maths 2019 (1 Viewer)

stressedadfff · Sep 27, 2021

hey guys i realised that theres many questions from the old syllabus from this paper does anyone know why, also are we suppose to know how to do 15b?...

Screen Shot 2021-09-27 at 11.03.56 am.png

cossine · Sep 27, 2021

Try finding AC and BC.

Then use pythgoras to equate AC, BC with AB

stressedadfff · Sep 27, 2021

cossine said:
Try finding AC and BC.

Then use pythgoras to equate AC, BC with AB

oh why would we equate them... im so confused

Drongoski · Sep 27, 2021

For 15b:

$h = \sqrt{pq} $ is equivalent to $ h^2 = pq $ or equivalently $ \frac{h}{p} = \frac {q}{h}$

It can be shown that triangles ADC and CDB are SIMILAR and therefore their corresponding sides are proportional.
In particular:

$\frac{CD}{AD} = \frac{DB}{DC}\\ \\ $i.e. $ \frac{h}{p} = \frac{q}{h}\\ \\ \therefore h^2 = pq $ and $ \therefore h = \sqrt{pq}$

A.void.that.thinks · Sep 27, 2021

I’m fairly sure that the 2019 paper is from the old course so that’s why a lot of the questions are part of the old syllabus

jimmysmith560 · Sep 27, 2021

The 2019 paper is indeed from the old syllabus, back when Mathematics Advanced was called Mathematics

stressedadfff · Sep 27, 2021

jimmysmith560 said:
The 2019 paper is indeed from the old syllabus, back when Mathematics Advanced was called Mathematics

ohhhhhh thank god!!

CM_Tutor · Sep 27, 2021

The new syllabus (2020 HSC onwards) has greatly reduced the inclusion of geometry, but it can still be asked as a part of something else as it is assumed knowledge. For example, as a MAX / MIN problem... holding the length AB constant, show that the maximum value of $h$ occurs when $p=q$ .

The formula given is easily found using similar triangles, as @Drongoski showed, or can be done using Pythagoras' Theorem, as @cossine suggested:

$\text{In } \Delta ADC: \quad AC^2 = AD^2 + DC^2 = p^2 + h^2 \quad \text{(Pyth)}$

$\text{In } \Delta BDC: \quad BC^2 = BD^2 + DC^2 = q^2 + h^2 \quad \text{(Pyth)}$

$\begin{align*} \text{In } \Delta ABC: \quad AB^2 &= AC^2 + BC^2 \quad \text{(Pyth)} \\ (p + q)^2 &= \big(p^2 + h^2\big) + \big(q^2 + h^2\big) \quad \text{using the results above for $AC^2$ and $BC^2$} \\ p^2 + 2pq + q^2 &= p^2 + q^2 + 2h^2 \\ 2pq &= 2h^2 \\ h &= \sqrt{pq} \quad \text{as $h$, $p$, and $q$ are all distances and thus are positive} \end{align*}$

For the context I suggested, a MAX/MIN problem, we'd have $p + q = D$ where $D$ is a positive constant, the length of side $AB$ .

We then seek to maximise $h = \sqrt{p(D - p)}$ :

$\begin{align*} h &= \sqrt{p(D - p)} \\ &= \sqrt{pD - p^2} \\ &= \left(pD - p^2\right)^{\frac{1}{2}} \\ \cfrac{dh}{dp} &= \cfrac{1}{2}\left(pD - p^2\right)^{-\frac{1}{2}} \times (D - 2p) \qquad \qquad \text{(Chain Rule)} \\ &= \cfrac{D - 2p}{2\sqrt{pD - p^2}} \\ \text{For a stationary point, put $\cfrac{dh}{dp} = 0$:} \quad \cfrac{D - 2p}{2\sqrt{pD - p^2}} &= 0 \\ D - 2p &= 0 \qquad \text{noting $\sqrt{pD - p^2} > 0$ as $0 < p < D$} \\ p &= \cfrac{D}{2} \end{align*}$

$\text{Now, if $0 < p < \cfrac{D}{2}$ then $D - 2p > 0$ and so $\cfrac{dh}{dp} > 0$ and $h$ is increasing in value.}$

$\text{Similarly, if $\cfrac{D}{2} < p < D$ then $\cfrac{dh}{dp} < 0$ and $h$ is decreasing.}$

$\text{So, $h$ has a maximum turning point when $p = \cfrac{D}{2}$.}$

$\text{At this point, $q = D - p = D - \cfrac{D}{2} = \cfrac{D}{2}$, and so this maximum value of $h$ occurs when $p = q$ $\left(\text{as both are}\ \cfrac{D}{2}\right)$.}$

Note: The calculus here is much easier if you get rid of the square root by defining another variable:

$\begin{align*} \text{Let} \quad V &= h^2 \qquad \qquad \qquad \implies \qquad \cfrac{dV}{dh} = 2h \\ &= \left(\sqrt{p(D - p)}\right)^2 \\ &= pD - p^2 \qquad \qquad \implies \qquad \cfrac{dV}{dp} = D - 2p \\ \text{Using the Chain Rule} \qquad \cfrac{dh}{dp} &= \cfrac{dh}{dV} \times \cfrac{dV}{dp} \\ &= \cfrac{1}{2h} \times (D - 2p) \\ &= \cfrac{D - 2p}{2h} \\ \text{For a stationary point, put $\cfrac{dh}{dp} = 0$:} \quad \cfrac{D - 2p}{2h} &= 0 \\ p &= \cfrac{D}{2} \qquad \text{as $0 < h < D$} \end{align*}$

This has allowed us to avoid differentiating the function $\sqrt{pD - p^2}$ .

Life'sHard · Sep 27, 2021

From my personal experience, geometry can show up quite a few times in vectors and a couple times in the trig sections (a bit less).

CM_Tutor · Sep 27, 2021

Life'sHard said:
From my personal experience, geometry can show up quite a few times in vectors and a couple times in the trig sections (a bit less).

Removing geometry was a bad idea, IMO - it teaches valuable skills in mathematical reasoning and proof.

You are correct that geometry issues also show up in vectors, though that is an MX1 topic. It's even more useful in parts of MX2, particularly some types of complex numbers problems, where some circle geometry theorems must be known (even though they are no longer in the MX1 or MX2 syllabi) or some problems become much more difficult.

ExtremelyBoredUser · Sep 27, 2021

CM_Tutor said:
Removing geometry was a bad idea, IMO - it teaches valuable skills in mathematical reasoning and proof.

You are correct that geometry issues also show up in vectors, though that is an MX1 topic. It's even more useful in parts of MX2, particularly some types of complex numbers problems, where some circle geometry theorems must be known (even though they are no longer in the MX1 or MX2 syllabi) or some problems become much more difficult.

Yeah I enjoyed geometry in Junior Years the most so the absence of geometry in Year 11 kinda weakened my geometry skills into Year 12 so when I'm doing complex numbers or vectors, which are my first 2 topics, the geometry that pops up feels significantly harder than any of the algebra involved - so at times I'm more comfortable with doing paragraphs of working out however I've been slowly easing back into the geometric concepts since it is more intuitive and interesting. My teacher occasionally went on rants about how Circle Geo or Conics shouldn't have been removed for statistics or other topics so I can understand his frustration now.

maths 2019 (1 Viewer)

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