Reply #933 on:** March 04, 2017, 08:26:19 pm »

how would you do this question?

Evaluate the following using integration by parts

RuiAce · « **Reply #934 on:** March 04, 2017, 09:10:45 pm »

Quote from: cutiepie30 on March 04, 2017, 08:26:19 pm

how would you do this question?

Evaluate the following using integration by parts

$\text{I have no idea why you would use integration by parts here}\\ \text{But you would split }\sin^2\theta\text{ into }\sin\theta\sin\theta$
$\begin{align*}\text{Let }I&=\int \sin^2\theta \, d\theta \tag{1}\\ \implies I&=\int \sin \theta\sin \theta\,d\theta\\ &= -\cos\theta\sin\theta-\int -\cos\theta\cos\theta\,d\theta\\ \implies I &= -\cos\theta\sin\theta+\int \cos^2\theta \, d\theta \tag{2}\end{align*}$
$\text{Adding (1) and (2), and combining the integrand:}\\ \begin{align*}2I&= -\cos \theta\sin\theta + \int (\sin^2\theta+\cos^2\theta)d\theta\\ &= -\cos\theta\sin\theta + \int d\theta \tag{Pythag identity}\\ &= -\cos\theta\sin\theta + \theta + 2C\\ \implies I&= \frac12(\theta - \cos\theta\sin\theta)+C\end{align*}$

Kle123 · « **Reply #935 on:** March 05, 2017, 07:51:57 am »

Rui, could you help me out with part ii). Thank you!

cutiepie30 · « **Reply #936 on:** March 05, 2017, 11:19:28 am »

Quote from: RuiAce on March 04, 2017, 09:10:45 pm

$\text{I have no idea why you would use integration by parts here}\\ \text{But you would split }\sin^2\theta\text{ into }\sin\theta\sin\theta$
$\begin{align*}\text{Let }I&=\int \sin^2\theta \, d\theta \tag{1}\\ \implies I&=\int \sin \theta\sin \theta\,d\theta\\ &= -\cos\theta\sin\theta-\int -\cos\theta\cos\theta\,d\theta\\ \implies I &= -\cos\theta\sin\theta+\int \cos^2\theta \, d\theta \tag{2}\end{align*}$
$\text{Adding (1) and (2), and combining the integrand:}\\ \begin{align*}2I&= -\cos \theta\sin\theta + \int (\sin^2\theta+\cos^2\theta)d\theta\\ &= -\cos\theta\sin\theta + \int d\theta \tag{Pythag identity}\\ &= -\cos\theta\sin\theta + \theta + 2C\\ \implies I&= \frac12(\theta - \cos\theta\sin\theta)+C\end{align*}$

Thanks so much RuiAce

!!

cutiepie30 · « **Reply #937 on:** March 05, 2017, 11:20:45 am »

how would you do this question?

Evaluate the following using integration by parts

jakesilove · « **Reply #938 on:** March 05, 2017, 11:34:57 am »

Quote from: cutiepie30 on March 05, 2017, 11:20:45 am

how would you do this question?

Evaluate the following using integration by parts

$\int \theta \sin^2 (\theta) d\theta$

$u=\theta$
$du=d\theta$

$dv=\sin^2 (\theta) d\theta = \frac{1}{2}(1-\cos(2\theta))d\theta$
$v=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

Using integration by parts

$\frac{\theta}{2}(\theta-\sin(\theta)\cos(\theta)) - \int \frac{1}{2}(\theta - \sin(\theta)\cos(\theta)) d\theta$
$\frac{\theta}{2}(\theta-\sin(\theta)\cos(\theta))-\frac{1}{2}(\frac{\theta^2}{2}-\frac{\sin^2(\theta)}{2})+C$

Then, you can use a bunch of trig identities to simply down to

$\frac{1}{8}(2\theta(\theta-\sin(2\theta))-\cos(2\theta))$

according to Wolfram Alpha. Don't necessarily think you need to do this in an exam situations. Also, I'm sure there are plenty of other ways to solve this question, some of which would have yielded the simplified answer above straight away.

cutiepie30 · « **Reply #939 on:** March 05, 2017, 12:21:47 pm »

Quote from: jakesilove on March 05, 2017, 11:34:57 am

$\int \theta \sin^2 (\theta) d\theta$

$u=\theta$
$du=d\theta$

$dv=\sin^2 (\theta) d\theta = \frac{1}{2}(1-\cos(2\theta))d\theta$
$v=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

Using integration by parts

$\frac{\theta}{2}(\theta-\sin(\theta)\cos(\theta)) - \int \frac{1}{2}(\theta - \sin(\theta)\cos(\theta)) d\theta$
$\frac{\theta}{2}(\theta-\sin(\theta)\cos(\theta))-\frac{1}{2}(\frac{\theta^2}{2}-\frac{\sin^2(\theta)}{2})+C$

Then, you can use a bunch of trig identities to simply down to

$\frac{1}{8}(2\theta(\theta-\sin(2\theta))-\cos(2\theta))$

according to Wolfram Alpha. Don't necessarily think you need to do this in an exam situations. Also, I'm sure there are plenty of other ways to solve this question, some of which would have yielded the simplified answer above straight away.

THanks jakesilove

but i dont get how you went from $dv=\sin^2 (\theta) d\theta = \frac{1}{2}(1-\cos(2\theta))d\theta$
$v=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

how did you get $v=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

Also how did you get the $- \int \frac{1}{2}( - \sin(\theta)\cos(\theta)) d\theta$ =
$-\frac{1}{2}(-\frac{\sin^2(\theta)}{2})+C$ in one go ?

jakesilove · « **Reply #940 on:** March 05, 2017, 12:40:18 pm »

Quote from: cutiepie30 on March 05, 2017, 12:21:47 pm

THanks jakesilove

but i dont get how you went from $dv=\sin^2 (\theta) d\theta = \frac{1}{2}(1-\cos(2\theta))d\theta$
$v=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

how did you get $v=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

Also how did you get the $- \int \frac{1}{2}( - \sin(\theta)\cos(\theta)) d\theta$ =
$-\frac{1}{2}(-\frac{\sin^2(\theta)}{2})+C$ in one go ?

Fair enough, I definitely didn't explain myself very well. We want to integrate dv, so

$\int \frac{1}{2}(1-\cos(2\theta))d\theta$

This integrates in the normal way

$\frac{1}{2}(\theta-\frac{\sin(2\theta)}{2})$

However, we can use trigonometric identities to get

$\frac{1}{2}(\theta-\frac{2\sin(\theta)\cos(\theta)}{2}=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

as required.

For the last one, I just used the reverse chain rule. Recall that

$\int f(x)f'(x)dx=\frac{1}{2}f(x)^2$

Using this, and noting that

$\frac{d}{dx}\sin(\theta)=\cos(\theta)$

we can recognise that

$\int \sin(\theta)\cos(\theta) d\theta=\frac{1}{2}\sin^2(\theta)$

And that's where I got it! If you can't do that in one go, that's no problem at all (I've had a lot of practice). Take your time, set out your working however you want. However, over time, integrations like this will become natural

cutiepie30 · « **Reply #941 on:** March 05, 2017, 01:05:15 pm »

Quote from: jakesilove on March 05, 2017, 12:40:18 pm

Fair enough, I definitely didn't explain myself very well. We want to integrate dv, so

$\int \frac{1}{2}(1-\cos(2\theta))d\theta$

This integrates in the normal way

$\frac{1}{2}(\theta-\frac{\sin(2\theta)}{2})$

Thanks so much for

However, we can use trigonometric identities to get

$\frac{1}{2}(\theta-\frac{2\sin(\theta)\cos(\theta)}{2}=\frac{1}{2}(\theta-\sin(\theta)\cos(\theta))$

as required.

For the last one, I just used the reverse chain rule. Recall that

$\int f(x)f'(x)dx=\frac{1}{2}f(x)^2$

Using this, and noting that

$\frac{d}{dx}\sin(\theta)=\cos(\theta)$

we can recognise that

$\int \sin(\theta)\cos(\theta) d\theta=\frac{1}{2}\sin^2(\theta)$

And that's where I got it! If you can't do that in one go, that's no problem at all (I've had a lot of practice). Take your time, set out your working however you want. However, over time, integrations like this will become natural

Thanks so much for clarifying jakesilove

Makes sense now !!

cutiepie30 · « **Reply #942 on:** March 05, 2017, 04:43:24 pm »

How would you do this question?
Use a substitution and an integration by parts to evaluate each of the following

jakesilove · « **Reply #943 on:** March 05, 2017, 05:32:27 pm »

Quote from: cutiepie30 on March 05, 2017, 04:43:24 pm

How would you do this question?
Use a substitution and an integration by parts to evaluate each of the following

I would use the substitution of theta=5x+2, then apply that throughout the integral, expand everything out and do integration by parts where necessary. Show us your working out, and if you're still stuck we can talk you through it!

RuiAce · « **Reply #944 on:** March 05, 2017, 07:58:40 pm »

Quote from: Kle123 on March 05, 2017, 07:51:57 am

Rui, could you help me out with part ii). Thank you!

Fair warning: My activity is reduced for this month. I won't be able to get to everything and when I do it might take a while to get there.

Supposedly it gets bizarre

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