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Re: Multivariable Calculus Well yeah it does need to be smooth enough so that we can switch the order of the mixed partials (which doesn't require C-infinity). But that was probably an implicit assumption of the question.

One geometric interpretation is that it tells us how much a linear transformation scales areas by. $\noindent For example, if $T$ is the linear map from $\mathbb{R}^{2} \to \mathbb{R}^{2}$, with $T\left(\bold{x}\right) = A\bold{x}$, then $\left |\det \left(A\right)\right |$ is the factor by...

A proof of 1) may be found here: http://community.boredofstudies.org/238/extracurricular-topics/350230/least-squares.html . For 2), it is known as the Gram-Schmidt orthonormalisation process, and the proof is typically done by induction.

Re: MX2 2016 Integration Marathon $\noindent Using the `border-flip' and skipping over some steps, we have$ $$\begin{align*}\int _0 ^{\frac{\pi}{4}}\frac{1-\sin 2x}{1+\sin 2x}\text{ d}x &= \int _0 ^{\frac{\pi}{4}}\frac{1-\cos 2x}{1+\cos 2x}\text{ d}x \quad (\text{Pythagorean trig. identity})...

$\noindent I think you mis-evaluated the integral. Note that if $a>0$, then$ $$\begin{align*} \int \frac{1}{\sqrt{a^2 u^2 -1}}\text{ d}u &= \frac{1}{a}\int \frac{1}{\sqrt{u^2 -\frac{1}{a^2}}}\text{ d}u \\ &= \frac{1}{a} \ln \left(u+\sqrt{u^2 -\frac{1}{a^2}}\right). \end{align*}$$ $\noindent...

$\noindent To get the answer back in terms of $x$, we can use the identity above of $\left |\tan\frac{\theta}{2} \right |=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$. Then use the following fact: if $\cos \theta > 0$ (which we can assume because we can impose $-\frac{\pi}{2}<\theta <...

$\noindent Yeah the integrals of sec and csc are not standard (for HSC purposes). But it's worth remembering them and how to quickly derive them (so that you can avoid half-angles). Here's the classic tricks:$ $\noindent We have$ $$\begin{align*}\int \sec x \text{ d}x &= \int \frac{\sec x...

Re: University Statistics Discussion Marathon $\noindent The mean of the sample mean is just the original mean. The standard deviation becomes $\frac{\sigma}{\sqrt{n}}$. These are standard facts about sample means.$

$\noindent Yeah, that's $\csc \theta$. The integral of this is well known to be $-\ln \left|\csc \theta + \cot \theta \right|$, or another way to give it is $\ln \left | \csc \theta - \cot \theta \right |$. This can be shown by differentiating these.$

E.g. "find the range of y = 1/√(x)". Answer by inspection is the set of all positive reals.

Well we didn't need that formula to do that integral in the original post. We could avoid half-angles as I showed. But if you did introduce half-angles, the way to get out of them is to use the half-angle formulas in some way, because these formulas tell us how to go from sin(a/2) and cos(a/2)...

Sure it said 100, but that doesn't mean it's necessarily correct. Why do you think that answer is correct other than "it says so" (if at all)?

s = ut + ½ a t2, with u = 30.0 m/s, t = 3 s, and a = 9.80 m/s2.

$\noindent To get that identity from the Quora link, note that $\sin^2 \frac{a}{2}=\frac{1}{2}-\frac{1}{2}\cos a$ and $\cos^2 \frac{a}{2} = \frac{1}{2} + \frac{1}{2}\cos a$. Division gives $\tan^2 \frac{a}{2} = \frac{1-\cos a}{1+\cos a}$, provided all denominators are non-zero. Now, taking...

I think davidgoes4wce had a source (maybe some past HSC exam book or something like that) that said the answer was 90 (which is the only reasonable answer given the info we are told about the tap turning off at the end).

Have you asked your teacher (or tutor if you go tutoring) for help with them?

Inspection.

I'm quite sure it's unreliable. And I'm not sure, check HSC papers from the previous century.

Which identity? You will need to leave the final answer in terms of x. For that integral, you could also use a trig. substitution of x = a*tan(theta). You end up with something involving integral of cosec(theta), which is: -ln(cosec(theta) + cot(theta)). Then you could convert this back to...

Sounds potentially unreliable (I doubt it's the hardest rates of change question ever).