Author Topic: 4U Maths Question Thread (Read 665270 times) Tweet Share

jamonwindeyer · « **Reply #255 on:** June 20, 2016, 03:10:27 pm »

Quote from: birdwing341 on June 20, 2016, 02:10:39 pm

Hello!

I think I remember somewhere that one moderator had a small checklist that they went through after completing each question (to check for stuff like units, constants, dxs etc.).

Would you be able to remind me what that is?

Thanks!

In a guide I wrote a while ago I had (and this is what I used):

SURD - Simplest Form, Units, Rationalised, Does it Make Sense?

However, I expanded on it in an article I wrote about the BIG mistakes students make in exams, which you can find here!

Integration has so many little traps, I should probably come up with a new acronym for it! I used SURD because I always forgot units, and I always forgot to rationalise

EDIT: How about SURDIC? Simplest Form, Units, Rationalised, Does it Make Sense, Integration Notation, Constants? Ahaha

amandali · « **Reply #256 on:** June 20, 2016, 03:43:50 pm »

ive found the equation chord for PQ shown below and Mpo * Mqo =-1
and im not sure how to progress

RuiAce · « **Reply #257 on:** June 20, 2016, 04:53:38 pm »

Quote from: amandali on June 20, 2016, 03:43:50 pm

ive found the equation chord for PQ shown below and Mpo * Mqo =-1
and im not sure how to progress

If this is one of the further questions in the Cambridge textbook then it might not be so easy to do it. These questions do not reflect the difficulty of the HSC (even though they are valid in reflecting question type) as the question is not broken down into parts for you - you have to discover everything yourself.
$\text{We're saying that }m_{PA}m_{QA}=-1,\text{ that is}\\ \begin{align*}\left(\frac{b\sin{\theta}}{a\cos{\theta}-a}\right)\left(\frac{b\sin{\phi}}{a\cos{\phi}-a}\right)&=-1 \\ \implies \left(\frac{\sin{\theta}}{\cos{\theta}-1}\right)\left(\frac{\sin{\phi}}{\cos{\phi}-1}\right)&=-\frac{a^2}{b^2} \end{align*}$
$\text{We are interesting in the }x\text{-intercept of the chord. Suppose this is the point }C(c,0)\text{. Then:}\\ \begin{align*} \frac{c}{a}\cos{\frac{\theta+\phi}{2}}+\frac{0}{b}\sin{\frac{\theta+\phi}{2}}&=\frac{\theta+\phi}{2}\\ \implies c&=a\sec{\frac{\theta+\phi}{2}}\cos{\frac{\theta-\phi}{2}}\end{align*}$
$\text{Our aim is essentially to show that }c\text{ is FIXED }\\ \text{i.e. }c\text{ is independent of }\theta, \phi\\ \text{Since }c\text{ is written in terms of half angles, attempt to reconstruct some of what we have.}$
$\begin{align*}\frac{\sin{\theta}}{\cos{\theta}-1}&=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{-2\cos^2{\frac{\theta}{2}}}\\&= -\cot{\frac{\theta}{2}}\end{align*}\\ \text{Hence in the earlier equation we now have }\cot{\frac{\theta}{2}}\cot{\frac{\phi}{2}}=-\frac{a^2}{b^2}$
$\text{And now rigorously decomposing }c\text{ up:}\\ \begin{align*}c&=a\left(\frac{\cos{\frac{\theta-\phi}{2}}}{\cos{\frac{\theta+\phi}{2}}}\right)\\ &= a\left(\frac{\cos{\frac{\theta}{2}}\cos{\frac{\phi}{2}}+\sin{{\frac{\theta}{2}}}\sin{\frac{\phi}{2}}}{\cos{{\frac{\theta}{2}}}\cos{\frac{\phi}{2}}-\sin{{\frac{\theta}{2}}}\sin{\frac{\phi}{2}}}\right)\\ &= a\left(\frac{1+\tan{\frac{\theta}{2}}\tan{\frac{\phi}{2}}}{1-\tan{\frac{\theta}{2}}\tan{\frac{\phi}{2}}}\right)\\ &= a\left(\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}\right)\end{align*}$
$\text{A lot of algebra can be done here just to tidy up this expression, however at this point it's}\\ \text{now quite obvious that the point }C\text{ does not depend on the position of }P\text{ or }Q$

jamonwindeyer · « **Reply #258 on:** June 20, 2016, 05:04:18 pm »

Quote from: RuiAce on June 20, 2016, 04:53:38 pm

If this is one of the further questions in the Cambridge textbook then it might not be so easy to do it. These questions do not reflect the difficulty of the HSC (even though they are valid in reflecting question type) as the question is not broken down into parts for you - you have to discover everything yourself.
$\text{We're saying that }m_{PA}m_{QA}=-1,\text{ that is}\\ \begin{align*}\left(\frac{b\sin{\theta}}{a\cos{\theta}-a}\right)\left(\frac{b\sin{\phi}}{a\cos{\phi}-a}\right)&=-1 \\ \implies \left(\frac{\sin{\theta}}{\cos{\theta}-1}\right)\left(\frac{\sin{\phi}}{\cos{\phi}-1}\right)&=-\frac{a^2}{b^2} \end{align*}$
$\text{We are interesting in the }x\text{-intercept of the chord. Suppose this is the point }C(c,0)\text{. Then:}\\ \begin{align*} \frac{c}{a}\cos{\frac{\theta+\phi}{2}}+\frac{0}{b}\sin{\frac{\theta+\phi}{2}}&=\frac{\theta+\phi}{2}\\ \implies c&=a\sec{\frac{\theta+\phi}{2}}\cos{\frac{\theta-\phi}{2}}\end{align*}$
$\text{Our aim is essentially to show that }c\text{ is independent of }\theta, \phi\\ \text{Since }c\text{ is written in terms of half angles, attempt to reconstruct some of what we have.}$
$\begin{align*}\frac{\sin{\theta}}{\cos{\theta}-1}&=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{-2\cos^2{\frac{\theta}{2}}}\\&= -\cot{\frac{\theta}{2}}\end{align*}\\ \text{Hence in the earlier equation we now have }\cot{\frac{\theta}{2}}\cot{\frac{\phi}{2}}=-\frac{a^2}{b^2}$
$\text{And now rigorously decomposing }c\text{ up:}\\ \begin{align*}c&=a\left(\frac{\cos{\frac{\theta-\phi}{2}}}{\cos{\frac{\theta+\phi}{2}}}\right)\\ &= a\left(\frac{\cos{\frac{\theta}{2}}\cos{\frac{\phi}{2}}+\sin{{\frac{\theta}{2}}}\sin{\frac{\phi}{2}}}{\cos{{\frac{\theta}{2}}}\cos{\frac{\phi}{2}}-\sin{{\frac{\theta}{2}}}\sin{\frac{\phi}{2}}}\right)\\ &= a\left(\frac{1+\tan{\frac{\theta}{2}}\tan{\frac{\phi}{2}}}{1-\tan{\frac{\theta}{2}}\tan{\frac{\phi}{2}}}\right)\\ &= a\left(\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}\right)\end{align*}$
$\text{A lot of algebra can be done here just to tidy up this expression, however at this point it's}\\ \text{now quite obvious that the point }C\text{ does not depend on the position of }P\text{ or }Q$

Damn, got about half as far as you did before you posted

these ellipse questions can actually be pretty nasty, I'm glad I didn't have to deal with them in 2014

RuiAce · « **Reply #259 on:** June 20, 2016, 05:17:08 pm »

Quote from: jamonwindeyer on June 20, 2016, 05:04:18 pm

Damn, got about half as far as you did before you posted these ellipse questions can actually be pretty nasty, I'm glad I didn't have to deal with them in 2014

Thing is, when someone asks you a question and it's clearly out of Cambridge 4U, all you have to do is decipher the worked solutions

jamonwindeyer · « **Reply #260 on:** June 20, 2016, 05:38:04 pm »

Quote from: RuiAce on June 20, 2016, 05:17:08 pm

Thing is, when someone asks you a question and it's clearly out of Cambridge 4U, all you have to do is decipher the worked solutions

Okay, sorts it, I need to buy Cambridge

Is that kind of the standard 4U text or are there other common ones?

amandali · « **Reply #261 on:** June 21, 2016, 06:12:50 pm »

1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3

2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres

RuiAce · « **Reply #262 on:** June 21, 2016, 06:37:05 pm »

Quote from: amandali on June 21, 2016, 06:12:50 pm

1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3

2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres

$\text{The first integral strictly equals 0 due to the symmetry of an odd function: }\int_{-a}^{a}{o(u)du}=0$
$\text{The second integral represents the area of a semicircle with radius 4, which equals }8\pi\\ \text{Thus, the answer does falls out.}$

Let me know if my handwriting is not legible.

RuiAce · « **Reply #263 on:** June 21, 2016, 07:10:18 pm »

Quote from: amandali on June 21, 2016, 06:12:50 pm

1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3

2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres

$\text{Assume that the parabolic cross-section can be sketched on the }X\text{-}Y\text{ plane.}\\ \text{For simplicity, assume that the parabola concaves up and has its stationary point at the height as shown in the diagram.}$

(The image makes the Y-intercept look like some number between 3 and 4. Just pretend that it's not there.)
$\text{Effectively, we want to consider the parabola }Y=aX^2-h\\ \text{where }h\text{ is the height}$
$\text{We are told that the height of the parabola is equal to its width.} \\ \text{Take a typical cross section with respect to the }x\text{-axis}$
$\text{The width of the parabola is therefore }w=2\sqrt{100-x^2}$
$\text{Hence, our parabola will have equation }Y=aX^2-2\sqrt{100-x^2}$
$\text{Now, if you think about it carefully, the }X\text{-intercepts of this parabola}\\ \text{will be at }X=\pm \sqrt{100-x^2}\\ \textit{This occurs because the parabola is symmetric about the }Y\text{-axis!}$
$\therefore 0=a\left(\pm \sqrt{100-x^2}\right)^2-2\sqrt{100-x^2}\\ \implies a=\frac{2}{\sqrt{100-x^2}}$
$\text{So the cross section has equation }\\ Y=\frac{2X^2}{\sqrt{100-x^2}}-2\sqrt{100-x^2}$
$\text{Hence finally, we have }\\ \begin{align*}A&=\left|\underbrace{\int_{-\sqrt{100-x^2}}^{\sqrt{100-x^2}}{\left(\frac{2X^2}{\sqrt{100-x^2}}-2\sqrt{100-x^2}\right)dX}}_{\text{even function integral}}\right| \\ &= 2 \left|\int_{0}^{\sqrt{100-x^2}}{\left(\frac{2X^2}{\sqrt{100-x^2}}-2\sqrt{100-x^2}\right)dX}\right| \\ &= 2\left| \left[\frac{2X^3}{3\sqrt{100-x^2}}-2X\sqrt{100-x^2}\right]_0 ^{\sqrt{100-x^2}}\right| \\ &= \frac{8\sqrt{100-x^2}}{3}\end{align*}$
$\textit{Note: The final answer obviously treated }y=\sqrt{100-x^2}\\ \textit{which whilst gets the same final result, is probably way neater.}$

This was a messy question and required some lateral thinking in considering the parabolic cross section in it's OWN X-Y plane. Tell me if something does not make sense.

amandali · « **Reply #264 on:** June 22, 2016, 03:41:19 pm »

need help with these ques

RuiAce · « **Reply #265 on:** June 22, 2016, 04:32:47 pm »

Quote from: amandali on June 22, 2016, 03:41:19 pm

(Image removed from quote.)
(Image removed from quote.)

need help with these ques

$\text{Analyse the cross section with respect to the }h\text{-axis}$
$\text{The equation of the ellipse is }\\ \frac{x^2}{a^2\lambda^2}+\frac{y^2}{b^2\lambda^2}=1\\ \text{So the length of the semi-major axis is }a\lambda\\ \text{and the length of the semi-minor axis is }b\lambda\\ \textit{or vice-versa}$
$\text{Take a typical cross section with respect to the }h\text{-axis.}\\ \text{The area of the elliptical cross section is }A=\pi (a\lambda)(b\lambda) = \pi a b \lambda^2$
(Note: The area of any ellipse is πAB, where A and B are the lengths of the semi-major and semi-minor axes respectively)
$\text{So the volume of a typical cross section is }\\ \delta V = \pi ab \left(1-\frac{2h^2}{H^2}+\frac{h^4}{H^4}\right)\delta h\\ \text{for some thickness }\delta h$
$\begin{align*}\therefore V&=\lim_{\delta h \to 0}\sum_{h=0}^{H}\pi ab \left(1-\frac{2h^2}{H^2}+\frac{h^4}{H^4}\right)\delta h\\ &= \pi ab \int_0^H{\left(1-\frac{2h^2}{H^2}+\frac{h^4}{H^4}\right)dh}\\ &= \left[h-\frac{2h^3}{3H^2}+\frac{h^5}{5H^4}\right]_0^H\\ &= \frac{8\pi abH}{15}\end{align*}$

RuiAce · « **Reply #266 on:** June 22, 2016, 05:02:20 pm »

Quote from: amandali on June 22, 2016, 03:41:19 pm

(Image removed from quote.)
(Image removed from quote.)

need help with these ques

$\text{This question is more straight-forward than meets the eye.}\\ \text{The height of the triangle is already given: }h=16-x^2$
$\text{Take a typical triangular cross-section perpendicular to the plane and with respect to the }x\text{-axis.}\\ \text{The equation of the semicircle is }y=\sqrt{16-x^2}\\ \text{This forms the base of the solid}$
$\text{Hence, the volume of a typical cross section is:}\\\begin{align*}\delta V&= \frac{1}{2} \times \sqrt{16-x^2} \times \left(16-x^2\right) \times \delta x\\ &=\frac{\left(16-x^2\right)^{\frac{3}{2}}}{2}\delta x\end{align*}\\ \text{for some thickness }\delta x$
$\begin{align*}\therefore V&=\lim_{\delta x \to 0}\sum_{x=-4}^{4}\frac{\left(16-x^2\right)^{\frac{3}{2}}}{2}\\ &= \underbrace{\int_{-4}^{4}\frac{\left(16-x^2\right)^{\frac{3}{2}}}{2}dx}_{\text{even function integral}}\\ &= \int_{0}^{4}\left(16-x^2\right)^{\frac{3}{2}}dx \tag{our desired result}\\ &=\underbrace{\int_0^{\frac{\pi}{2}}\left(16\cos^2{\theta}\right)^{\frac{3}{2}}(4\cos{\theta}\,d\theta)}_{x=4\sin{\theta} \implies dx=4\cos{\theta}\, d\theta}\\ &= \int_0^{\frac{\pi}{2}}{256\cos^4{\theta}\,d\theta}\\ &=256\left(\frac{3}{4}\right) \int_0^{\frac{\pi}{2}}{\cos^2{\theta}\,d\theta} \tag{reduction formulae}\\ &=192\left(\frac{1}{2}\right)\int_0^{\frac{\pi}{2}}{d\theta} \tag{reduction formulae}\\ &=96[\theta]_0^{\frac{\pi}{2}}\\ &= 48\pi\end{align*}$
__________________________________________

$\text{You can tell that the hard part of this question was the actual integration.}\\ \text{To derive the reduction formula:}\\ \begin{align*}I_n&=\int_0^{\frac{\pi}{2}}\cos^n{x}\,dx\\ &=\int_0^{\frac{\pi}{2}}\cos{x}\cos^{n-1}{x}\,dx\\ &= \left[\sin{x}\cos^{n-1}{x}\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}}\sin{x}(n-1)\sin{x}\cos^{n-2}{x}\, dx \tag{integration by parts}\\ &= 0-(n-1)\int_0^{\frac{\pi}{2}}\left(1-\cos^2{x}\right)\cos^{n-2}{x}\,dx \tag{Pythagorean identity}\\ &= -(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}{x}+(n-1)\int_0^{\frac{\pi}{2}}\cos^n{x}\,dx\\ &=-(n-1)I_{n-2}+(n-1)I_n\\ \implies -n I_n &= -(n-1)I_{n-2} \\ \implies I_n&=\frac{n-1}{n}I_{n-2}\end{align*}$

The more exhaustive method to integrating cos⁴(x) is to rewrite as (cos²(x))², apply the double angle formula, expand the whole thing out, and reapply the double angle formula

katherine123 · « **Reply #267 on:** June 23, 2016, 01:03:10 am »

how do i do part (i)

jamonwindeyer · « **Reply #268 on:** June 23, 2016, 10:35:51 am »

Quote from: katherine123 on June 23, 2016, 01:03:10 am

how do i do part (i)

I'll let Rui correct me if this isn't quite 4 Unit friendly, but this is actually very simple if we know the formula for a sphere in three dimensions (x,y,z):

$x^2+y^2+z^2=R^2$

At the cross section, the height above the xy plane (the z-value) is h, so we can substitute:

$x^2+y^2+h^2=R^2 \quad \therefore x^2+y^2=R^2-h^2$

What we notice here is that this is actually the formula for a circle. We've just found the formula for the outer circle of the cross section, with that term on the right being the square of the radius like normal. So:

$A=\text{Area of Bigger Circle} - \text{ Area of Smaller Circle} \\ = \pi(R^2-h^2)-\pi r^2 \\ = \pi(R^2-h^2-r^2)$

Let me know if this makes sense! And Rui might come along and give a better method, I'm not sure if 3-dimensional surface equations are what would be the normal approach

katherine123 · « **Reply #269 on:** June 23, 2016, 11:41:13 am »

how do i do part iii)

i got
Part i) h=[(root3)*a]/2
part ii) ZY=xa/b

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