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

appleibeats · Jan 12, 2016

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

I thought the numerator had to be 1 to be able to use the standard formula to get tan^-1 ..... ( the one in the reference sheet)

Also the question said to use the reverse chain rule, which is what I am confused about.

InteGrand · Jan 12, 2016

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

appleibeats said:
I thought the numerator had to be 1 to be able to use the standard formula to get tan^-1 ..... ( the one in the reference sheet)

Also the question said to use the reverse chain rule, which is what I am confused about.

$\noindent The numerator just needed to be the derivative of $\mathrm{e}^x$ for us. I.e. it was (leaving out the limits) $\int \frac{\mathrm{e}^{x}}{1+\left(\mathrm{e}^{x}\right)^2}\text{ d}x =\int \frac{f^\prime (x)}{1+\left(f(x)\right)^2}\text{ d}x $, where $f(x) = \mathrm{e}^{x}$, so $f^\prime (x) = \mathrm{e}^{x}$. So this integrates to (using the reverse chain rule) $\tan^{-1} \left( f(x)\right) = \tan^{-1} \left( \mathrm{e}^{x}\right)$, as I did in my post.$

InteGrand · Jan 12, 2016

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

$\noindent To make it even clearer, we can let $u=\mathrm{e}^{x}\Rightarrow \mathrm{d}u = \mathrm{e}^{x}\text{ d}x$, so the integral is $\int \frac{\mathrm{d}u}{1+u^2}=\tan^{-1} u = \tan^{-1} \left(\mathrm{e}^{x}\right)$.$

appleibeats · Jan 12, 2016

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

Thanks, I got it.

Blitz_N7 · Jan 13, 2016

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

In part a, I found the tangent to be:

From the graph we see that the tangent is to the 'right' of the curve y=arcsinx x=siny but the y-intercept ends up to be pi/3-sqrt3 and subbing in in the integral doesnt give me 1/4. Not sure where I'm wrong :/

InteGrand · Jan 14, 2016

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

Blitz_N7 said:
In part a, I found the tangent to be:

From the graph we see that the tangent is to the 'right' of the curve y=arcsinx x=siny but the y-intercept ends up to be pi/3-sqrt3 and subbing in in the integral doesnt give me 1/4. Not sure where I'm wrong :/

$\noindent The shaded region below the $x$-axis becomes a cone when rotated, and the volume of this is $V_\text{cone} =\frac{1}{3}\pi r^2 h$, where $r = \frac{\sqrt{3}}{2} -\frac{\pi}{6}$ (the $x$-intercept of the tangent line) and $h= \sqrt{3}-\frac{\pi}{3}$ (the negative of the $y$-intercept; we need to take the negative of it in order to make the height positive, as it should be).$

$\noindent The volume of revolution of the upper part of the region (the part above the $x$-axis) is $V_\text{upper} =\pi \int _{0}^{\frac{\pi}{3}}\left( \ell (y)^2 - \sin ^{2}y \right) \text{ d}y$, where $x=\ell (y)$ is the equation of the tangent line. So evaluate this integral and the cone volume, and then the total volume is the sum of these.$

Blitz_N7 · Jan 14, 2016

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

Thank you

appleibeats · Jan 15, 2016

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

I don't get question 5 b) Mathematic Induction Q from this HSC Paper

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97MAT3U.PDF

Also if you go down to page 56 question 5b in the notes for this paper here

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97EN_MAT.PDF

it says relationship between r! and (r + 1)! . What is the relationship?

InteGrand · Jan 15, 2016

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

appleibeats said:
I don't get question 5 b) Mathematic Induction Q from this HSC Paper

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97MAT3U.PDF

Also if you go down to page 56 question 5b in the notes for this paper here

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97EN_MAT.PDF

it says relationship between r! and (r + 1)! . What is the relationship?

$\noindent The relationship is $(r+1)! =(r+1)r!$ for any integer $r \geq 0$.$

InteGrand · Jan 15, 2016

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

appleibeats said:
I don't get question 5 b) Mathematic Induction Q from this HSC Paper

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97MAT3U.PDF

Also if you go down to page 56 question 5b in the notes for this paper here

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97EN_MAT.PDF

it says relationship between r! and (r + 1)! . What is the relationship?

$\noindent Here's how to do the induction question.$

$\noindent \textbf{Proposition.} For any integer $n\geq 3$, we have $\sum _{j=3}^{n}\binom{j-1}{2} = \binom{n}{3}$.$

$\noindent \textbf{Proof by induction.}$

$\noindent For $n=3$, we have $\binom{n}{3}=\binom{3}{3}=1$ and $\sum _{j=3}^{n}\binom{j-1}{2} = \sum _{j=3}^{3}\binom{j-1}{2} = \binom{3-1}{2}=\binom{2}{2}=1$, so the proposition holds for $n=3$.$

$\noindent Now suppose that the proposition holds for some particular integer $k\geq 3$, so $\sum _{j=3}^{k}\binom{j-1}{2} = \binom{k}{3} \quad (*)$.$

$\noindent We need to show that $\sum _{j=3}^{k+1}\binom{j-1}{2} = \binom{k+1}{3}$.$

$\noindent We have$

$\begin{align*}\sum _{j=3}^{k+1}\binom{j-1}{2} &= \sum _{j=3}^{k}\binom{j-1}{2} +\binom{(k+1)-1}{2} \\ &= \binom{k}{3}+\binom{k}{2} \quad (\text{using the inductive hypothesis } (*)) \\ &= \binom{k+1}{3} \quad (\text{from the result of part (b) (i)}),\end{align*}$

$\noindent as required.$

$\noindent It follows by the Principle of Mathematical Induction that the proposition is true for all integers $n\geq 3$. $\blacksquare$$

appleibeats · Jan 15, 2016

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

Sorry, I also don't get part i) of the question

Also, should I know that relationship? Don't remember learning it.

InteGrand · Jan 15, 2016

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

appleibeats said:
Sorry, I also don't get part i) of the question

Also, should I know that relationship? Don't remember learning it.

Yes, that's a crucial relationship to know. For part (i), use the factorial definition of the binomial coefficients and take common denominators and simplify etc. (you'll need to use that relation).

Blitz_N7 · Jan 17, 2016

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

Part a asks to find the maximum velocity and it is clear that it occurs when t=10 i.e. 20m/s but I'm not sure how to do a proper working out :/
Also I'm stuck on figuring out the total distance.

InteGrand · Jan 17, 2016

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

Blitz_N7 said:
Part a asks to find the maximum velocity and it is clear that it occurs when t=10 i.e. 20m/s but I'm not sure how to do a proper working out :/
Also I'm stuck on figuring out the total distance.

$\noindent For the first 10 seconds, the car is accelerating at a constant rate of $2\text{ m}/\text{s}^2$ (so it is constantly picking up speed during this time). Since $\Delta v = a \Delta t$, where $a$ is the constant acceleration, $\Delta t$ is the time this acceleration takes place for, and $\Delta v$ is the change in velocity, we have in the first ten seconds $\Delta v = 2 \times 10 = 20\text{ m}/\text{s}$. As it is constantly increasing speed in the first ten seconds, its fastest speed is at $t=10$, with a speed of 20 metres per second (as it started off stationary). After this time it never speeds up (it only has zero acceleration or a deceleration, i.e. slowing down, and it slows down uniformly to a speed of 0 at the end).$

$\noindent During the first ten seconds, the acceleration is a constant $a$, so its distance travelled is $s=\frac{1}{2}at^2 = \frac{1}{2}\times 2 \times 10^2 = 100$ m. It then travels for 30 seconds at a constant speed of $20\text{ m}/\text{s}$, so it travels $20\times 30 = 600$ m during this time. It then slows down from $20 \text{ m}/\text{s}$ at $1 \text{ m}/\text{s}^2$ for 20 seconds, so the distance travelled during this time is $s=ut+\frac{1}{2}at^2 = 20\times 20 + \frac{1}{2}\cdot (-1)\cdot 20^2 = 200$ m. So the overall distance travelled is $100 + 600 + 200 = 900$ m.$

Blitz_N7 · Jan 18, 2016

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

Thank you, but is there a way to do it using calculus instead of physics formulae?

InteGrand · Jan 18, 2016

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

Blitz_N7 said:
Thank you, but is there a way to do it using calculus instead of physics formulae?

$\noindent Yes, you essentially end up deriving the physics formulae. Like when the acceleration is a constant 2 (all units in SI units), then $\ddot{x}=2$. Integrating gives $\dot{x} = 2t + c_1$. When $t=0, \dot{x}$ is given to be 0, so $c_1 = 0$. So $\dot{x} = 2t$. Hence $x = t^2 + c_2$ (integrating again). Defining displacement to be 0 at time 0, we have $c_2 = 0$. Hence $x = t^2$ for the first ten seconds. This lets us find the distance travelled in the first ten seconds. Also $v=2t$ for the first ten seconds as we found, so at $t=10$, $v = 20$, etc.$

$\noindent (Actually in the HSC, you're not meant to use the physics formulae, you have to derive all the relevant equations by integration etc.)$

appleibeats · Jan 18, 2016

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

Find the indefinite integral of the following;

Integral of cos^2 (x/2) dx

The answer says = 1/2 Integral of (cos x + 1) dx

I don't get this part of the working.

InteGrand · Jan 18, 2016

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

appleibeats said:
Find the indefinite integral of the following;

Integral of cos^2 (x/2) dx

The answer says = 1/2 Integral of (cos x + 1) dx

I don't get this part of the working.

$\noindent This is due to the following trig. identity: $\cos^2 \theta = \frac{1}{2}\left(1+\cos 2\theta \right)$.$

Blitz_N7 · Jan 26, 2016

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

I'm trying to do part c however the answer I get is 41.268... and the answer at the back of the book is 50.
My working out is the following:
http://imgur.com/7PO10sn
Thank you.

InteGrand · Jan 26, 2016

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

Blitz_N7 said:
I'm trying to do part c however the answer I get is 41.268... and the answer at the back of the book is 50.
My working out is the following:
http://imgur.com/7PO10sn
Thank you.

$\noindent Note that as $\alpha$ is acute, $\sin \alpha,\cos \alpha >0$. Since $\tan \alpha = \frac{2}{5}$, we can easily see by constructing a right-angled triangle that $\sin \alpha = \frac{2}{\sqrt{29}}$ and $\cos \alpha = \frac{5}{\sqrt{29}}$.$

$\noindent Let $X$ be the ball's horizontal displacement left of the wall and $Y$ its vertical displacement from the ground, with $t$ being the time since the ball hits the wall. Then the equations of motion for the ball after the rebound are $X=0.2v t\cos \alpha $ and $Y=52.5+0.2vt\sin \alpha -5t^2$, where $\alpha = \tan^{-1} \frac{2}{5}$ and $v=5\sqrt{58}$ (all units SI Units). These equations are easy to derive via calculus. When the ball hits the ground, we have $t>0$ and $Y=0$, so at this time$

$\begin{align*}52.5+0.2vt\sin \alpha -5t^2 &= 0 \quad (\text{setting }Y=0) \\ \Longleftrightarrow 5t^2 -0.2vt \sin \alpha -52.5 &= 0 \\ \Rightarrow t &= \frac{0.2v\sin \alpha \pm \sqrt{0.04v^2 \sin^2 \alpha -4 \cdot 5 \cdot (-52.5)}}{2\cdot 5} \quad (\text{quadratic formula}) \\ &= \frac{0.2 \cdot 5\sqrt{58} \cdot \frac{2}{\sqrt{29}}+\sqrt{0.04 \cdot 5^2 \cdot 58 \cdot \frac{4}{29} + 1050}}{10} \quad (\text{taking the positive solution and using }v=5\sqrt{58}) \\ &= \frac{2\sqrt{2}+\sqrt{1058}}{10}.\end{align*}$

$\noindent At this time, we have$

$\begin{align*}X&= 0.2 v \cos \alpha \times \frac{2\sqrt{2}+\sqrt{1058}}{10} \quad (\text{substituting into }X\text{ equation})\\ &= \sqrt{58}\cdot \frac{5}{\sqrt{29}}\times \frac{2\sqrt{2}+\sqrt{1058}}{10} \quad (\text{substituting values for known quantities }v \text{ and }\cos \alpha) \\ &= 5\sqrt{2} \times \frac{2\sqrt{2}+\sqrt{1058}}{10} \\ &= \frac{1}{2}\times \left(2\times 2 + \sqrt{1058\times 2}\right) \\ &= \frac{1}{2}\left(4 + 46 \right) \\ &= \frac{1}{2}\times 50 \\ &= 25.\end{align*}$

$\noindent So the ball lands 25 m from the wall, which is 50 m from Gee Ming.$

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