HSC 2017-2018 Maths Marathon (1 Viewer)

1729 · Apr 9, 2017

Re: HSC 2017 Maths Marathon

davidgoes4wce said:

$$\noindent \textbf{(a)} Now $0.52^{\circ} \equiv \frac{13\pi}{4500}$ radians. If we construct an altitude from earth viewpoint, this angle is halved, i.e. $\frac{13\pi}{9000}$ radians. \\\\ Now $\tan{\left(\frac{13\pi}{9000}\right)} = \frac{r_{moon}}{385000} \Rightarrow r_{moon} = 385000\tan{\left(\frac{13\pi}{9000}\right)}$. And so we have $r_{moon} = 1700$ km (correct to 2 s.f.) \\\\ Similarly, $\tan{\left(\frac{13\pi}{9000}\right)} = \frac{r_{sun}}{150000000} \Rightarrow r_{sun} = 150000000\tan{\left(\frac{13\pi}{9000}\right)}$. And so $r_{sun} = 680000$ km (correct to 2 s.f.)$

$$\noindent \textbf{(b)} $3.8$ cm per year $\Rightarrow (3.8 \times 10^9)$ cm per $10^9$ years. That is, the moon moves 38000 kilometres away from earth viewpoint, a total of 423000 km. Noting that $r_{moon}$ is constant, half of the new angle subtended by the moon $\frac{\theta}{2}$ is given by $\tan{\left(\frac{\theta}{2}\right)} = \frac{r_{moon}}{423000}$. And so $\frac{\theta}{2} \approx 0.237^{\circ} \Rightarrow \theta = 0.47^{\circ}$ (to 2 s.f.)$

$$\noindent \textbf{(c)} $15$ cm per year $\Rightarrow (15 \times 10^9)$ cm per $10^9$ years. That is, the earth moves 150000 kilometres away from the sun, a total of 150150000 km. Noting that $r_{sun}$ is constant, half of the new angle subtended by the sun $\frac{\psi}{2}$ is given by $\tan{\left(\frac{\psi}{2}\right)} = \frac{r_{sun}}{150150000}$. And so $\frac{\psi}{2} \approx 0.259^{\circ} \Rightarrow \psi = 0.52^{\circ}$ (to 2 s.f.)$

$$\noindent \textbf{(d)} The radius of the area obscured by the moon $r$ is given by $\tan{\left(\frac{0.47^{\circ}}{2}\right)} = \frac{r}{150150000}$. That is, $r = 150150000\tan{\left(\frac{47\pi}{36000}\right)}.$ Now call this obscured area $A_1$ and call the area of the disc represented by the sun $A_2.$ Well $A_1 = \pi r^2 = \pi \left[150150000 \tan{\left(\frac{47 \pi}{36000}\right)}\right]^2$ and $A_2 = \pi (r_{sun})^2 = \pi \left[150000000 \tan{\left( \frac{13 \pi}{9000}\right)}\right]^2$ \\\\ Now $\frac{A_1}{A_2} = \frac{\pi \left[150150000 \tan{\left(\frac{47 \pi}{36000}\right)}\right]^2}{\pi \left[150000000 \tan{\left( \frac{13 \pi}{9000}\right)}\right]^2} \approx 0.818$. Thus, $82\%$ of the disc represented by the sun is obscured by the moon (to 2 s.f.)$

Mathew587 · Apr 18, 2017

Re: HSC 2017 Maths Marathon

what happened to this thread

o
and how do you guys know if you thoroughly know a topic in math?

HeroWise · Dec 8, 2017

Im going to Year 10 in 2018 and am learning differential calculus, to answer OP's question I sketched the graph, any advanced criticism is appreciated as I am learning

HoldingOn · Dec 13, 2017

HeroWise said:
Im going to Year 10 in 2018 and am learning differential calculus, to answer OP's question I sketched the graph, any advanced criticism is appreciated as I am learning

When you do working in 2u make sure to write out obvious statements, they may seem stupid but they are necessary to get the marks, particularly for curve sketching. For example above- include in your working a phrase like "testing nature," and a concluding statement "therefore horizontal point of inflexion at P(X,Y) etc.

fan96 · Dec 13, 2017

HeroWise said:
Im going to Year 10 in 2018 and am learning differential calculus, to answer OP's question I sketched the graph, any advanced criticism is appreciated as I am learning

It doesn't matter if your graph's scale isn't perfect, you don't need to measure out centimetres on your ruler. As long as you sketch a reasonable graph, showing any necessary features (including any that were asked for) e.g. intercepts, maxima/minima, pts. of inflexion you'll get the marks. The intent is not to test you on your measuring skills.

Just as a quick example, I would sketch it something like this. Just note that this graph shows the y-intercept more clearly. It's possible to find any x intercepts but this isn't expected in the 2U course, although it should be in the 3U course. I was taught to label my points of inflexion with a line going through the point but I don't think it matters as long as you make it clear that the point IS one of inflexion. In an exam I would draw the graph larger, though.

A point of inflexion means a change in concavity, i.e. from positive to negative or vice versa. so to prove that a point is one of inflexion, you need to show that the sign of the second derivative changes around that point.

for your curve:

$y'' = 18x - 12 \\ x = \frac{2}{3}, y'' = 0 \\ x = 1, y'' = 6 \\ x = 0, y ''= -12$

Because $y''$ changes signs around $x=\frac{2}{3}$ , a point of inflexion exists at $x = \frac{2}{3}$ .

Just to elaborate on why you need to test the second derivative around the point - first note that any quadratic, such as $y = x^2$ , has no points of inflexion because $y''$ is always a constant (in this case, $2$ ) and so the equation $y'' = 0$ has no solution.

Consider the curve $y = x^4$ . The curvature of the graph is identical to $y = x^2$ (you can graph it digitally to check). Here, $y'' = 12x^2$ , so the equation $y'' = 0$ has a solution at $x = 0$ . But from the previous paragraph, clearly this curve can't have a point of inflexion. This is because $y'' \geq 0$ for all real $x$ , so the second derivative never changes signs and because of this, no points of inflexion exist on this curve.

pikachu975 · Dec 14, 2017

fan96 said:
It doesn't matter if your graph's scale isn't perfect, you don't need to measure out centimetres on your ruler. As long as you sketch a reasonable graph, showing any necessary features (including any that were asked for) e.g. intercepts, maxima/minima, pts. of inflexion you'll get the marks. The intent is not to test you on your measuring skills.

Just as a quick example, I would sketch it something like this. Just note that this graph shows the y-intercept more clearly. It's possible to find any x intercepts but this isn't expected in the 2U course, although it should be in the 3U course. I was taught to label my points of inflexion with a line going through the point but I don't think it matters as long as you make it clear that the point IS one of inflexion. In an exam I would draw the graph larger, though.

A point of inflexion means a change in concavity, i.e. from positive to negative or vice versa. so to prove that a point is one of inflexion, you need to show that the sign of the second derivative changes around that point.

for your curve:

$y'' = 18x - 12 \\ x = \frac{2}{3}, y'' = 0 \\ x = 1, y'' = 6 \\ x = 0, y ''= -12$

Because $y''$ changes signs around $x=\frac{2}{3}$ , a point of inflexion exists at $x = \frac{2}{3}$ .

Just to elaborate on why you need to test the second derivative around the point - first note that any quadratic, such as $y = x^2$ , has no points of inflexion because $y''$ is always a constant (in this case, $2$ ) and so the equation $y'' = 0$ has no solution.

Consider the curve $y = x^4$ . The curvature of the graph is identical to $y = x^2$ (you can graph it digitally to check). Here, $y'' = 12x^2$ , so the equation $y'' = 0$ has a solution at $x = 0$ . But from the previous paragraph, clearly this curve can't have a point of inflexion. This is because $y'' \geq 0$ for all real $x$ , so the second derivative never changes signs and because of this, no points of inflexion exist on this curve.

Just a tiny tiny comment but make sure your horizontal point of inflection looks really horizontal so that if you drew a tangent to it the tangent would be horizontal.

HeroWise · Jan 4, 2018

pikachu975 said:
Just a tiny tiny comment but make sure your horizontal point of inflection looks really horizontal so that if you drew a tangent to it the tangent would be horizontal.

I asked my teacher he says about 3mm run in HPOI is good
Well in reality no one is gonna actually measure for 3mm run, around that should bbe fine

HSC 2017-2018 Maths Marathon (1 Viewer)

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