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Re: HSC 2017 MX2 Integration Marathon You can also try reduction formulas.

$\noindent That curved inequality sign is being used to denote a partial order $(\geq )$ on the given set.$

Depends which ones. You can say things like "associativity of addition" etc.

Add the negative of z to both sides and use associativity of addition.

There are various ways to show it. It comes out essentially immediately with a bit of Linear Algebra knowledge.

What is w? There won't exist such real a and b if w is real, but if w is non-real, then there will.

$\noindent The decomposition is of the form $\frac{x^2 -8x + 2}{(x-2)\left(x^2 + 1\right)} \equiv \frac{A}{x-2} + \frac{Bx + C}{x^2 + 1}$. What progress did you make?$

$\noindent 1) Assume the point of tangency is at $\left(t, e^{3t}\right)$. On the line, the $y$-value is $mt$, so we must have $mt = e^{3t}$. The slope here on the curve $y = e^{3x}$ is $3e^{3t}$ and the line's slope is $m$, so we must have $m = 3e^{3t}$. We have established two equations in two...

$\noindent Rewrite the quadratic in a logarithm in your answer as $(x-2)(x-3)$, then use log laws to split both logarithms and it should simplify down to the right answer.$

It's a quadratic over a quadratic, so use long division first and then the "remainder part" may be integrated using partial fraction decomposition (the quotient will just be a polynomial, which is easy to integrate).

They would do it by rote learning the answer beforehand maybe (or using intuition). Also not that many marks in the HSC Physics paper are devoted to calculations/derivations etc. (Majority of marks should be, but it's actually a minority.)

Re: Several Variable Calculus $\noindent (Note that $\left(\mathbf{r}\times \mathbf{v}\right)\cdot \mathbf{v} = 0$, recalling that in general a scalar triple product like this is always zero if two of the vectors are the same.)$

Re: Several Variable Calculus $\noindent Since there's a cross product, we are in $\mathbb{R}^{3}$ here. We have $\mathbf{F} (t) = m\mathbf{v}'(t) = \lambda \mathbf{r}(t)\times \mathbf{v}(t)$. Dotting both sides with $\mathbf{v}(t)$ yields $m\mathbf{v}'(t) \cdot \mathbf{v}(t) = 0$, which...

Prove what?

The current HSC Physics course is relatively unrelated to Maths Extension 2.

Re: Several Variable Calculus You appear to have miscalculated the velocity vectors when subbing in the values of t and/or s (check the first components, noting you're subbing in s = 1 (not 2) and t = -1 (not -2)).

Re: Several Variable Calculus $\noindent So e.g. the first curve is $\mathbf{r}_{1}(t) = \left(t^{2} -t, t^{2} + t\right)$, so the derivative is $\mathbf{r}_{1}^{\prime}(t) = \left(2t-1, 2t+1\right)$. This direction of this essentially tells us the direction of the curve at $t$ (it's basically...

Re: Several Variable Calculus Find the s and t values at the points of intersection and plug them into the derivatives of the parametric curves. This will give us the "direction vectors" of the curves at the points of intersection. Find the angle between these vectors.

Re: Several Variable Calculus Solve the system of equations t^2 - t = s + s^2 (1) t^2 + t = s - s^2 (2).

No, it's not.