HSC 2016 Maths Marathon (archive) (1 Viewer)

InteGrand · Nov 2, 2015

Re: HSC 2016 2U Marathon

Rathin said:
View attachment 32636 might of missed a couple of steps. am i correct?

$For your method, you need to write in the quotients that it's the limit as $\Delta x \to 0$. (And avoid using `d$x$' for that since $\mathrm{d}x$ has another meaning, use something else, like $\delta x$ or $h$ or $\Delta x$ etc.)$

Green Yoda · Nov 2, 2015

Re: HSC 2016 2U Marathon

InteGrand said:
$I didn't check the details of what you did (just glanced at the general method), but there is a faster method: just use the chain rule. If $f$ is even, then $f(x)=f(-x)$ for all $x$ in the domain. Differentiating both sides with respect to $x$, we have$

$\begin{align*} f^\prime (x) &=f^\prime (-x)\cdot \frac{\mathrm{d}}{\mathrm{d}x}(-x) \quad (\text{chain rule})\\ &=f^\prime (-x)\cdot (-1)\\ &=-f^\prime (-x), \end{align*}$

$and so $f^\prime $ is odd, as required.$

oh yeah..shit. and i need to start learning how to use latex.

Green Yoda · Nov 2, 2015

Re: HSC 2016 2U Marathon

Next Question:
The gradient of a curve is y' = x^2 − 4x + 3 and the curve passes through the origin. Find the equation of the curve.

Drsoccerball · Nov 2, 2015

Re: HSC 2016 2U Marathon

Rathin said:
Next Question:
The gradient of a curve is y' = x^2 − 4x + 3 and the curve passes through the origin. Find the equation of the curve.

$y= \frac{x^3}{3} -2x^2 +3x$

davidgoes4wce · Nov 2, 2015

Re: HSC 2016 2U Marathon

That was my working out:

Speed6 · Nov 2, 2015

Re: HSC 2016 2U Marathon

Heeeey! It's Dr Xie in the thread

davidgoes4wce · Nov 2, 2015

Re: HSC 2016 2U Marathon

Is Doctor Xie? FrankXie by any chance?

Speed6 · Nov 2, 2015

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Is Doctor Xie? FrankXie by any chance?

Yeah that's him, I rarely see him on here, maybe once in a blue moon?

Drsoccerball · Nov 2, 2015

Re: HSC 2016 2U Marathon

davidgoes4wce said:
That was my working out:

$$ You can also do $ \int_0^y(y')dy' = \int_0^x(x^2-4x+3)dx$

InteGrand · Nov 2, 2015

Re: HSC 2016 2U Marathon

Drsoccerball said:
$$ You can also do $ \int_0^y(y')dy' = \int_0^x(x^2-4x+3)dx$

$It should be: $y=f(x)=\int _0 ^x f^\prime (t) \text{ d}t$ (where $f^\prime (t) = t^2-4t+3$ is given).$

Drsoccerball · Nov 2, 2015

Re: HSC 2016 2U Marathon

InteGrand said:
$It should be: $y=f(x)=\int _0 ^x f^\prime (t) \text{ d}t$ (where $f^\prime (t) = t^2-4t+3$ is given).$

Not really... When doing a definite integral in mechanics it is absolutely necessary to have both sides definite integral. Usually its just t on the LHS and therefore it doesnt really appear as a definite integral but it is. An example of this is the rocket launch question I can post if you like?

InteGrand · Nov 2, 2015

Re: HSC 2016 2U Marathon

Drsoccerball said:
Not really... When doing a definite integral in mechanics it is absolutely necessary to have both sides definite integral. Usually its just t on the LHS and therefore it doesnt really appear as a definite integral but it is. An example of this is the rocket launch question I can post if you like?

What do you mean by "not really"? The one I wrote is just a restatement of the fundamental theorem of calculus pretty much (since f(0) = 0).

InteGrand · Nov 2, 2015

Re: HSC 2016 2U Marathon

Drsoccerball said:
Not really... When doing a definite integral in mechanics it is absolutely necessary to have both sides definite integral. Usually its just t on the LHS and therefore it doesnt really appear as a definite integral but it is. An example of this is the rocket launch question I can post if you like?

$I realised what you meant by yours; but it should be this on the LHS: $\int _{y(0)} ^{y(x)} \mathrm{ d}y$ (that's why I said that was wrong; and this just simplifies to what I said to use). And yeah, I did that in the mechanics derivation of escape velocity on one of the Physics threads.$

$For your LHS integral, it would give this:$

$\begin{align*}\int _0 ^y y^\prime \text{ d}y^\prime &= \left[\frac{\left(y^\prime\right)^2}{2} \right] _{y^\prime = 0} ^{y^\prime = y} \\ &= \frac{y^2}{2}\end{align*}$

$which is not what we want the LHS to be. I think you just miswrote what you intended (shouldn't be $\mathrm{d}y^\prime$ in the integral; or maybe you just meant $y^\prime$ as a dummy variable of integration (might be better to just use $\gamma$ or something, to avoid confusion with the derivative $y^\prime$). If this is what you meant, there shouldn't be $y^\prime$ in the integrand).$

InteGrand · Nov 2, 2015

Re: HSC 2016 2U Marathon

$In general, if $y=f(x)$ and $f$ is differentiable, with $f(x_0) = y_0$, we have the following: $y-y_0 = \int _{x_0}^{x} f^\prime (t) \text{ d}t$. The LHS here is just $\int _{y_0} ^y \mathrm{ d}\gamma$ (alternatively, view it via the Fundamental Theorem of Calculus applied to the RHS integral; this gives the LHS).$

Speed6 · Nov 3, 2015

Re: HSC 2016 2U Marathon

Unlike Trebla's 4U marathon where former students are generally discouraged from posting answers to the questions posed, on this thread and also the 3U thread, there are no restrictions on who can and cannot post!

leehuan · Nov 3, 2015

Re: HSC 2016 2U Marathon

Rathin said:
oh yeah..shit. and i need to start learning how to use latex.

I told calambe the same thing. I never learnt LaTeX, I always used Daum Equation Editor.

Drsoccerball said:
$$ You can also do $ \int_0^y(y')dy' = \int_0^x(x^2-4x+3)dx$

I love this method. But personally I reckon keep it out of the 2u student's mindset because it's more confusing.

EDIT: Also, lol yeah what InteGrand said.

leehuan · Nov 3, 2015

Re: HSC 2016 2U Marathon

NEXT QUESTION:
$\\$The area in the first quadrant bound between the coordinate axes and the curves $ y=\sin { x } $, $y=\cos { x } $ is rotated around the $x$-axis. \\ a) On the unit circle, plot any arbitrary points A, and B. Let O be the origin, and C be the point (0,1). Let angle AOC = a and angle BOC = b. \\ (i) Find the coordinates of A and B. \\ (ii) Use the distance formula to find an expression for AB. \\ (iii) Use the cosine rule to find another expression for AB and hence deduce that $\cos { \left( a-b \right) } =\cos { a } \cos { b } +\sin { a } \sin { b }$ \\ (iv) Find an expression for $\cos { 2x }$ \\ b) Hence, calculate the volume of the solid.$

davidgoes4wce · Nov 3, 2015

Re: HSC 2016 2U Marathon

By first quadrant Im assuming it has to be between the domain of [0, pi/2]

davidgoes4wce · Nov 3, 2015

Re: HSC 2016 2U Marathon

'Arbitrary points' could you select (pi/4, 1/sqrt(2)) as Point A and call say point B (pi/6, 1/2)? (Given that C is (0,1))

I think of arbitrary as just random points

davidgoes4wce · Nov 3, 2015

Re: HSC 2016 2U Marathon

I can't be bothered Googling but I am slightly confused about the domain for an absolute function.

Would the below setting be right for the domain of a function over that interval?

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