Cambridge HSC MX1 Textbook Marathon/Q&A (5 Viewers)

InteGrand · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
How common is Volumes in the HSC Exams? I rarely see them

Halp Please

http://i.imgur.com/kzIHB2w.jpg

Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.

si2136 · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.

Yes, the volume. And are feet and foot the same (dumb question)?

InteGrand · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
Yes, the volume. And are feet and foot the same (dumb question)?

Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.

This means viewing everything with origin A, the equation of the inner parabola is y² = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).

We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).

si2136 · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.

This means viewing everything with origin A, the equation of the inner parabola is y² = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).

We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).

Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge

InteGrand · Apr 19, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge

I doubt they'd ask it in the HSC without providing a conversion rate though, so don't worry.

davidgoes4wce · Apr 20, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
We can set the mean position to be x = 0.

So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).

(If we have centre of motion (mean position) x₀ instead, the formula becomes v^2 = n^2 (A^2 - (x - x₀)^2).)

We already know A = 5 (given), so v^2 = n^2 (25 - x^2).

Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x₀ have opposite signs at all times.)

So -4 = -n^2 * 4, so n = 1.

So v^2 = 25 - x^2.

Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.

Is there something wrong with the solution for that question because it shows the speed to be $v=5 \sqrt{2} \ and \ 3 \sqrt{2}$

InteGrand · Apr 20, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Is there something wrong with the solution for that question because it shows the speed to be $v=5 \sqrt{2} \ and \ 3 \sqrt{2}$

Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).

davidgoes4wce · Apr 20, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).

I fixed up your mistake and redid the solution and was able to get the speeds.

Usernameismyname · Apr 22, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Hey guys,

Sorry I'm a bit new to using this.

Could I get some help with Exercise 10I question 10???

There are two round tables, one oak and one mahogany, each with five seats. In how many
ways may a group of ten people be seated?

Thanks

jathu123 · Apr 22, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

There are (5-1)! ways to arrange the 5 people in the first table then another (5-1)! ways to arrange in the other table. But for the first table, any 5 of the 10 people can be chosen and the remaining 5 chosen for the 2nd table.
So the answer is 10C5*4!*4! = 145152

davidgoes4wce · Apr 23, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Cambridge EX 10 D Q 19

$Two dice are rolled. A three appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than seven.$

Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).

Am I missing something?

porcupinetree · Apr 23, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Cambridge EX 10 D Q 19

$\\ $ Two dice are rolled. A three appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than seven. $$

Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).

Am I missing something?

$The favourable outcomes are as follows:$ $$(3,5) , (3,6), (5,3), (6,3)$$ $(Thus there are four favourable outcomes.) The number of possible outcomes (where a three appears on at least one of the dice) is 6^2 - 5^2 = 11.$
$\\ $Hence the probability is $\frac{4}{11}.$

davidgoes4wce · Apr 23, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Can understand it a bit better now, my thinking initially was 4/36 but can see why its 4/11, since the total possible combinations with a 3 are: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3) and (6,3).

Only 4 of those have combined value above 8 or more.

davidgoes4wce · Apr 25, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Ex 3A Q 9 (h)
$A particle moves according to $ x=10 \ cos \ \frac{\pi \ t}{12}, $ in units of metres and seconds$

$Use the graph and table of values to find when the particle is more than 15 metres from its initial position$

$I got the following graph :$

Should the solution be between the range?
8 < t < 16, 32 < t < 40, 54 < t< 60 ?

The books answer is:

8 < t < 16

leehuan · Apr 25, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Ex 3A Q 9 (h)
$A particle moves according to $ x=10 \ cos \ \frac{\pi \ t}{12}, $ in units of metres and seconds$

$Use the graph and table of values to find when the particle is more than 15 metres from its initial position$

$I got the following graph :$

Should the solution be between the range?
8 < t < 16, 32 < t < 40, 54 < t< 60 ?

The books answer is:

8 < t < 16

Since they only wanted you to consider when you could use the graph AND the table of values...

davidgoes4wce · Apr 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

$Cambridge Ex 3A Q 13 (c)$

$Two engines, Thomas and Henry, move on closer parallel tracks. They start at the origin, and are together again at time t=e-1. Thomas' displacement-time equation, in units of metres and minutes, is x=300 log (t+1), and Henry's is x=kt, for some constant k.$

$I understand everything including the exact values but my question is how to convert it to a non-exact value form$

$300 \ log (e-1)-\frac{300(e-2)}{e-1} $ in which case they get 37 metres. I was a bit unsure how they converted t=e-2 to 43'' $$

davidgoes4wce · Apr 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

37 metres and 43'' I dont quite get.

I get a value of -410.36 instead of 37 metres

With the time part , I went (e-2) x 60 (remembering these units were initially in minutes)= which gave me approximately 8 seconds

davidgoes4wce · Apr 27, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

$Part (c) of the question asked : Use calculus to find the maximum distance between Henry and Thomas during the first e-1 minutes, and the time when it occurs (in exact form, and then correct to the nearest meter or the nearest second$

davidgoes4wce · Apr 28, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

^^^ Don't worry I sorted it out , with my scientific calculator I needed to type 'e1' for it to read it as 'e'

davidgoes4wce · Apr 30, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

$Cambridge Exercise 3C Q 4 (b)$

$The answer was a bit unclear so I wanted to clear it up here$

$$ A stone is thrown downwards from the top of a 120 metre building, with an initial speed of 25 m/s. Take g $ =10m/s^{2}, $ and take upwards as positive, so that $ \ddot{x}=-10$

$Rework part (a) with the origin at the top of the building, and downwards positive$

$I set the initial conditions as t=0, v=25 and x=-120.$

$\ddot{x}=10,\ \dot{x}=10t+25,\ x=5t^{2}+25t-120$

$Could someone confirm if those 3 equations are right?$

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