Search results

In this one, left is positive because the circle centre is to left (like the centre of the turn). In general, for banked track Q's, it's good to make the positive horizontal direction the direction pointing towards the centre of the turn.

Well the mu*N arrow is parallel to the large triangle's (the track's) hypotenuse, so if we draw in a horizontal there, the angle between the horizontal and the mu*N arrow is just theta. For the other one (angle for the normal force vector), since that comes up basically on every banked track...

For convenience of anyone who wants to see the answers from BOSTES, here they are (see page 23): https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-maths-ext-2.pdf . $\noindent It's $N\sin \theta$ (not $N\cos \theta$ as you wrote) that points horizontally. And it points...

$\noindent That's basically it. Since the second derivative is $f''(x) = 12x^{-4}>0$ for all $x\neq 0$, $f$ is concave up everywhere other than $x=0$.$

It's useful to note in general that if the particle's velocity and acceleration are of opposite signs, it's slowing down (speed decreasing), whilst if the velocity and acceleration are of same sign, it's speeding up.

Re: HSC 4U Integration Marathon 2017 $\noindent Find $\int \frac{1}{x^2 -3x +2}\, \mathrm{d}x$.$

$\noindent Let the equation be $y=ax^2 + bx+c$. Since $(0,-2)$ lies on the parabola, we have $c = -2$ ($y$-intercept). So $y=ax^2 + bx -2$. Sub. in $(1,0)$: $a+b-2 = 0$. Sub. in $(3,-8)$: $9a +3b-2 = -8$. Since $a + b-2=0$, we have $a = 2-b$. So $9a + 3b -2 = -8 \Rightarrow 9(2-b) + 3b =...

Re: HSC 4U Integration Marathon 2017 I guess he considered that 'masochistic'? (Not really sure how it is, but anyway.) Anyway, just note that by point symmetry, the area under the curve (and hence the integral) is half the area of the rectangle enclosing the arc in question, i.e. 1/2 *...

Re: MATH1231/1241/1251 SOS Thread They got the kernel of that matrix. The kernel can be found by inspection (as they did), but if you aren't comfortable with that, note that further row-reduction makes the matrix into: [2 5] [0 0]. So set x_2 = 2*alpha say (since second column is...

Re: HSC 4U Integration Marathon 2017 Worth exposing 2017'ers to since they may not have followed the older Marathons. (Wasn't intended as a joke integral.)

Re: HSC 4U Integration Marathon 2017 $\noindent Find $\int \sqrt{\tan x}\, \mathrm{d}x$.$

$\noindent Explanation of how to set up the integral (I'm assuming you've sketched the region or have visualised it): recall that the area of th region (call it $\mathcal{R}$) is $\iint _{\mathcal{R}} 1\,\mathrm{d}A$ (i.e. the integral over the region of the constant function $1$. Note that...

$\noindent $\int_{x = 0}^{x =1} \int _{y=x^3}^{y=x^2} 1 \, \mathrm{d}y \, \mathrm{d}x$$ (I wrote like "x = " and "y = " in the bounds for the integrals, but you don't need to write them, I just put them in because it may help with visualisation/understanding and keeping track of which variable...

Ah. Yeah that was probably just a typo/blunder (the sample solutions have some disclaimer at the start saying that they may not be complete or even correct, if I recall correctly).

The answers actually did up to 2n+1 games, they just wrote to go up to 2n games by accident I think. The last term in their series in the first line was (2n choose n)*(1/2^{2n+1}). This is indeed the probability that A wins in 2n+1 games, because we fix the last game as an A win, then we choose...

Re: HSC 2017 4U Marathon $\noindent Or alternatively, for any complex number $\zeta$, $|\Re \left(\zeta\right)| \leq |\zeta|$, because $|\zeta|^2 = \left( \Re \left(\zeta\right) \right)^2 + \left(\Im \left(\zeta\right)\right)^2 \geq \left(\Re \left(\zeta\right) \right)^2 \Rightarrow |\zeta|...

The middle terms add up to 1 times a log. E.g. with k = 6, it would be: (1/2)*(ln(1) + ln(2)) + (1/2)*(ln(2) + ln(3)) + (1/2)*(ln(3) + ln(4)) + (1/2)*(ln(4) + ln(5)) + (1/2)*(ln(5) + ln(6)) = ln(2) + ln(3) + ln(4) + ln(5) + (1/2)*ln(6) = ln(5!) + (1/2)*ln(6) = ln((k-1)!) + (1/2)*ln(k). The...

Why did they do that?

It's always 4 – x (in this question). Firstly, for the outer radius, the x is always on the left side of the plane, so x is negative. Secondly, it doesn't actually matter, even if x were positive, it'd still be 4 – x. As long as x < 4 (which it always is for this Q.) the distance from x to 4 is...

Note that x is negative (except at the origin, where it's 0) for the outer radius (since the curve is in the left half of the plane), so by doing 4 – x, we are adding on a positive number to 4 (since -x is positive, so it's 4 + (-x) > 4). In general, if x < X, then the distance along the real...