HSC 2015 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Oct 1, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$No, you need to vary the angle of projection $\theta$, as well as the ``sideways'' angle ($\phi$ say).$

VATTTT D: I dont see it

InteGrand · Oct 1, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
VATTTT D: I dont see it

$(Make sure to draw a diagram, in 3D.)$

$Realised that the question tells us that the segment is parabolic, so we don't need to vary the sideways angle $\varphi$, we just need to find three points on this parabola and then we can find its equation, and then the area. The easiest three to find are the two roots and the $Y$-intercept (where the $X-Y$ plane is the wall, with the origin $O_{X-Y}$ at the base of the wall in line with the nozzle, the $Y$-axis being perpendicular to the ground on the wall, and the $X$-axis being parallel to the ground along the base of the wall).$

$Recall that the maximum range of a projectile projected with speed $V$ and angle of projection $\theta$ is $R_{\text{max.}}=\frac{V^2}{g}$ (need to derive this in the exam). Note that as $V=2\sqrt{gd}$, we have $R_{\text{max.}} = 4d$ and $d=\frac{1}{4}R_{\text{max.}}$.$

$Let $N$ be the point where the nozzle is, and $A$ and $B$ be the furthest points on the bottom of the wall that the sprinkler can reach. Note that $O_{X-Y}A = O_{X-Y}B$ by symmetry. Also, $NA=NB=4d$ (since the points $A$ and $B$ are points of max. range for the sprinkler). By Pythagoras' theorem in $\triangle O_{X-Y}AN$, we have $r:=O_{X-Y}A = \sqrt{(4d)^2 - d^2}=\sqrt{15}d$. Hence the parabola is of the form $Y=a\left(X-r \right)\left(X+r \right) = a\left( X^2 -15d^2\right)$, where $a<0$ We need to determine $a$.$

$When the nozzle shoots straight ahead, the furthest point up the wall it will reach is when its peak is at the wall (this is because the distance to the wall, $d$, is less than $R_{\text{max.}}$ for the sprinkler). It is easy to derive the following formula for the height of a projectile $y$ when it has travelled a horizontal distance $d$: $y = -\frac{g\sec^2 \theta}{2V^2}d^2 + d\tan \theta$ (again, need to derive in exam). As $V^2 = 4gd$, this simplifies to the following for us:$

$y=-\frac{1}{8}\sec^2 \theta d + d\tan \theta$

$\Rightarrow y = -\frac{1}{8}(\tan^2 \theta +1) d + d\tan \theta$

$$\Rightarrow y = -\frac{d}{8}T^2 + d\cdot T - \frac{d}{8}$, where $T\equiv \tan \theta$. Hence the height is maximised at $d$ when the angle satisfies $T=-\frac{d}{2\times \left(-\frac{d}{8} \right)}=4$. Plugging in to our formula, $y_{\text{max.}}=-\frac{d}{8}\times 4^2 + 4d - \frac{d}{8}=\frac{15d}{8}$.$

$So on the parabola $Y = a\left( X^2 - 15d^2\right)$, we have $Y = \frac{15d}{8}$ when $X=0$. Substituting these values and solving for $a$, we find that $a=-\frac{1}{8d}$. Therefore, the required area $\mathcal{A}$ is:$

$\mathcal{A}=\int _{-\sqrt{15}d} ^{\sqrt{15}d}-\frac{1}{8d}\left(X^2 - 15d^2 \right) \text{ d}X$

$=2\int _{0} ^{\sqrt{15}d}-\frac{1}{8d}\left(X^2 - 15d^2 \right) \text{ d}X$

$=-\frac{1}{4d}\int _{0} ^{\sqrt{15}d}\left(X^2 - 15d^2 \right) \text{ d}X$

$=-\frac{1}{4d}\left[\frac{X^3}{3}-15d^2 X \right]_{X=0} ^{X=\sqrt{15}d}$

$=-\frac{1}{4d}\left(\frac{15\sqrt{15}}{3}d^3-15d^2 \times \sqrt{15}d \right)$

$=-\frac{1}{4d}\left(5\sqrt{15}-15\sqrt{15} \right)d^3$

$=\frac{1}{4}\left(10\sqrt{15} \right)d^2$

$$=\frac{5}{2}\sqrt{15}d^2$, as required.$

lita1000 · Oct 1, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$(Make sure to draw a diagram, in 3D.)$

$Realised that the question tells us that the segment is parabolic, so we don't need to vary the sideways angle $\varphi$, we just need to find three points on this parabola and then we can find its equation, and then the area. The easiest three to find are the two roots and the $Y$-intercept (where the $X-Y$ plane is the wall, with the origin $O_{X-Y}$ at the base of the wall in line with the nozzle, the $Y$-axis being perpendicular to the ground on the wall, and the $X$-axis being parallel to the ground along the base of the wall).$

$Recall that the maximum range of a projectile projected with speed $V$ and angle of projection $\theta$ is $R_{\text{max.}}=\frac{V^2}{g}$ (need to derive this in the exam). Note that as $V=2\sqrt{gd}$, we have $R_{\text{max.}} = 4d$ and $d=\frac{1}{4}R_{\text{max.}}$.$

$Let $N$ be the point where the nozzle is, and $A$ and $B$ be the furthest points on the bottom of the wall that the sprinkler can reach. Note that $O_{X-Y}A = O_{X-Y}B$ by symmetry. Also, $NA=NB=4d$ (since the points $A$ and $B$ are points of max. range for the sprinkler). By Pythagoras' theorem in $\triangle O_{X-Y}AN$, we have $r:=O_{X-Y}A = \sqrt{(4d)^2 - d^2}=\sqrt{15}d$. Hence the parabola is of the form $Y=a\left(X-r \right)\left(X+r \right) = a\left( X^2 -15d^2\right)$, where $a<0$ We need to determine $a$.$

$When the nozzle shoots straight ahead, the furthest point up the wall it will reach is when its peak is at the wall (this is because the distance to the wall, $d$, is less than $R_{\text{max.}}$ for the sprinkler). It is easy to derive the following formula for the height of a projectile $y$ when it has travelled a horizontal distance $d$: $y = -\frac{g\sec^2 \theta}{2V^2}d^2 + d\tan \theta$ (again, need to derive in exam). As $V^2 = 4gd$, this simplifies to the following for us:$

$y=-\frac{1}{8}\sec^2 \theta d + d\tan \theta$

$\Rightarrow y = -\frac{1}{8}(\tan^2 \theta +1) d + d\tan \theta$

$$\Rightarrow y = -\frac{d}{8}T^2 + d\cdot T - \frac{d}{8}$, where $T\equiv \tan \theta$. Hence the height is maximised at $d$ when the angle satisfies $T=-\frac{d}{2\times \left(-\frac{d}{8} \right)}=4$. Plugging in to our formula, $y_{\text{max.}}=-\frac{d}{8}\times 4^2 + 4d - \frac{d}{8}=\frac{15d}{8}$.$

$So on the parabola $Y = a\left( X^2 - 15d^2\right)$, we have $Y = \frac{15d}{8}$ when $X=0$. Substituting these values and solving for $a$, we find that $a=-\frac{1}{8d}$. Therefore, the required area $\mathcal{A}$ is:$

$\mathcal{A}=\int _{-\sqrt{15}d} ^{\sqrt{15}d}-\frac{1}{8d}\left(X^2 - 15d^2 \right) \text{ d}X$

$=2\int _{0} ^{\sqrt{15}d}-\frac{1}{8d}\left(X^2 - 15d^2 \right) \text{ d}X$

$=-\frac{1}{4d}\int _{0} ^{\sqrt{15}d}\left(X^2 - 15d^2 \right) \text{ d}X$

$=-\frac{1}{4d}\left[\frac{X^3}{3}-15d^2 X \right]_{X=0} ^{X=\sqrt{15}d}$

$=-\frac{1}{4d}\left(\frac{15\sqrt{15}}{3}d^3-15d^2 \times \sqrt{15}d \right)$

$=-\frac{1}{4d}\left(5\sqrt{15}-15\sqrt{15} \right)d^3$

$=\frac{1}{4}\left(10\sqrt{15} \right)d^2$

$$=\frac{5}{2}\sqrt{15}d^2$, as required.$

Pretty sure the question asks you to PROVE that it is a parabolic segment

InteGrand · Oct 1, 2015

Re: HSC 2015 4U Marathon

lita1000 said:
Pretty sure the question asks you to PROVE that it is a parabolic segment

$Oh right, misread that part. In that case, if we go a horizontal displacement $X$ along the wall's base from $O_{X-Y}$ (call this point $P_X$), and $D$ is the length $NP_X$, then from$

$$y=-\frac{g}{2V^2}D^2 T^2 + DT - \frac{g}{2V^2}D^2$ (height at wall of water sprinkler's jet, where $T\equiv \tan \theta$, where $\theta$ is the angle of projection), we achieve the maximal height $Y=y_\text{max.}$ by choosing $\theta$ so that $T=\frac{V^2}{gD}$ (everything lower than this height can also be achieved by taking a smaller $T$, which can always be done). So substituting this in gives us $Y=-\frac{g}{2V^2}\times \frac{V^4}{g^2 D^2}D^2 + D\times \frac{V^2}{gD}-\frac{g}{2V^2}D^2\Rightarrow Y = \frac{V^2}{2g}-\frac{g}{2V^2}D^2.$ Since $D^2 = X^2 + d^2$ (Pythagoras's theorem), $Y$ is a quadratic function of $X$, so the required segment is parabolic as required. With a bit more algebra, we can see that it comes out to the same parabola as before: $Y=\frac{g}{2V^2}\left\{\left[ \left(\frac{V^2}{g} \right)^2 -d^2 \right] - X^2 \right \}$ (sub. $V^2 = 4gd$).$

braintic · Oct 1, 2015

Re: HSC 2015 4U Marathon

lita1000 said:
$$A tall building stands on level ground. The nozzle of a water sprinkler is positioned at a point P on the ground at a distance d from a wall of the building. Water sprays from the nozzle with speed V and the nozzle can be pointed in any direction from P. Suppose that $V= 2(gd)^{1/2}. $Show that the portion of the wall that can be sprayed with water is a parabolic segment with area$ (5/2)(d^2 * (15)^{1/2})$

Isn't this a 60s or 70s HSC question?

InteGrand · Oct 1, 2015

Re: HSC 2015 4U Marathon

braintic said:
Isn't this a 60s or 70s HSC question?

Yes, I remember it from a past HSC from those times, but can't remember the exact year, but would surely be between 1960s to 1980.

Drsoccerball · Oct 2, 2015

Re: HSC 2015 4U Marathon

braintic said:
Isn't this a 60s or 70s HSC question?

Please don't tell me this was your HSC question when you did it

Carrotsticks · Oct 2, 2015

Re: HSC 2015 4U Marathon

braintic said:
Isn't this a 60s or 70s HSC question?

InteGrand said:
Yes, I remember it from a past HSC from those times, but can't remember the exact year, but would surely be between 1960s to 1980.

It also features in the Extension section of the projectile motion chapter in of the 3 unit Cambridge book, if I recall correctly.

braintic · Oct 2, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Please don't tell me this was your HSC question when you did it

Ah ... no. I just found it in the '74 paper. At that stage I was probably practising my times tables.

Drsoccerball · Oct 2, 2015

Re: HSC 2015 4U Marathon

Okay i made this question lets see how bad it turns out:

$i) $ For $ 0<x<1 $ prove via induction : $ sin(x) \geq x -\frac{x^3}{3!}+ \frac{x^5}{5!}- \frac{x^7}{7}+...+(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)!} $ for even integers $ n\geq 2$

$ii) $ Hence, or otherwise prove that: $ 0= -\pi + \frac{\pi ^3}{3!} - \frac{\pi ^5}{5!}+...$

glittergal96 · Oct 3, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Okay i made this question lets see how bad it turns out:

$i) $ For $ 0<x<1 $ prove via induction : $ sin(x) \geq x -\frac{x^3}{3!}+ \frac{x^5}{5!}- \frac{x^7}{7}+...+(-1)^{n+1}\frac{x^{2n-1}}{n!} $ for $ n\geq 2$

$ii) $ Hence, or otherwise prove that: $ 0= -\pi + \frac{\pi ^3}{3!} - \frac{\pi ^5}{5!}+...$

EDIT: Let me fix up the domain gimme a minute.

$i) Define:\\ \\ $P_n(x)=\sum_{k=0}^{n} (-1)^k\frac{x^{2k+1}}{(2k+1)!}.$\\ \\ We claim: \\ \\$P_{2n+1}(x)\leq \sin(x) \leq P_{2n}(x)$ for all non-negative integers $n$ and $x\geq 0$. \\ \\ It is well known that $\sin(x)\leq x=P_0(x)$ for all $x\geq 0$.\\ \\ If we have $P_{2n}(t)-\sin(t)\geq 0$ for $t\geq 0$ then we also have\\ \\ $\sin(x)-P_{2n+1}(x)=\int_0^{x}\int_0^s P_{2n}(t)-\sin(t) \, dt \, ds\geq 0 \quad (1).$\\ \\ On the other hand, if we have $P_{2n-1}(t)-\sin(t)\leq 0$ for $x\geq 0$ then we also have\\ \\ $\sin(x)-P_{2n}(x)=\int_0^{x}\int_0^s P_{2n-1}(t)-\sin(t) \, dt \, ds\leq 0 \quad (2).$\\ \\ Alternating use of facts (1) and (2) complete the proof via induction.\\ \\ii) We also have $|P_{n}(x)-\sin(x)|\leq |P_{n}(x)-P_{n+1}(x)|=\frac{x^{2n+3}}{(2n+3)!}\rightarrow 0$ for all $x\geq 0$ as factorials dominate powers. \\ Hence we can conclude that \\$RHS=\lim_{n\rightarrow\infty}-P_n(\pi)=-\sin(\pi)=0.$$

Drsoccerball · Oct 3, 2015

Re: HSC 2015 4U Marathon

glittergal96 said:
$i) Define:\\ \\ $P_n(x)=\sum_{k=0}^{n} (-1)^k\frac{x^{2k+1}}{(2k+1)!}.$\\ \\ We claim: \\ \\$P_{2n+1}(x)\leq \sin(x) \leq P_{2n}(x)$ for all non-negative integers $n$ and $x\geq 0$. \\ \\ It is well known that $\sin(x)\leq x=P_0(x)$ for all $x\geq 0$.\\ \\ If we have $P_{2n}(t)-\sin(t)\geq 0$ for $t\geq 0$ then we also have\\ \\ $\sin(x)-P_{2n+1}(x)=\int_0^{x}\int_0^s P_{2n}(t)-\sin(t) \, dt \, ds\geq 0 \quad (1).$\\ \\ On the other hand, if we have $P_{2n-1}(t)-\sin(t)\leq 0$ for $x\geq 0$ then we also have\\ \\ $\sin(x)-P_{2n}(x)=\int_0^{x}\int_0^s P_{2n-1}(t)-\sin(t) \, dt \, ds\leq 0 \quad (2).$\\ \\ Alternating use of facts (1) and (2) complete the proof via induction.\\ \\ii) We also have $|P_{n}(x)-\sin(x)|\leq |P_{n}(x)-P_{n+1}(x)|=\frac{x^{2n+3}}{(2n+3)!}\rightarrow 0$ for all $x\geq 0$ as factorials dominate powers. \\ Hence we can conclude that \\$RHS=\lim_{n\rightarrow\infty}-P_n(\pi)=-\sin(\pi)=0.$$

I changed the question to even numbers since its not true for odd. Also you could've just subbed in negative pi

glittergal96 · Oct 3, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I changed the question to even numbers since its not true for odd. Also you could've just subbed in negative pi

What I wrote is 100% true, the even partial sums are upper bounds and the odd partial sums are lower bounds. Also, if you only proved the claim about upper bounds this alone would not establish convergence to the sine function.

And because of the way I set it up, I only proved these bounds for non-negative x (of course the flipped bounds hold for negative x as both sin and the polynomials involved are odd), this is why I substituted a positive number in.

asdfghjklqwerty · Oct 3, 2015

Re: HSC 2015 4U Marathon

hey can someone do part b for me? im not quite sure how to do part b. thanks

Paradoxica · Oct 3, 2015

Re: HSC 2015 4U Marathon

Consider a small slice of the volume desired. Approximate this volume by taking one of the cross-sectional planes of the volume at some value of $y$ , and extending the section out by some value of $y$ , which we shall call $\delta y$ . Then this volume will approximate a small slice of the volume.

The base of this has a length that is dependent upon the x-co-ordinates corresponding with the value of $y$ we chose. Solving for $x$ , we get $x=\pm \frac{a}{b}\sqrt{b^2-y^2}$ .

Because we want the length of the base, we take the positive value and double it. Then the length of the base is simply $2\frac{a}{b}\sqrt{b^2-y^2}$

The shape of the cross-sectional slice is similar to the shape from a). Scaling down this shape allows us to find the area of the cross section. We express the area of the slice as a percentage of the original shape in a). This percentage is found by dividing the cross-sectional base value by the base value of the original shape. So the percentage is $\frac{2\frac{a}{b}\sqrt{b^2-y^2}}{2a}=\frac{1}{b}\sqrt{b^2-y^2}$

The area of the slice is then the percentage times the original area, i.e. $ab\left(\sqrt{2}+\ln{(1+\sqrt{2})}\right ) \cdot \frac{1}{b}\sqrt{b^2-y^2}=a\left(\sqrt{2}+\ln{(1+\sqrt{2})}\right )\sqrt{b^2-y^2}$

We now have the area of the slice, and hence, the volume of the slice is $a\left(\sqrt{2}+\ln{(1+\sqrt{2})}\right )\sqrt{b^2-y^2}\delta y$
Summing this from $-b$ to $b$ will approximate the entire volume. However, observe that the solid is symmetrical along the x-axis. This means we only have to double the volume of half the solid, i.e, we can sum from $0$ to $b$ instead.

$2\sum_{y=0}^{b}a\left(\sqrt{2}+\ln{(1+\sqrt{2})} \right )\sqrt{b^2-y^2}\delta y = 2a\left(\sqrt{2}+\ln{(1+\sqrt{2})} \right )\sum_{y=0}^{b}\sqrt{b^2-y^2}\delta y$ since $a\left (\sqrt{2}+\ln{(1+\sqrt{2})} \right )$ is constant.

As we make $\delta y$ smaller and smaller, the volume approximates the true volume of the solid. In the limit, it becomes equal to the true volume (Rigorous justification beyond High School level

).

i.e. ignoring the outside constant for now, we have $\lim_{\delta y \rightarrow 0}\sum_{y=0}^{b}\sqrt{b^2-y^2}\delta y = \int_0^b{\sqrt{b^2-y^2}}dy$

This integral can now be evaluated to obtain the volume, which matches the given answer exactly.

asdfghjklqwerty · Oct 4, 2015

Re: HSC 2015 4U Marathon

A 3 metre string AB has a mass of 5kg attached at point B. The string is rotated in a horizontal circle about A and breaks as soon as it exceeds a speed of rotation of 45 revolutions per minute. Find the maximum possible tension in the string.

Answer: T=135pi/4 N.
THANKS

Drsoccerball · Oct 4, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
A 3 metre string AB has a mass of 5kg attached at point B. The string is rotated in a horizontal circle about A and breaks as soon as it exceeds a speed of rotation of 45 revolutions per minute. Find the maximum possible tension in the string.

Answer: T=135pi/4 N.
THANKS

Yes! can someone do this I forgot how to do these type of questions

asdfghjklqwerty · Oct 4, 2015

Re: HSC 2015 4U Marathon

Dw i got it the answer was wrong its ok ahahaha

InteGrand · Oct 4, 2015

Re: HSC 2015 4U Marathon

$The four complex numbers $z_1, z_2, z_3, z_4$ are represented on the complex (Argand) plane by the points $A, B, C, D$ respectively. If $z_1 - z_2 + z_3 - z_4 = 0$ and $z_1 - iz_2 - z_3 + iz_4 = 0$, determine the possible shape(s) for the quadrilateral $ABCD$. (This question is from the 1970 HSC 4U paper.)$

Drsoccerball · Oct 4, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
Dw i got it the answer was wrong its ok ahahaha

Still can you show what you did

...

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