Author Topic: 4U Maths Question Thread (Read 667963 times) Tweet Share

Ali_Abbas · « **Reply #1335 on:** July 17, 2017, 12:25:39 am »

Quote from: RuiAce on July 17, 2017, 12:09:59 am

Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...

Overkill is my middle name

pikachu975 · « **Reply #1336 on:** July 17, 2017, 12:37:03 am »

Quote from: RuiAce on July 17, 2017, 12:09:59 am

Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...

Is the irrational conjugate theorem in HSC? Never heard of it

RuiAce · « **Reply #1337 on:** July 17, 2017, 09:15:34 am »

Quote from: pikachu975 on July 17, 2017, 12:37:03 am

Is the irrational conjugate theorem in HSC? Never heard of it

Yes and it works exactly like the complex conjugate root theorem, which states that if a polynomial has real coefficients but has a complex root then another root must be its complex conjugate.

Also note that the HSC introduces theorems, but more often than not chooses not to give their names (e.g. intermediate value theorem subtly being placed in the 3U estimation of roots topic)

BradMate · « **Reply #1338 on:** July 17, 2017, 11:26:40 am »

Thanks Rui and AA. Much appreciated.

chelseam · « **Reply #1339 on:** July 19, 2017, 11:42:43 pm »

Hi! Could someone please help me with this question? I don't understand why the answers say that dV=2y(4-x)dx. I've tried to use similar triangles but I'm not sure if that's even relevant here... Thank you

RuiAce · « **Reply #1340 on:** July 20, 2017, 12:36:16 am »

Quote from: chelseam on July 19, 2017, 11:42:43 pm

Hi! Could someone please help me with this question? I don't understand why the answers say that dV=2y(4-x)dx. I've tried to use similar triangles but I'm not sure if that's even relevant here... Thank you

$2y\text{ is the base of the rectangular cross section}\\ \text{This can be immediately inferred from looking at the diagram;}\\ \text{The base goes from }-y\text{ to }y$
$\text{The }4-x\text{ comes from dealing with the height, which is trickier}\\ \text{Consider the significance of the }45^\circ$
This is actually easier to explain if they taught 3D coordinate geometry with the x-y-z plane.
$\text{The height of the rectangle is dependent on that }45^\circ\\ \textbf{as well}\text{ as the actual }x\text{-coordinate}$
$\text{The height is }4-x\\ \text{Attempt to visualise the front view. What's actually happening}\\ \text{is that if we imagine an }x\text{-}z\text{ plane}\\ \text{The line }z=4-x\text{ is being sketched out.}\\ \text{Note how the }45^\circ\text{ gives us the gradient of the line}\\ \text{and we figure out that the gradient is also negative from just looking at the diagram}\\ \textit{Exercise: Try to also figure out where the 4 came from.}$
$\text{Of course, combining this makes it more obvious that }\delta V = 2y(4-x)\delta x$ Can understand if that didn't make sense but I'm too sleepy now so if something doesn't puzzle up I'll attend to it tomorrow

chelseam · « **Reply #1341 on:** July 20, 2017, 12:50:23 am »

Quote from: RuiAce on July 20, 2017, 12:36:16 am

Can understand if that didn't make sense but I'm too sleepy now so if something doesn't puzzle up I'll attend to it tomorrow

Thanks so much Rui, I think I get it now!

Is the 4 coming from the dotted line in the diagram?

RuiAce · « **Reply #1342 on:** July 20, 2017, 09:10:17 am »

Quote from: chelseam on July 20, 2017, 12:50:23 am

Thanks so much Rui, I think I get it now! Is the 4 coming from the dotted line in the diagram?

Pretty much that's one way of thinking about it

bluecookie · « **Reply #1343 on:** July 20, 2017, 09:40:10 am »

In the diagram the circles XYPS and XYRQ intersect at the point X and Y, and PXQ, PYR, QSY, PST and QTR are straight lines.
(i) Explain why ∠STQ=∠YRQ+∠YPS.
(ii) Show that ∠YRQ+∠YPS+∠SXQ=pi.
(iii) Deduce that STQX is a cyclic quadrilateral.
(iv) Let ∠QPY=a and ∠PQY=b. Show that ∠STQ=a+b.

Can I have help with iv?

bluecookie · « **Reply #1344 on:** July 20, 2017, 09:42:45 am »

Find V if
v=integral from 0 to h, of invcos(x/h)-(x/h)root(1-(x/h)^2) dx given that integral of invcos(theta)d(theta)=thetainvcos(theta)-root(1-theta^2) and let theta=x/h.

RuiAce · « **Reply #1345 on:** July 20, 2017, 10:09:20 am »

Quote from: bluecookie on July 20, 2017, 09:42:45 am

Find V if
v=integral from 0 to h, of invcos(x/h)-(x/h)root(1-(x/h)^2) dx given that integral of invcos(theta)d(theta)=thetainvcos(theta)-root(1-theta^2) and let theta=x/h.

This was difficult to read. In the future, when +'s and -'s are present please use an extra space so that the terms don't look clumped. Same with "thetainvcos" - a space is beneficial. Also, please use sqrt for the square root, and preferably "arccos" for invcos.

Additionally, at times like this please resort to the modify function instead of double posting.
$\text{Let }x=\theta h \implies dx = h \, d\theta$
$\text{Observe that:}\\\begin{align*}\int \cos^{-1}\theta\ d\theta &= \theta \cos^{-1}\theta -\sqrt{1-\theta^2}\\ \implies\int \cos^{-1}\theta\ d\theta&= \theta \cos^{-1}\theta + \int \frac{\theta}{\sqrt{1-\theta^2}}d\theta\\ \implies \int \left(\cos^{-1}\theta - \frac{\theta}{\sqrt{1-\theta^2}}\right)d\theta &=\theta \cos^{-1}\theta \end{align*}$
$\text{Hence, applying the substitution}\\ \begin{align*}\int_0^h \left(\cos^{-1}\frac{x}{h}-\frac{\frac{x}h}{\sqrt{1-\left(\frac{x}h\right)^2}}\right)dx&=\int_0^1 \left(\cos^{-1}\theta-\frac{\theta}{\sqrt{1-\theta^2}}\right)h\,d\theta\\ &= h\left[\theta \cos^{-1}\theta\right]_0^1\\ &= 0\end{align*}$

seventeenboi · « **Reply #1346 on:** July 20, 2017, 12:26:18 pm »

hiii

I'm 100% lost on how to do this question please help
thanks!

RuiAce · « **Reply #1347 on:** July 20, 2017, 12:33:33 pm »

Quote from: bluecookie on July 20, 2017, 09:40:10 am

In the diagram the circles XYPS and XYRQ intersect at the point X and Y, and PXQ, PYR, QSY, PST and QTR are straight lines.
(i) Explain why ∠STQ=∠YRQ+∠YPS.
(ii) Show that ∠YRQ+∠YPS+∠SXQ=pi.
(iii) Deduce that STQX is a cyclic quadrilateral.
(iv) Let ∠QPY=a and ∠PQY=b. Show that ∠STQ=a+b.

Can I have help with iv?

The diagram they offered was terrible in the fact that it did not make QSY appear to be a straight line.

RuiAce · « **Reply #1348 on:** July 20, 2017, 12:34:27 pm »

Quote from: seventeenboi on July 20, 2017, 12:26:18 pm

hiii
I'm 100% lost on how to do this question please help
thanks!

Hint: You will have to assume that the area of an ellipse is $\text{Area}=\pi AB$, where A and B are the lengths of the major and minor axes.
$\text{The relevant integral to compute is}\\ V=\int_0^H \pi ab\left(1-\frac{h^2}{H^2}\right)dh$

Natmo243 · « **Reply #1349 on:** July 20, 2017, 07:53:04 pm »

Could somebody please help me with this one? I'm trying to get an answer independent of a (their answer is sqrt(3)+pi/3). I've tried the substitution x=asinӨ and integration by parts. Any help would be much appreciated

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