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If you are ranked first internally and externally, then your overall HSC mark will just be equal to whatever your external mark (examination mark) is. (Ignoring any complicating factors like being equal first or something.) More generally, if you're ranked (outright) first internally, your...

Re: MATH1131 help thread There are multiple ways of expressing the conditions required, but they all express the same thing, just in a different way. It's like how expressing a plane in parametric vector form can be done in many different ways, or how a system of simultaneous equations can be...

Re: MATH1131 help thread It is geometrically clear that the (principal) argument is -3pi/4. I.e. Draw the point (-2,-2) in the x-y plane. Draw the ray from the origin through this point. The angle formed between this and the positive x-axis is clearly 135 degrees (since it lies on the line y...

$\noindent The first one is a circle with diameter $6$ centred on the $y$-axis, tangential to the $x$-axis; its Cartesian equation is $x^2 + \left(y-3\right)^2 = 9$. The second is a circle centred on the $x$-axis, with diameter 2, tangential to the $y$-axis; its Cartesian equation is...

Re: Multivariable Calculus $\noindent What I used is basically the multivariate chain rule -- have you learnt about this?$

Re: Multivariable Calculus $\noindent This requires use of a \textsl{total derivative}. The volume $V$ is given as a function of $h$ (height) and $r$ (radius) as $V=f\left(h,r\right) = \pi r^2 h$. Now, $r=r(t)$ and $h=h(t)$ themselves are functions of time $t$, so we can talk about the total...

Re: Multivariable Calculus $\noindent In terms of finding partials of $z$, yes, we could use implicit differentiation. Note $x^2 + y^2 + z^2 = 3$ and that $z$ is defined as a differentiable function of $x,y$ near $\vec{a}=\left(1,1,1\right)$. Using implicit differentiation, we have for instance...

Re: Multivariable Calculus $\noindent In general, if we're given a smooth surface of the form $F\left(x,y,z\right) = C$ (where $C$ is a constant), a normal to the surface at a point $\vec{a}$ on the surface is given by $\nabla F \left(\vec{a}\right)$. Once we have the normal from this, since we...

Re: Multivariable Calculus $\noindent In this case it can be done by inspection. The surface is a sphere centred at the origin (its radius is $\sqrt{3}$ incidentally, though that's not important here), so a normal at a point $\vec{a}$ on the sphere is just $\vec{a}$ itself (this is essentially...

Why did you throw them away?

$\noindent This is another one that will be based on symmetry. Here's a sketch of the method. Write $I(a) = \int _{\frac{1}{a}}^{a} \frac{\tan^{-1} x}{x} \text{ d}x$, for $a\in \mathbb{R}$. (Note $a\neq 0$.) First deal with the case where $a>0$. Use a substitution of $x=\frac{1}{u}$ and use the...

The ln(2)/2 is just a constant, so integrating that gives (pi/4)*(ln(2)/2) = pi*(ln(2)/8). $\noindent In general, if $c$ is a constant, then $\int _a ^b c \text{ d}x = c\left(b-a \right).$

$\noindent The answer is based on using symmetry essentially (rather than finding an antiderivative). There's a lot of integrals that we don't have elementary functions for for the primitive, but can still evaluate the definite integral. In this one, let $I = \int_0 ^a...

OK yeah, I guess quite a lot of people do say it that way. But imagine a sum, people say like a sum from 1 to N (with 1 being the lower limit and N the upper limit). So similar thing with integrals I guess (integrals are basically continuous sums).

Yeah. But some people say "2 to 0" to mean the bottom one to be to 0. It should be said "0 to 2".

The original question had limits from 0 to 2. This makes it doable. Without the limits, there's no elementary solution.

This was answered here: http://community.boredofstudies.org/238/extracurricular-topics/350045/dot-product.html#post7148468 .

$\noindent Diagonal matrices are an important case where they always commute. I.e. if $D_1,D_2$ are same-sized (so that they are compatible) diagonal matrices, then $D_1 D_2 = D_2 D_1$.$

$\noindent Also, the first formula doesn't follow \textsl{straight} from $\Delta s = v \Delta t$, since $v$ is varying in when we have a constant acceleration (unless it's a trivial 0 acceleration). The correct formula is: $\Delta s = \bar{v} \Delta t$, where $v$ is the average velocity over a...

$\noindent The first one is helpful to find the (change in) displacement given a time value if we know the object has a constant acceleration and we know its initial velocity $u$. Examples of problems we can solve with this are problems involving for instance how long it'll take a dropped ball...