Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

appleibeats · Jun 25, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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Stuck on part c). Not sure how to use part a) to help.

InteGrand · Jun 25, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33288

Stuck on part c). Not sure how to use part a) to help.

$\noindent From part (a), we know that $\theta = \tan ^{-1}t + \frac{\pi}{4}$, for $t\geq 0$.$

$\noindent Then since the inverse tan function has range $0\leq \tan ^{-1} t< \frac{\pi}{2}$ for all $t\geq 0$, clearly $0+\frac{\pi}{4} \leq \theta < \frac{\pi}{2}+ \frac{\pi}{4}\Rightarrow \frac{\pi}{4}\leq \theta < \frac{3\pi}{4}$. Since the particle starts at $\theta = \frac{\pi}{4}$, and asymptotically approaches $ \theta = \frac{3\pi}{4}$ as $t\to \infty$, the particle never rotates through an angle more than: $\text{limiting angle}- \text{initial angle}=\frac{3\pi}{4} - \frac{\pi}{4}=\frac{\pi}{2}$.$

appleibeats · Jun 27, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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Don't understand what the question is asking me.

Answers says 4 units

leehuan · Jun 27, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33292

Don't understand what the question is asking me.

Answers says 4 units

Midpoints.
4/2=2
2/2=1
1/2=1/2
etc.
$S_\infty=2+1+\frac{1}{2}+\dots$

si2136 · Jun 27, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33292

Don't understand what the question is asking me.

Answers says 4 units

Not too sure either, but let one of the values be x and the other be x + 4. The distances between them are 4. Now because the midpoint would be x+2, the distances to each side is 2, so because it goes left to x and right to x + 4, the total distance is 4, since 2 + 2 = 4.

Therefore 4 units.

InteGrand · Jun 27, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33292

Don't understand what the question is asking me.

Answers says 4 units

It's asking for total distance moved using halving the interval.

E.g. Say we're on the interval [0,4] and we start at 0. Say the root happens to be at 1.245 (but we don't know this).

First we'd move to point 2. Then we'd find we've overshot and we'll go to 1. Then we'll see we're too low and we'll move to 1.5. Then 1.25, then 1.125, and so on.

We're asked to find the total distance we've travelled.

(Note that if the actual root turns out to be either 0, 2, 3 or 1, 3.5 or 1.5, … , then the distance travelled will be less than 4, because we'll land exactly on the root. As an exercise, you may like to try and prove or disprove that such values are the only ones where we'll land exactly on the root after a finite number of steps.

)

leehuan · Jun 27, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
Not too sure either, but let one of the values be x and the other be x + 4. The distances between them are 4. Now because the midpoint would be x+2, the distances to each side is 2, so because it goes left to x and right to x + 4, the total distance is 4, since 2 + 2 = 4.

Therefore 4 units.

I am quite positive that it is just a series question that applies the numerical estimation of roots topic to it.

InteGrand · Jun 27, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Yes, if we know that the process goes on indefinitely (i.e. the root wasn't 0, or 2, or 3, or 1, or … (exercise: classify exactly which numbers will result in a finite run only)), then the total distance travelled is just the total length of the interval (which is 4 here). This follows from the observation that the length of the step halves each time and the first step is length 2, so the total distance given there is an infinite number of steps is 2 + 1 + 1/2 + 1/4 + … = 4.

In other words, it was mainly just a series and sequences Q. as leehuan has said.

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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Unsure how to answer part a). I get part b).

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

The answer is 1/2^(1/4)

InteGrand · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33301

Unsure how to answer part a). I get part b).

appleibeats said:
The answer is 1/2^(1/4)

$\noindent Note that the fifth year is $\color{blue}{4}$ years after the first year. Let the common ratio be $r$, then since the fifth year's sales are \color{red}{half} \color{black}{that} of the first year, we must have $r^{\color{blue}{4}} = \color{red}{\frac{1}{2}}$, whence $r = \sqrt[4]{\frac{1}{2}}$.$

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Two bulldozers are sitting in a construction site facing each other. Bulldozer A is at x = 0, and bulldozer B is 36 metres away at x = 36. A bee is sitting on the scoop at the very front of bulldozer A. At 7:00am the workers start up both bulldozers and start them moving towards each other at the same speed V m/s. The bee is disturbed by the commotion and flies at twice the speed of the bulldozers to land on the scoop of bulldozer B.
(a) Show that the bee reaches bulldozer B when it is at x = 24.
(b) Immediately the bee lands, it takes off again and flies back to bulldozer A. Where is
bulldozer A when the two meet?
(c) Assume that the bulldozers keep moving towards each other and the bee keeps flying
between the two, so that the bee will eventually be squashed.
(i) Where will this happen? (ii) How far will the bee have flown?

InteGrand · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Two bulldozers are sitting in a construction site facing each other. Bulldozer A is at x = 0, and bulldozer B is 36 metres away at x = 36. A bee is sitting on the scoop at the very front of bulldozer A. At 7:00am the workers start up both bulldozers and start them moving towards each other at the same speed V m/s. The bee is disturbed by the commotion and flies at twice the speed of the bulldozers to land on the scoop of bulldozer B.
(a) Show that the bee reaches bulldozer B when it is at x = 24.
(b) Immediately the bee lands, it takes off again and flies back to bulldozer A. Where is
bulldozer A when the two meet?
(c) Assume that the bulldozers keep moving towards each other and the bee keeps flying
between the two, so that the bee will eventually be squashed.
(i) Where will this happen? (ii) How far will the bee have flown?

We can immediately say the answer to the very last part is just the initial separation between the bulldozers (in this case 36 m) doing essentially no calculation. Can you see why?

Also by symmetry, for c) (i), we conclude with almost no calculation that the bee gets squashed halfway between the bulldozers' starting points, i.e. x = 18.

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

No

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Is it something to do with the limiting sum of a GP?? Also I don't get how to answer part a)

InteGrand · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
No

$\noindent To see why the answer to the last part is just the initial separation $d$ (in this case $d = 36$ m), note the following key observations:$

$\noindent 1) The bee travels for the entire time $T$ at a constant speed of $2V$, where $T$ is the time taken for the bulldozers to crash into each other.$

$\noindent 2) For this \emph{same length of time} $T$, using the principle of \emph{relative velocity}, wrt the second bulldozer, the first bulldozer travels at a constant velocity of $2V$ right. Since the bulldozers meet after a time of $T$ and were distance $d$ apart to start with, this tells us that an object moving at speed $2V$ for a time of $T$ travels \underline{distance $d$}.$

$\noindent It follows immediately from 1) and 2) that the distance the bee travels is precisely $d$, the initial bulldozer separation.$

InteGrand · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Is it something to do with the limiting sum of a GP?? Also I don't get how to answer part a)

Yes the GP method is the other way to do it, and was probably the way the book expected students to do it. But this requires (slightly) tedious calculation, whereas the answer can be obtained with essentially no calculation using the method above.

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

how do you show x = 24 for part a) ??

davidgoes4wce · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
how do you show x = 24 for part a) ??

$Part (a)$

$Since the bee started together with bulldozer A on one end with velocity 2V m/s, it will be completing the journey with bulldozer B coming from the other end at velocity V m/s. So they will both complete 36 metres at 3V m/s, Which will take them 36/(3V)=12/V seconds. Hence the bee travelled:$

$2V \times \frac{12}{V}=24 \ metres$

InteGrand · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Also I don't get how to answer part a)

$\noindent Note that we are told that the bee is moving twice as fast as the bulldozer. Therefore, at the time the bee meets Bulldozer B, it has travelled twice as far as Bulldozer B. Since the bee is going right from $x=0$ and the Bulldozer B is going left from $x=36$, at the time of meeting, the bee must have gone two-thirds of the way towards the endpoint $x=d$, and Bulldozer B come one-third of the way from there towards $x=0$ (so that distance travelled by bee is twice that travelled by Bulldozer B). Hence they meet two-thirds of the way across from $x=0$, which is at $x=24$.$

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