[Teacher POV] 2023 HSC Extension 1 Solution Speedrun Send Help (1 Viewer)

epicmaths · Oct 20, 2023

Hiya,

In the mood to write up the solutions - the immediate feedback from my students was "hard". They probably overstudied the recent HSCs.

Backseating my poor algebra is welcome!

youtube: epicmaths

Link to First Draft of the solutions is on the youtube live page for that stream!

Luukas.2 · Oct 20, 2023

For the volume question, the bottom part is just a cylinder with $V = \pi r^2h=\pi(10)^2 \times 4 = 400\pi\ \text{cu units}$ , so skip integration.

For the top part, we can make the numbers a lot easier to deal with...

$\begin{align*} V &= \int_4^{12} \pi x^2\ \!dy \qquad \qquad \qquad \text{where $x = \frac{60}{y} - 5 = 5\left(\frac{12}{y} - 1\right)$} \\ &= (5)^2\pi \int_4^{12} \left(\frac{12}{y} - 1\right)^2\ \!dy \\ &= 25\pi \int_{y = 4}^{y = 12} \left(\frac{12}{4u} - 1\right)^2\ 4\,du \qquad \text{on substituting $y = 4u \implies dy = 4\,du$} \\ &= 100\pi \int_{1}^{3} \left(\frac{3}{u} - 1\right)^2\ \!du \\ &= 100\pi \int_{1}^{3} \frac{9}{u^2} - \frac{6}{u} + 1\ \!du \\ &= 100\pi \left[\frac{9u^{-1}}{-1} - 6\ln{|u|} + u\right]_1^3 \\ &= 100\pi \left[\frac{-9}{3} - 6\ln{3} + 3 - \left(\frac{-9}{1} - 6\ln{1} + 1\right)\right]_1^3 \\ &= 100\pi \left(-6\ln{3} + 9 + 0 - 1\right) \\ \\ \text{Thus, the combined volume } &= 400\pi + 100\pi(8 - 6\ln{3}) = 600\pi(2 - \ln{3})\ \text{cubic units} \end{align*}$

And, the question could have been asked as: Show that the volume is

$V = 600\pi\ln{\frac{e^2}{3}}\ \text{cubic units}$

=)(= · Oct 20, 2023

Luukas.2 said:
For the volume question, the bottom part is just a cylinder with $V = \pi r^2h=\pi(10)^2 \times 4 = 400\pi\ \text{cu units}$ , so skip integration.

For the top part, we can make the numbers a lot easier to deal with...

$\begin{align*} V &= \int_4^{12} \pi y^2\ \!dx \qquad \qquad \qquad \text{where $y = \frac{60}{x} - 5 = 5\left(\frac{12}{x} - 1\right)$} \\ &= (5)^2\pi \int_4^{12} \left(\frac{12}{x} - 1\right)^2\ \!dx \\ &= 25\pi \int_{x = 4}^{x = 12} \left(\frac{12}{4u} - 1\right)^2\ 4\,du \qquad \text{on substituting $x = 4u \implies dx = 4\,du$} \\ &= 100\pi \int_{1}^{3} \left(\frac{3}{u} - 1\right)^2\ \!du \\ &= 100\pi \int_{1}^{3} \frac{9}{u^2} - \frac{6}{u} + 1\ \!du \\ &= 100\pi \left[\frac{9u^{-1}}{-1} - 6\ln{|u|} + u\right]_1^3 \\ &= 100\pi \left[\frac{-9}{3} - 6\ln{3} + 3 - \left(\frac{-9}{1} - 6\ln{1} + 1\right)\right]_1^3 \\ &= 100\pi \left(-6\ln{3} + 9 + 0 - 1\right) \\ \\ \text{Thus, the combined volume } &= 400\pi + 100\pi(8 - 6\ln{3}) = 600\pi(2 - \ln{3})\ \text{cubic units} \end{align*}$

And, the question could have been asked as: Show that the volume is

$V = 600\pi\ln{\frac{e^2}{3}}\ \text{cubic units}$

wait if i did ln12/4 and didnt simplfy would i lose a mark ( ik its stupid i was stressed)

Luukas.2 · Oct 20, 2023

For 14(c), Let p be the projection of a onto b. Then the vector a - p is the the altitude of the triangle with base as b, and so

area of triangle, A = |b||a - p| / 2

It's another "otherwise" approach...

epicmaths · Oct 20, 2023

[[drive.google.com/drive/folders/1TWaMpXobf7MHLrhus5L3YmUx5wcR69nH?usp=sharing]]

The pdf from the past 2 hours.
Aye I'll collate some of the better ways for 14c as they come in.

Luukas.2 · Oct 20, 2023

For 14(c)(ii), the product rule can be avoided. Working towards the form of A that I have here may be quicker, too.

$\begin{align*} A &= Rr|\sin{t}(1 + \cos{t})| = Rr|\sin{t} + \sin{t}\cos{t}| = \frac{Rr}{2}\left|2\sin{t} + 2\sin{t}\cos{t}\right| = \frac{Rr}{2}\left|2\sin{t} + \sin{2t}\right| \\ \\ \frac{dA}{dt} &= \frac{Rr}{2}\left|2\cos{t} + 2\cos{2t}\right| = Rr\left|\cos{t} + 2\cos^2{t} - 1\right| = Rr\left|(2\cos{t} - 1)(\cos{t} + 1)\right| \end{align*}$

There are two solutions, $t = \pm\frac{\pi}{3}$ , because the domain is $t \in \left[-\pi,\,\pi\right]$ and the function is even. The stationary points at $t = \pm\pi$ are the minima.

Luukas.2 · Oct 20, 2023

Also, 11e is easy by sketching y = sin x and y = cos x and adding... the solution x = 0, pi/2, 2pi drops immediately.

tywebb · Oct 20, 2023

Luukas.2 said:
For 14(c), Let p be the projection of a onto b. Then the vector a - p is the the altitude of the triangle with base as b, and so

area of triangle, A = |b||a - p| / 2

It's another "otherwise" approach...

Here is yet another "otherwise" approach.

$\text{Area of triangle spanned by } \overrightarrow{OA},\overrightarrow{OB}\text{ is}$

$\textstyle\frac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}|=\frac{1}{2}\left|\begin{vmatrix}\bold i&\bold j&\bold k\\a_1&a_2&0\\b_1&b_2&0\end{vmatrix}\right|=\frac{1}{2}|(a_1b_2-a_2b_1)\bold k|=\frac{1}{2}|a_1b_2-a_2b_1|$

Luukas.2 · Oct 20, 2023

tywebb said:
Here is yet another "otherwise" approach.

$\text{Area of triangle spanned by } \overrightarrow{OA},\overrightarrow{OB}\text{ is}$

$\textstyle\frac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}|=\frac{1}{2}\left|\begin{vmatrix}\bold i&\bold j&\bold k\\a_1&a_2&0\\b_1&b_2&0\end{vmatrix}\right|=\frac{1}{2}|(a_1b_2-a_2b_1)\bold k|=\frac{1}{2}|a_1b_2-a_2b_1|$

Yes, I saw it, thanks. It's definitely the easy way so long as you have covered cross products of 3D vectors, which is beyond MX2 content and well beyond MX1.

The approach that @epicmaths took, using $ab\sin{C} / 2$ , and Lucid's solutions, are both "otherwise" methods too.

I really wonder about the intended method using the information given...

tywebb · Oct 20, 2023

Luukas.2 said:
It's definitely the easy way.

Well in maths I like what Kennedy didn’t say.

He didn’t say “We don’t do it because it hard. We do it because it’s easy”

Luukas.2 · Oct 20, 2023

I figured out what they wanted for 14(c)(i), the non-otherwise approach.

Take vectors OA and OB as having lengths A and B, respectively. Without loss of generality, assume that OA has the larger angle between itself and the x-axis than has OB.

Take OB' = b₂i - b₁j so (from the information given) it is perpendicular to OB and still has magnitude of B.

Let phi be the angle between OA and OB', so that cos(phi) = OA . OB' / AB = |a₁b₂ - a₂b₁| / AB.

The absolute value takes care of the problem if points A and B are not in quadrant 1, and for the change in sign if my assumption is wrong and so we need OA' rather than OB'.

Now, taking theta as the angle between OA and OB, the required area is

ABsin(theta) / 2 = ABcos(90 - theta) / 2 = ABcos(phi) / 2 = AB|a₁b₂ - a₂b₁| / 2AB = |a₁b₂ - a₂b₁| / 2

as required.

Like @epicmaths, I kept wanting to use the given information to get at a half times base times height approach, but it is actually giving us a complementary angle.

2azn4u · Oct 21, 2023

Dropbox

www.dropbox.com

Here are my solutions. I did q14 area of triangle using the hint.

epicmaths · Oct 21, 2023

I've updated my solutions on the YT also, similar hints based solution to Luukas for Q14. I like Mok's also.

The hint is awkward and I get too paranoid about cases to want to do that under exam conditions.

Luukas.2 · Oct 21, 2023

epicmaths said:
I've updated my solutions on the YT also, similar hints based solution to Luukas for Q14. I like Mok's also.

The hint is awkward and I get too paranoid about cases to want to do that under exam conditions.

I agree that the hint is awkward. I like the way Mok has used it, although the existence of cases is not treated. I kept thinking of how the perpendicular vector gas the direction of the altitude and trying to adjust its magnitude to match the height of the triangle. I suspect it will have led many to think about A = bh / h, and consequently I wonder if it was really that helpful.

tywebb · Oct 21, 2023

Here is yet another otherwise method by Âmîňńñ

tywebb · Oct 21, 2023

Here is a modification of Âmîňńñ's method which is a bit easier (again "do it not because it's hard but because it's easy")

$\text{Area}(\triangle OAB)=|\text{Area}(\triangle OAC)+\text{Area}(\text{trapezium }ACDB)-\text{Area}(\triangle OBD)|$
$=\textstyle|\frac{1}{2}a_1a_2+\frac{1}{2}(a_2+b_2)(b_1-a_1)-\frac{1}{2}b_1b_2|$
$=\textstyle\frac{1}{2}|a_1a_2+a_2b_1+b_2b_1-a_2a_1-b_2a_1-b_1b_2|$
$=\textstyle\frac{1}{2}|a_2b_1-b_2a_1|$
$=\textstyle\frac{1}{2}|a_1b_2-a_2b_1|$

tywebb · Oct 21, 2023

Roch solutions are here: https://community.boredofstudies.or...on-1-predictions-thoughts.405500/post-7515646

These ones are typed up

epicmaths · Oct 21, 2023

The Armin and Tywebb version of 14ci is a work of art! I also tried doodling components and acting like I can prove pythagoras for a few minutes and gave up!
Mok's is probably what the writing committee intended - and boy that took projections to a whole new level, Luukas and I both missed what to really do with the perpendicular - I tried really hard to rescale it to the height also.

tywebb · Oct 21, 2023

These are essentially the same diagram rotated.

So you can do it a similar way - but with just 1 diagram

Screen Shot 2023-10-21 at 12.01.23 pm.png

Fix the solitary student for which there are 5 possiblilities.

Then there are 3! ways for the teachers and 4! ways for the other students

Answer is therefore 3!4!x5=3!5! .......... B

tywebb · Nov 14, 2023

Luukas.2 said:
I figured out what they wanted for 14(c)(i)

Here are nesa's solution:

from the mg at https://educationstandards.nsw.edu....xam-pack+++++++++++++++++++++++++++++++++++++

[Teacher POV] 2023 HSC Extension 1 Solution Speedrun Send Help (1 Viewer)

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