Author Topic: 4U Maths Question Thread (Read 665220 times) Tweet Share

jakesilove · « **Reply #225 on:** May 27, 2016, 12:03:52 pm »

Quote from: jamonwindeyer on May 27, 2016, 11:51:02 am

Hey Katherine! Do you mean what steps they used to get to the end of each line?

I'll pull the binomial-style term out of each one to show you the process. I'll just do the outline, I might skip a step here or there, let me know if anything doesn't make sense!

Here are the first two:

$(a\sec{\theta}-a\cos{\theta})^2=a^2(\sec{\theta}-\cos{\theta})^2=\frac{a^2(1-\cos^2{\theta})^2}{\cos^2{\theta}}=\frac{a^2\sin^4{\theta}}{\cos^2{\theta}}=a^2\cos^2{\theta}\tan^4{\theta}$

$(b\tan{\theta}+b\cot{\theta})^2=\frac{b^2(\sin^2{\theta}+\cos^2{\theta})^2}{\sin^2{\theta}\cos^2{\theta}}=\frac{b^2}{\sin^2{\theta}\cos^2{\theta}}=b^2\csc^2{\theta}\sec^2{\theta}$

I was just posting a reply! Damn you and your superior LaTex skills

jamonwindeyer · « **Reply #226 on:** May 27, 2016, 01:22:26 pm »

Quote from: jakesilove on May 27, 2016, 12:03:52 pm

I was just posting a reply! Damn you and your superior LaTex skills

Damn sorry man! I think we need a system between Rui, you and myself

jakesilove · « **Reply #227 on:** May 27, 2016, 02:41:33 pm »

Quote from: jamonwindeyer on May 27, 2016, 01:22:26 pm

Damn sorry man! I think we need a system between Rui, you and myself

Aha no problem at all, usually we at least get out different methods, which can be useful for people learning the tricks of the trade!

amandali · « **Reply #228 on:** May 29, 2016, 09:16:20 am »

how do i find volume generated using cylindrical shell method
when area bounded by y=1-x^2 and y=1-x is rotated About x axis

RuiAce · « **Reply #229 on:** May 29, 2016, 10:28:00 am »

Quote from: amandali on May 29, 2016, 09:16:20 am

how do i find volume generated using cylindrical shell method
when area bounded by y=1-x^2 and y=1-x is rotated About x axis

$\text{If we wish to rotate the given curve about the }x\text{-axis using the shells method, we need to rearrange the equations.}\\ x=1-y\\ x=\sqrt{1-y}\\ \text{Note that we take the positive root because we consider the right portion of the parabola}$
$\text{Take a typical shell with respect to the }x\text{-axis. Note that the square root curve lies above the line.}$
$\text{The shell has:}\\\text{Radius: }r=y\\ \text{Height: }h=\sqrt{1-y}-(1-y)$
$\text{We have the volume of a typical shell:}\\ \delta V=2\pi y\left(\sqrt{1-y}-(1-y)\right)\delta y\\ \text{for some thickness }\delta y$
$\text{Thus, }\\\begin{align*} V&=\lim_{\delta y \to 0}{\sum_{y=0}^1{2\pi y\left(\sqrt{1-y}-(1-y)\right)\delta y}}\\&= \int_0^1{2\pi y\left(\sqrt{1-y}-(1-y)\right)\, dy}\\ &=2\pi \int_0^1{(1-y)(\sqrt{y}-y) \, dy}\quad \text{by the f(a-x) border-flip trick} \end{align*}$
$\text{The remainder of the question and the proof of the property }\int_0^a{f(a-x)\,dx}=\int_0^a{f(x)\,dx}\\ \text{are left as exercises.}$

katherine123 · « **Reply #230 on:** May 30, 2016, 10:39:45 pm »

1. find the volume generated using cylindrical shells when region between y=2x^2 and y=x^4-2x^2 is revolved around y axis
whats the height?

2.find volume generated by using cylindrical shell method when area enclosed by x^2/25+ y^2/16=1 is rotated about x=8

RuiAce · « **Reply #231 on:** May 31, 2016, 01:51:56 pm »

Quote from: katherine123 on May 30, 2016, 10:39:45 pm

1. find the volume generated using cylindrical shells when region between y=2x^2 and y=x^4-2x^2 is revolved around y axis
whats the height?

2.find volume generated by using cylindrical shell method when area enclosed by x^2/25+ y^2/16=1 is rotated about x=8

The height for the first one is just h=2x² - (x⁴-2x²) by inspection of the graph

$\text{To rotate about the line }x=8\text{ which is parallel to the }y\text{-axis, make }y\text{ the subject}\\ \frac{x^2}{25}+\frac{y^2}{16}=1 \Leftrightarrow y=\pm 16\sqrt{1-\frac{x^2}{25}}$
$\text{The radius of the shell will be given }r=8-x\\ \text{By inspection of the graph of the ellipse, the top half corresponds to }y=16\sqrt{1-\frac{x^2}{25}}\\ \text{and the bottom half corresponds to }y=-16\sqrt{1-\frac{x^2}{25}}\\ \text{Hence the height of the shell is }h=\left(16\sqrt{1-\frac{x^2}{25}}\right)-\left(-16\sqrt{1-\frac{x^2}{25}}\right)=32\sqrt{1-\frac{x^2}{25}}$
$\text{The volume of a typical cylindrical shell is}\\ \delta V = 2\pi r h = 2\pi (8-x) \left(32\sqrt{1-\frac{x^2}{25}}\right)\delta x\\ \text{for some thickness }\delta x$
$\text{Thus}\\\begin{align*} V&=\lim_{\delta x \to 0}{\sum_{x=-5}^{5}{ 64\pi (8-x) \left(\sqrt{1-\frac{x^2}{25}}\right)\delta x }} \\ &= \int_{-5}^{5}{64\pi (8-x) \left(\sqrt{1-\frac{x^2}{25}}\right)\, dx} \\ &= 512\pi \underbrace{\int_{-5}^{5}{\frac{1}{5}\sqrt{25-x^2}dx}}_{\text{area of a semicircle with radius 5}} - 64\pi \underbrace{\int_{-5}^{5}{x\sqrt{1-\frac{x^2}{25}}dx}}_{\text{odd function}}\\ &= \frac{512\pi}{5} \left(\frac{25\pi}{2}\right) - 0 \\ &= 6400\pi^2\end{align*}$

Subject to inaccuracy

katherine123 · « **Reply #232 on:** June 01, 2016, 11:45:10 pm »

the region bounded by y=e^x, x=1 and y=1 is rotated about line x=1. Find the volume of solid

i used the slicing method
and got
(1-ln(y)) as the radius

v=integral of π(1-ln(y))^2 dy

is this correct or does cylindrical method only apply for this ques

the answer is 2π(e-2) units^3

amandali · « **Reply #233 on:** June 02, 2016, 12:34:08 am »

how to find the radius of smaller and bigger circle using slicing method.
the aim is to find the volume when rotated about y axis

RuiAce · « **Reply #234 on:** June 02, 2016, 08:36:23 am »

Quote from: katherine123 on June 01, 2016, 11:45:10 pm

the region bounded by y=e^x, x=1 and y=1 is rotated about line x=1. Find the volume of solid

i used the slicing method
and got
(1-ln(y)) as the radius

v=integral of π(1-ln(y))^2 dy

is this correct or does cylindrical method only apply for this ques

the answer is 2π(e-2) units^3

$\textbf{I am of the belief that the answers are WRONG.}$

$\text{If we were to apply the discs (slicing - circles) method, then skipping a load of steps we do indeed take:}\\ R=1-\ln{y}\\ \text{Hence our integral becomes the same one you gave:} \\ \pi \int_1^e{\left(\ln{y}\right)^2 dy} \\ \text{This is best evaluated using reduction formulae. But this evaluates to }\pi(2e-5)$

Cylindrical shells may make it somewhat easier. If we just take the radius and height to be:
r=(1-x)
h=(e^x)-(1)
This integral just requires one application of integration by parts to evaluate:
$V=2\pi \int_0^1{(1-x)e^x\,dx}$
$\text{Yet this also evaluates to }\pi(2e-5) \\ \text{Which makes sense, because you want both the shells and disc methods to reconcile}$

$\text{Interesting worthwhile mention: }2\pi \int_0^1{xe^x dx}=\pi (2e-4)$

RuiAce · « **Reply #235 on:** June 02, 2016, 09:17:45 am »

Quote from: amandali on June 02, 2016, 12:34:08 am

(Image removed from quote.)

how to find the radius of smaller and bigger circle using slicing method.
the aim is to find the volume when rotated about y axis

$\text{This is WAY FAR easier done with shells. To use washers (slicing - annuli) we rearrange the equation:}\\ \begin{align*} x^4-4x^2+4y&=0 \\ \Leftrightarrow x^2&=\frac{4\pm \sqrt{16-16y}}{2} \\ \Leftrightarrow x^2&=2 \pm \sqrt{4-4y} \\ \Rightarrow x&=\sqrt{2 \pm \sqrt{4-4y}} \end{align*}$
$\text{In the last step, we are safe to take the positive root as the graph is symmetrical about the }y\text{-axis.}\\ \text{Therefore we only consider the right branch of the quartic.}$

$\text{Now, there is one massive problem with the other plus or minus. Observe the next truncated graph.}$

$\textbf{THE GRAPH SPLITS}$
$y=x^2-\frac{x^4}{4}\text{ attains a local maximum at }\left(\sqrt{2},1\right) \\ \text{Therefore:} \\ \text{The portion corresponding to }x=\sqrt{2+\sqrt{4-4y}}=\sqrt{2}\sqrt{1+\sqrt{1-y}}\text{ is to the right of the maximum}\\\text{The portion corresponding to }x=\sqrt{2-\sqrt{4-4y}}=\sqrt{2}\sqrt{1-\sqrt{1-y}}\text{ is to the left of the maximum}$

$\text{With the hard part done, to compute the volume through a rotation about the }y\text{-axis we have}\\ \text{Outer radius: }R=\sqrt{2}\sqrt{1+\sqrt{1-y}}\\ \text{Inner radius: }r=\sqrt{2}\sqrt{1-\sqrt{1-y}}$
$\text{The volume of a typical slice is therefore}\\ \delta V=\pi \left(R^2-r^2\right) \delta y \\ =\pi \left( 2\left(1+\sqrt{1-y}\right) - 2\left(1-\sqrt{1-y}\right) \right) \delta y \\ = 4\pi \sqrt{1-y} \, \delta y \\ \text{for some thickness }\delta y$
$\text{So the volume is}\\ V=\pi \lim_{\delta y \to 0}{\sum_{x=0}^1{4\sqrt{1-y} \, \delta y}}\\ = \pi \int_0^1{4\sqrt{1-y}\, dy} \\ = \frac{8\pi}{3}$

This is easily subject to inaccuracies. Especially given that I haven't been able to reconcile the answer with the shells method yet.

Edit: I have now. Hence this should be a fully accurate solution.

relativity1 · « **Reply #236 on:** June 06, 2016, 02:09:18 pm »

How do you find the restriction in this question
The hyperbola H has equation xy = 16.
(a) Sketch this hyperbola and indicate on your diagram the positions and coordinates
of all points at which the curve intersects the axes of symmetry.
(b) P(4p, ), where p > 0, and Q(4q, ), where q > 0, are two distinct arbitrary
points on H. Find the equation of the chord PQ.
(c) Prove that the equation of the tangent at P is x + p2y = 8p.
(d) The tangents at P and Q intersect at T. Find the coordinates of T.
(e) The chord PQ produced passes through the point N(0,

.
(i) Find the equation of the locus of T.
(ii) Give a geometrical description of this locus.

I only need part ii) i think the restriction was x can only be between 0 and 4

RuiAce · « **Reply #237 on:** June 06, 2016, 04:16:44 pm »

Quote from: relativity1 on June 06, 2016, 02:09:18 pm

How do you find the restriction in this question
The hyperbola H has equation xy = 16.
(a) Sketch this hyperbola and indicate on your diagram the positions and coordinates
of all points at which the curve intersects the axes of symmetry.
(b) P(4p, ), where p > 0, and Q(4q, ), where q > 0, are two distinct arbitrary
points on H. Find the equation of the chord PQ.
(c) Prove that the equation of the tangent at P is x + p2y = 8p.
(d) The tangents at P and Q intersect at T. Find the coordinates of T.
(e) The chord PQ produced passes through the point N(0, .
(i) Find the equation of the locus of T.
(ii) Give a geometrical description of this locus.

I only need part ii) i think the restriction was x can only be between 0 and 4

If you only needed part e) (ii) then you should've included the answers to the previous parts, as now I have to find an answer to a lof of the answers from scratch. Please remember to do so next time so my life is easier

$\text{The answer to part b appears to be }x+pqy=4(p+q)$
$\text{The answer to part d appears to be the point }T\left(\frac{8pq}{p+q},\frac{8}{p+q}\right)$
$\text{Part e) (i) reveals the condition that }8pq=4(p+q)\\ \text{Note that the }x\text{- component of }T\text{ is given }\\ x=\frac{8pq}{p+q}$
$\text{Hence, by substituting in our new result we have something very convenient:}\\ x=\frac{4(p+q)}{p+q} \Leftrightarrow x=4$
(Note that p can't equal to q here or else the tangents at P and Q would coincide and thus T is not a strictly defined point.)
$\text{Hence the answer to part e) (ii) is just that the locus of }T\text{ is the line }x=4$

If one was to be more cautious we can take things a step further. (This is probably what you mean by restriction)
$\text{Then, consider the }y\text{- component. We have }\\ y=\frac{8}{p+q}=\frac{8}{2pq}=\frac{4}{pq}$
$\text{Since }p>0, q>0 \text{ we immediately have }y>0$
$\begin{align*}\text{From }2pq&=p+q\\ \Leftrightarrow q(2p-1)&=p \\ \Rightarrow q&=\frac{p}{2p-1}\end{align*}$
$\text{Hence }y=\frac{4(2p-1)}{p^2}$
$\text{It can be shown that this function attains a global maximum at }(1,4) \\ \text{Hence the allowable values for }y\text{ are } 0 < y < 4\\ \text{Note: We reject }y=4\\ \text{ as this would imply }p=q\text{ which can't happen, or }T\text{ is not strictly defined!}$

I highly doubt this is the fastest method, but it was all I could come up with for now.
________

The non calculus approach:

$pq=\frac{p+q}{2}\ge \sqrt{pq}\quad \text{ AM-GM inequality}\\ \text{Which implies }pq \ge 1 \implies \frac{4}{pq} \le 4$
$\text{By the assumption of the AM-GM inequality, the equality case holds when }p=q\text{ which we have to reject.}$
Proof of the AM-GM inequality for two terms should be known by every MX2 student.
$\begin{align*}(p-q)^2&\ge 0 \\ \Rightarrow p^2-2pq+q^2 &\ge 0 \\ \Rightarrow p^2+2pq+q^2 &\ge 4pq \\ \Rightarrow (p+q)^2 &\ge 4pq \\ \Rightarrow p+q \ge 2\sqrt{pq} \\ \Rightarrow \frac{p+q}{2} \ge \sqrt{pq} \end{align*}$

birdwing341 · « **Reply #238 on:** June 08, 2016, 05:36:19 pm »

This is a more general question (with no specifics) - but its basically on how to do well at circle geometry.

Do you have any particular tips that could help in an exam situation when I'm stumped by the question. I've heard in general that there is a lot of cyclic quads, angles standing on the same arc etc. but I was wondering more if you had a particular process you would go through.

Thanks!!

RuiAce · « **Reply #239 on:** June 08, 2016, 06:23:34 pm »

Quote from: birdwing341 on June 08, 2016, 05:36:19 pm

This is a more general question (with no specifics) - but its basically on how to do well at circle geometry.

Do you have any particular tips that could help in an exam situation when I'm stumped by the question. I've heard in general that there is a lot of cyclic quads, angles standing on the same arc etc. but I was wondering more if you had a particular process you would go through.

Thanks!!

You should be decently skilled (if not already mastered) circle geometry at the Extension 1 level before attempting problems that target Extension 2 students.

At the Extension 1 level, I was already familiar with what my theorems "LOOKED LIKE". What does this mean? Examples:
- Alternate angles look like Z angles on parallel lines
- Angles standing on same arc theorem looked like a nice M shape to me
- Alternate segment theorem involves a tangent and a triangle.
However how you visualise it may differ.

When I tackle an Extension 2 question, I don't label 50 thousand things at once if it's way too irrelevant. I work in one direction.

That is, I look at what I am trying to prove (or find).

Then I try to either work forwards, or backwards. That is, I look at the LHS or the RHS, then I start looking for an angle/side that equals to THAT ONLY. THEN, I keep going.

At the Extension 2 level you should also be prepared to manipulate lots of things. Similar triangles and cyclic quadrilaterals are examples of cliches, however even base angles of isosceles triangle are important. Equal radii is something I always keep at the back of my head regardless of if I need to use it.

Progressively, I build up to the final answer. But in doing so, I only consider RELEVANT information, not EXTRANEOUS information that may be useful for another part but not this present part.

A rare trap is the occurrence of trigonometry in circle geometry. When that happens, always look out for any right angled triangles BEFORE you attempt sine/cosine rule.

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