Author Topic: 4U Maths Question Thread (Read 665489 times) Tweet Share

RuiAce · « **Reply #450 on:** August 17, 2016, 05:09:52 pm »

Quote from: znaser on August 17, 2016, 03:20:52 pm

I tried following the answers but yh this question is something. Thanks in advance

What's the confusion? Is there a particular part about the answers that makes no sense? (Sorry, but your question is generic and I don't know where to focus my response on.)

Line 1: Cosine double angle formula: cos(2x) = 2cos²(x) - 1 so cos(x) = 2cos²(x/2) - 1
Line 1 second part: Expand out the compound angle

Line 2: As with line 1 albeit with minus not plus

Line 3: Sums the above. This makes sense - we are just deriving something similar to product to sum. The inspiration to use squares and not to the power of 1 is simply because the result to prove involves squares.

Line 4: They tell you what they did. This isn't exactly classic in the HSC (classic for me) but you should anticipate it. Because how else do you turn an equation into an inequality

Line 5: Rearrange
_________

(ii) is using the inequality in (i) over and over again. This is not uncommon: The proof of the AM-GM inequality for 4 variables is just the AM-GM for 2 variables using the same method.

I.e. (x+y)/2 ≤ √(xy) and (a+b+c+d)/4 ≤ ⁴√(abcd)
_________

There's hints in (iii) to use part (ii). What was perhaps not obvious was that 1/3(a+b+c) also had to be there. But look at it closely - you proved something involving multiplying 4 different cosines. You had to bring it out somehow.

Their typeface is pretty tedious to read though; I'll give that.

massive · « **Reply #451 on:** August 17, 2016, 08:55:55 pm »

guys how do you do this, been stuck on it for ages

znaser · « **Reply #452 on:** August 18, 2016, 12:14:47 pm »

Sorry should've been more specific. But yeah it was how to tackle part 2. Thanks, really helped

RuiAce · « **Reply #453 on:** August 18, 2016, 02:33:47 pm »

Quote from: znaser on August 18, 2016, 12:14:47 pm

Sorry should've been more specific. But yeah it was how to tackle part 2. Thanks, really helped

If part 2 was the hard part (which is understandable), I recommend you try attempting the example I provided. See if you can use a similar method to prove the 4 variable AM-GM.

massive · « **Reply #454 on:** August 18, 2016, 06:30:16 pm »

Btw did anyone manage to figure out the two volumes questions I posted previously, I still don't get them

. thanks!

jakesilove · « **Reply #455 on:** August 18, 2016, 09:18:57 pm »

Quote from: massive on August 18, 2016, 06:30:16 pm

Btw did anyone manage to figure out the two volumes questions I posted previously, I still don't get them . thanks!

Just because of the volume (lol) of questions you've been asking, do you think you could show us some working out before we tackle the questions you post? Just so we can see where we can help you specifically

massive · « **Reply #456 on:** August 19, 2016, 01:56:23 pm »

Quote from: jakesilove on August 18, 2016, 09:18:57 pm

Just because of the volume (lol) of questions you've been asking, do you think you could show us some working out before we tackle the questions you post? Just so we can see where we can help you specifically

Oh yeah fair enough, i'll start doing that from now on. Its just that for those 2 questions i posted I don't even know where to start

.

RuiAce · « **Reply #457 on:** August 20, 2016, 07:25:32 pm »

I can't guarantee that this answer is correct.

I would like to know what is the source of these questions?

RuiAce · « **Reply #458 on:** August 20, 2016, 11:42:39 pm »

Quote from: massive on August 17, 2016, 04:13:56 pm

Guys how do you even do this question? :S

You need to draw diagrams.
(i)
$\triangle CDE\text{ is isosceles with }\angle CDE = \angle CED = \theta\\ \text{To prove }\angle CDE=\theta\text{ just use corresponding angles on parallel lines.}\\ \text{To prove }\angle CED=\theta\text{ use the angle sum of }\triangle POB\text{ to get }\angle PBO=180^\circ - 2\theta\\ \text{Then more parallel lines and angle sum of }\triangle CDE$
$\text{Since }\triangle CDE\text{ is isosceles, the altitude bisects the base. So using right-angled trigonometry:}\\ DE=R-x\\ \therefore \sin \theta = \frac{\frac{R-x}{2}}{DC}\iff DC = \frac{R-x}{2\sin \theta}$
$\text{Fitting the parabola is the hard part. So try to use the same process I showed you in the other question}\\ \text{That being said, you might need to construct the }y\text{ and }z\text{ axes...}$
(ii)
This is a question you should've done despite the difficulty of part (i) because it is much more standard.
$\text{Volume of a typical cross section is }\delta V =A (\cos \theta\, \delta x)\\ \text{Which if you just sub in what you get in part a) you'll ultimately arrive at }\\ V=\frac{2}{3\tan \theta}\int_0^R (R-x)\sqrt{R^2-x^2}dx$
(iii)
$\text{Because you previously studied the volume of a torus before, you know to just split it up and use tricks}\\ \frac{3V\tan \theta}{2}=R\int_0^R \underbrace{\sqrt{R^2-x^2}dx}_{\text{1/4 area of circle}} - \underbrace{\int_0^R x\sqrt{R^2-x^2}dx}_{\text{odd function integral}}$
$\text{To bring in }H\text{, note that in }\triangle POE\\ \tan \theta = \frac{R}{H}$
Not doing (iv). That's definitely not in the syllabus (not that slanted slicing ever was). But I reckon it can be described by visualising the slice in your head.

amandali · « **Reply #459 on:** August 25, 2016, 02:48:12 pm »

need help with these ques
a)a disc is free to rotate in a horizontal plane about an axis through its centre O. A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.

b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min

RuiAce · « **Reply #460 on:** August 25, 2016, 05:24:30 pm »

Quote from: amandali on August 25, 2016, 02:48:12 pm

need help with these ques
a)a disc is free to rotate in a horizontal plane about an axis through its centre O. A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.

b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min

b) looks like a question that just involves tackling with units.

The radius is 6cm, so the angular velocity is given
ω = v/r = 147/6 rad s^-1

Recall that 1 radian is only 1/(2π) of a revolution, and that there are 60 seconds in a minute. Hence, we multiply by 60/2π = 30/π to convert to rev/min

ω = 735/π rev min^-1 (which you can plug into your calculator)

jakesilove · « **Reply #461 on:** August 25, 2016, 05:53:45 pm »

Quote from: amandali on August 25, 2016, 02:48:12 pm

need help with these ques
a)a disc is free to rotate in a horizontal plane about an axis through its centre O. A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.

b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min

Hey!

For a) I understood this question to mean the following:

The particle is on spinning disk, at radius 0.2m
The friction keeping the particle from slipping is 0.5 the normal force

From these two pieces of information, we can easily equate forces. The force 'inwards' (ie. keeping the particle in place) is going to be the friction force. The force pushing the particle 'outwards' (ie. making the particle slip) is going to be centripetal force. Therefore, the particle will be just about to slip when the inwards force equals the outwards force.

$F_{friction}=F_{centripetal}$
$0.5mg=\frac{mv^2}{r}$
$v^2=0.5gr=0.5*9.8*0.2=0.98$
$v=0.99 m/s$

Hopefully I interpreted the question correctly! I'm not sure what else you could do.

Jake

znaser · « **Reply #462 on:** August 26, 2016, 09:25:31 pm »

Hi. So i follow this question (with ofc additional workout since the solution skips a lot of steps) right up till this fraction: [ (a+b)^2 - [(a+b)^2]/2 ]/[...]
I get wats going on with the denominator since: (a^1/2 - b^1/2)^2 ≥ 0 ... (a^2 + b^2) ≥ 2ab ... [(a^2 + b^2)^2] / 4 ≥ (ab)^2
therefore [a^2 + b^2] / (ab)^2 ≥ (a^2 + b^2) / [(a^2 + b^2)^2 / 4] since the denominator of the former is smaller than the latter. But I don't get wats going on with the numerator of the first fraction I typed. Thanks

RuiAce · « **Reply #463 on:** August 26, 2016, 09:34:39 pm »

Quote from: znaser on August 26, 2016, 09:25:31 pm

Hi. So i follow this question (with ofc additional workout since the solution skips a lot of steps) right up till this fraction: [ (a+b)^2 - [(a+b)^2]/2 ]/[...]
I get wats going on with the denominator since: (a^1/2 - b^1/2)^2 ≥ 0 ... (a^2 + b^2) ≥ 2ab ... [(a^2 + b^2)^2] / 4 ≥ (ab)^2
therefore [a^2 + b^2] / (ab)^2 ≥ (a^2 + b^2) / [(a^2 + b^2)^2 / 4] since the denominator of the former is smaller than the latter. But I don't get wats going on with the numerator of the first fraction I typed. Thanks

$\begin{align*}a+b&\ge 2\sqrt{ab}\\ \implies \frac{a+b}{\sqrt{2}} &\ge \sqrt{2ab}\\ \implies \frac{(a+b)^2}{2}&\ge 2ab\\ \implies -\frac{(a+b)^2}{2}&\le -2ab\end{align*}\\\text{Then just sub it in.}$
$\text{It works exactly like the denominator. The difference is that the denominator goes in this direction instead.}\\ \frac{a+b}{2}\ge ab \implies \frac{(a+b)^2}{4} \ge \sqrt{ab} \implies \left(\frac{(a+b)^2}{4}\right)^2 \ge (ab)^2\\ \text{And of course, we use subtraction, not division.}$

znaser · « **Reply #464 on:** August 26, 2016, 09:41:57 pm »

Quote from: RuiAce on August 26, 2016, 09:34:39 pm

$\begin{align*}a+b&\ge 2\sqrt{ab}\\ \implies \frac{a+b}{\sqrt{2}} &\ge \sqrt{2ab}\\ \implies \frac{(a+b)^2}{2}&\ge 2ab\\ \implies -\frac{(a+b)^2}{2}&\le -2ab\end{align*}\\\text{Then just sub it in.}$
$\text{It works exactly like the denominator. The difference is that the denominator goes in this direction instead.}\\ \frac{a+b}{2}\ge ab \implies \frac{(a+b)^2}{4} \ge \sqrt{ab} \implies \left(\frac{(a+b)^2}{4}\right)^2 \ge (ab)^2\\ \text{And of course, we use subtraction, not division.}$

Thank you!

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