Author Topic: 4U Maths Question Thread (Read 665360 times) Tweet Share

RuiAce · « **Reply #105 on:** March 25, 2016, 11:55:15 pm »

Quote from: Happy Physics Land on March 25, 2016, 11:54:34 pm

Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ...

Ok go home. You're drunk.

jamonwindeyer · « **Reply #106 on:** March 26, 2016, 02:38:03 pm »

Quote from: Happy Physics Land on March 25, 2016, 11:54:34 pm

Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ...

Wow HPL, these puns are just unLAWful...

Happy Physics Land · « **Reply #107 on:** March 30, 2016, 05:47:23 pm »

Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!!

(I hope rui doesnt hit me for asking this question)

The chord AB of the hyperbola xy=c² subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.

I couldnt really start this one because Im struggle to even draw the diagram.

jakesilove · « **Reply #108 on:** March 30, 2016, 06:46:45 pm »

Quote from: Happy Physics Land on March 30, 2016, 05:47:23 pm

Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!! (I hope rui doesnt hit me for asking this question)

The chord AB of the hyperbola xy=c² subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.

I couldnt really start this one because Im struggle to even draw the diagram.

Hey! Below is my answer. Hope it makes sense!

Jake

RuiAce · « **Reply #109 on:** March 30, 2016, 06:56:42 pm »

Quote from: Happy Physics Land on March 30, 2016, 05:47:23 pm

Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!! (I hope rui doesnt hit me for asking this question)

The chord AB of the hyperbola xy=c² subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.

I couldnt really start this one because Im struggle to even draw the diagram.

You're off for this question cause it's still the point of half yearly exams and this is a bit tricky for an average 4U student. If it were HSC study time and you couldn't do this THEN I'd hit you.

But yeah, pretty sure I don't need to input anything else. Jake's method is probably what I would've used

Happy Physics Land · « **Reply #110 on:** March 31, 2016, 12:14:38 am »

Quote from: jakesilove on March 30, 2016, 06:46:45 pm

Hey! Below is my answer. Hope it makes sense!

(Image removed from quote.)

Jake

Thank you so much Jake! I understand it a lot better now! Im just wondering, were the coordinates meant to be A = (ca, c/a), B = (cb, c/b) and P = (cp, c/p)? Because I've always learnt that you let x=ct and y = c/t, where t is the variable. If Im the one getting confused here please correct me Jake!!! Thank you very much! I would have loved to contribute more to the forum but because half-yearlies are on currently I'm a bit tight with time, sorry for dropping the burden on you guys! I shall be back to join you guys soon!

))))

Best Regards
Happy Physics Land

jakesilove · « **Reply #111 on:** March 31, 2016, 12:30:15 am »

Quote from: Happy Physics Land on March 31, 2016, 12:14:38 am

Thank you so much Jake! I understand it a lot better now! Im just wondering, were the coordinates meant to be A = (ca, c/a), B = (cb, c/b) and P = (cp, c/p)? Because I've always learnt that you let x=ct and y = c/t, where t is the variable. If Im the one getting confused here please correct me Jake!!! Thank you very much! I would have loved to contribute more to the forum but because half-yearlies are on currently I'm a bit tight with time, sorry for dropping the burden on you guys! I shall be back to join you guys soon! ))))

Best Regards
Happy Physics Land

You're absolutely right about the co-ordinates, I have no idea how I managed to write out the complete mess that I did in the first line. Honestly it's like I had an aneurysm or something; anyway, thanks for picking that up!

Don't even worry, you contribute literally all the time. The community wouldn't be the same without you, and we totally understand that there will be times where you'll be quite busy and unable to do the same amount as usual! Never feel pressured to be on the forums

Jake

katherine123 · « **Reply #112 on:** April 01, 2016, 05:39:35 pm »

Quote from: jamonwindeyer on March 25, 2016, 12:51:12 pm

Hey Katherine! That is not the approach I would use, though it might work! I'll show how I would do it with integration by parts:

$I = \int sec^3(x)dx$

$u=\sec{x} \quad u'=\sec{x}\tan{x} \quad v=\tan{x} \quad v'=\sec^2{x}$

Through integration by parts:

$I = \sec{x}\tan{x} - \int \sec{x}tan^2{x}dx \\ = \sec{x}\tan{x}-\int(\sec^3{x}-\sec{x})dx \\ =\sec{x}\tan{x} - I + \int\sec{x}dx$

Now the integration of sec(x) is another job in itself. There is a common trick for figuring out this integral. Transform the integral like so:

$\int\sec{x}dx = \int\sec{x}\frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}}dx$

Then, use a substitution:

$u = \sec{x}+\tan{x}$

Follow this process to find that:

$\int\sec{x}dx=\ln|\sec{x}+\tan{x}|+C$

Taking the formula from above and adding this, and rearranging, will give us the final solution:

$I = \frac{1}{2}\sec{x}\tan{x}+\frac{1}{2}\ln|\sec{x}+\tan{x}|+C$

Let me know if anything here is unclear!! It is a long and difficult question

is this method the same as :
Let say I= integral (x*lnx)
let u=x v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)

jakesilove · « **Reply #113 on:** April 01, 2016, 06:14:14 pm »

Quote from: katherine123 on April 01, 2016, 05:39:35 pm

is this method the same as :
Let say I= integral (x*lnx)
let u=x v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)

Hey Katherine!

Yep, that's the integration by parts method! It's fairly common in 4U, so I would have a good working knowledge of it. Try a bunch of practice questions, including the one Jamon wrote up (which is a pretty difficult one, if I do say so myself!).

Best of luck!

Jake

RuiAce · « **Reply #114 on:** April 01, 2016, 07:56:51 pm »

Quote from: katherine123 on April 01, 2016, 05:39:35 pm

is this method the same as :
Let say I= integral (x*lnx)
let u=x v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)

Integration by parts is essentially the reverse of the product rule. One way of describing it is indeed, what you demonstrated.

Many questions will require you to demonstrate knowledge of this integration technique. E.g.
$\int{x\sin^{2}{x}\,dx}$

Note - The example I gave is a tiny bit harder, but not as hard as that for the integral of sec³(x)

katherine123 · « **Reply #115 on:** April 03, 2016, 07:52:47 am »

why isnt this B

RuiAce · « **Reply #116 on:** April 03, 2016, 09:30:10 am »

Quote from: katherine123 on April 03, 2016, 07:52:47 am

why isnt this B

$\text{As this is a multiple choice, rather than formally deriving the equations for either the washers or shells method}\\\text{I will proceed to quote the formulae required}$
$\text{I immediately notice that we are integrating from 0 to 4, with respect to }y\\\text{This implies that I will be using the washers method. Therefore, rearrange the given equation:}\\x=\frac{{y}^{2}}{8}$
$\text{Then I realise that because of where we are rotating it, we don't need washers. We only need the discs method here.}$
Note - You may have accidentally thought we were rotating about the y-axis, not the line x=2
This could potentially explain, for you, why the answer is not B
$\text{The formula for the discs method is simply}\\\pi \int_{a}^{b}{{r}^{2}\,dy}\\\text{where }r\text{ is the radius of the disc.}$
$\text{Draw a rectangle from the line }x=2\text{ (yes you should always use the diagram to aid you) to determine the radius.}\\\text{It would appear that }\\r=2-x=2-\frac{{y}^{2}}{8}$
$\begin{align*}\therefore V&=\pi \int_{-4}^{4}{{\left(2-\frac{{y}^{2}}{8}\right)}^{2}\,dy}\\&=2\pi \int_{0}^{4}{{\left(2-\frac{{y}^{2}}{8}\right)}^{2}\,dy}\end{align*}\\\text{using properties of even functions, but I can see you already knew that.}$

amandali · « **Reply #117 on:** April 03, 2016, 02:41:09 pm »

how do you do this ques?
i know that the angle will be 90

RuiAce · « **Reply #118 on:** April 03, 2016, 03:21:13 pm »

Quote from: amandali on April 03, 2016, 02:41:09 pm

how do you do this ques?
i know that the angle will be 90

$\text{Using what you said we have}\\\arg{\left(\frac{z-1+i}{z+1-i}\right)}=\pm\frac{\pi}{2}\\\text{So we just break this argument of a fraction down using the rule }\arg{\left(\frac{a}{b}\right)}=\arg{a}-\arg{b}\\\arg{(z-1+i)}-\arg{(z+1-i)}=\pm\frac{\pi}{2}\\\arg{(z-(1-i))}-\arg{(z-(-1+i))}=\pm\frac{\pi}{2}$
$\text{Without loss of generality, consider the diagram.}$
In this context, wlog just implies ignore the plus or minus sign as all we care is that the angle is indeed, 90 degrees.

$\text{Let }A\text{ represent }{z}_{1}=1-i \text{ and }B\text{ represent }{z}_{2}=-1+i\\\text{and }P\text{ represent }z=x+iy$
C, D and E exist for the sole purpose of labelling angles
$\text{We require }\arg{(z-1+i)}-\arg{(z+1-i)}=-\frac{\pi}{2}\\\text{which is equivalent to saying }\\\angle PBE-\angle EAD=-90deg\\\text{and since }BC \parallel AD \text{ using corresponding angles}\\\angle PBE-\angle PEC=-90deg$
Notice that all I am using here is the definition of arg(z-z₀) where z₀ is some complex number a+ib. That is, on the Argand diagram, you draw out the ray, inclined at an angle of theta to the horizontal.

And now comes the tricky part.
$\text{According to the exterior angle of }\Delta BPE\\\angle PEC=\angle BPE+\angle PBE\\ \Leftrightarrow\angle PBE-\angle PEC=-\angle BPE$
$\text{This means that }\angle BPE=\angle BPA=90deg\text{ from the above information!}\\\text{On your diagram, this means that the points P and E actually coincide!}$

$\text{Lastly, if P and E are to coincide, we now know that }BE \bot EA\\ \text{So it becomes quite obvious that }E\text{, and therefore }P\text{ represent the complex number }1+i\\\text{keeping in mind that the line }BC\text{ has equation }y=1\text{, and AD is }y=-1$
$\text{So }|z|=|1+i|=\sqrt{2}$

Maz · « **Reply #119 on:** April 03, 2016, 05:01:34 pm »

`hey, can someone please help me with this...i have little/to no idea how to do it...

find d/dx of:
$\int_{1}^{1+{x}^{2}}{\frac{1}{\sqrt{1+{e}^{t}}}\,dt}$

thankyou so much in advance

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