stressedadfff
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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?...
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maths 2019 (1 Viewer)
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stressedadfff
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oh why would we equate them... im so confused
A.void.that.thinks
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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
The 2019 paper is indeed from the old syllabus, back when Mathematics Advanced was called Mathematics
stressedadfff
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ohhhhhh thank god!!The 2019 paper is indeed from the old syllabus, back when Mathematics Advanced was called Mathematics
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 
occurs when 
.
The formula given is easily found using similar triangles, as @Drongoski showed, or can be done using Pythagoras' Theorem, as @cossine suggested:
})
})
} \\ (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 
where 
is a positive constant, the length of side 
.
We then seek to maximise})
:
} \\ &= \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*})
$.})
Note: The calculus here is much easier if you get rid of the square root by defining another variable:
}\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
.
The formula given is easily found using similar triangles, as @Drongoski showed, or can be done using Pythagoras' Theorem, as @cossine suggested:
We then seek to maximise
Note: The calculus here is much easier if you get rid of the square root by defining another variable:
This has allowed us to avoid differentiating the function
Life'sHard
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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
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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.