Author Topic: [Specialist] - Maths questions of the week thread! (Read 9015 times) Tweet Share

RuiAce · « **Reply #15 on:** July 24, 2016, 10:25:56 am »

Quote from: insanipi on July 24, 2016, 10:20:35 am

Q3.
Am I on the right track here?

Seems good. The main idea of this question would be to show the relationship by taking partial derivatives instead of using implicit differentiation on a multivariable function equal to 0. I'll let Steve comment further

sweetiepi · « **Reply #16 on:** July 24, 2016, 10:29:06 am »

Quote from: RuiAce on July 24, 2016, 10:25:56 am

Seems good. The main idea of this question would be to show the relationship by taking partial derivatives instead of using implicit differentiation on a multivariable function equal to 0. I'll let Steve comment further

Partial derivatives seem unfamiliar to me, but I'll sure look it up! Thanks for the feedback, Rui!

RuiAce · « **Reply #17 on:** July 24, 2016, 10:54:05 am »

Quote from: insanipi on July 24, 2016, 10:29:06 am

Partial derivatives seem unfamiliar to me, but I'll sure look it up! Thanks for the feedback, Rui!

I'll briefly explain how they work

Of course, when we have a function of just one variable, that variable is the only one that can change. But when we have a multivariable function, who knows how many variables are changing at the same time. We consider the two extreme cases:

1. Partial derivatives - Keep every other variable fixed, but allow just one of them to change. The partial derivative measures the rate of change of just that one variable in question. These are easy to compute because since everything else is fixed, they become constants. So we use our standard derivative methods.

2. Total derivatives - Allow all variables to change. The total derivative thus effectively measures the overall rate of change of the variable, and is typically computed with the aid of the multivariable chain rule.

zsteve · « **Reply #18 on:** July 24, 2016, 01:47:06 pm »

Rui's summed it up well. The point of the question was to show the method of finding derivatives of implicit relations using a method of partial derivatives (this is a result of the multivariable chain rule)
Your working looks fine! Well done

This is a good way to check implicit differentiation in Specialist maths exams (I used it in 2015 Exam 1

)

RuiAce · « **Reply #19 on:** July 27, 2016, 07:36:38 pm »

Kinda surprised nobody's attempted Q1 yet. Hint on the second proof:

Q1

Literally use all you know. Just keep in mind things such as $\cos 4x = \frac{e^{4ix}+e^{-4ix}}{2}$

_{Not sure about Q2 though - Ellipse is the word used, but y²=4ax is a parabola Seems like this got fixed}

lzxnl · « **Reply #20 on:** July 27, 2016, 09:24:20 pm »

Quote from: qazser on July 18, 2016, 07:47:03 am

Well, i guess i'll start this one off. Might be wrong, just learnt this on Khan Academy

LaTeX looks scrappy, don't know how to format well

Question 1

Using U-Substitution

Let $x=sin(u)=arcsin(x)$

Derivative of $sin(x)=cos(u)$ TYPOOOOOOOOOOOO capitalised to get your attention

$\int cos(u)\sqrt{(-sin^2(u)+1} du$

Using "fundamental' Pythagorean Identity

$sin^2(x)+cos^2(x)=1$

$\int cos(u)\sqrt{cos^2(u)} du$

$\int cos^2(u) du$

$\int \frac{cos(u)sin(u)}{2} + \frac{1}{2}u$

Substitute $arcsin(x)$ back into equation

$\dfrac{\arcsin\left(x\right)+x\sqrt{1-x^2}}{2}$

Alternatively, the CAS calculator states: $\frac{sin^{-1}(x)}{2}+\frac{x\sqrt(1-x^2)}{2}$
Let me know if working is wrong, i'm in Year 11, don't know how to set it out correctly

Quote from: zsteve on July 24, 2016, 01:47:06 pm

Rui's summed it up well. The point of the question was to show the method of finding derivatives of implicit relations using a method of partial derivatives (this is a result of the multivariable chain rule)
Your working looks fine! Well done

This is a good way to check implicit differentiation in Specialist maths exams (I used it in 2015 Exam 1 )

Yeah me too. It's so easy to do and the proof is fairly intuitive.
If you have a function of multiple variables, f(x,y,z), then the derivative df/dx measures the total change of f as x changes.
There are three ways f can change: because x changes, because y changes and because z changes. The change due to the change in x is the partial $\frac{\partial f}{\partial x}$
The change due to the change in y is given by the 1D chain rule expression $\frac{\partial f}{\partial y}\times\frac{\partial y}{\partial x}$ and similarly for z. Add all three for the total derivative.

I'll post one here, because I think this trick really is pretty awesome. Yes, you don't need anything other than spesh knowledge to solve this, forbidding as it may seem.

Evaluate the definite integral:
$\int_{0}^{\pi/2}\frac{\sin^{2016}{x}}{\sin^{2016}{x} + \cos^{2016}{x}} dx$

If you want another challenge, try $\int_{0}^{\pi/2} \ln{\sec{x}} dx$

In fact, I challenge any maths student to solve these two. They're solvable using VCE spesh techniques, but it's not easy to see how. If you want hints, just ask, because you may well need them.

Wolfram won't give you exact answers for these (I checked).

RuiAce · « **Reply #21 on:** July 27, 2016, 09:51:13 pm »

$\text{Suppose }a \in \mathbb{R}\text{. then}\\ \begin{align*}\int_0^{\frac{\pi}{2}}\frac{\sin^a x}{\sin^a x + \cos^a x}dx &=\int_0^{\frac{\pi}{2}}\frac{\sin^a \left(\frac{\pi}{2}-x\right)}{\sin^a \left(\frac{\pi}{2}-x\right)+ \cos^a \left(\frac{\pi}{2}-x\right)}dx \\ &= \int_0^{\frac{\pi}{2}}\frac{\cos^a x}{\sin^a x + \cos^a x}dx \\ \implies 2\int_0^{\frac{\pi}{2}}\frac{\sin^a x}{\sin^a x + \cos^a x}dx &= \int_0^{\frac{\pi}{2}}\frac{\sin^a x+\cos^a x}{\sin^a x + \cos^a x}dx\\ \iff \int_0^{\frac{\pi}{2}}\frac{\sin^a x}{\sin^a x + \cos^a x}dx &= \frac{1}{2}\int_0^{\frac{\pi}{2}}dx\\ &= \frac{\pi}{4}\end{align*}$
$\text{To prove the border-flip substitution }\int_0^a f(x)dx=\int_0^a f(a-x)dx\\ \text{Let }u=a-x \implies -du = dx\\ \begin{align*}LHS&=\int_0^a f(x)dx\\ &= \int_a^0 f(a-u)(-du)\\ &= \int_0^a f(a-u)du\\ &= \int_0^a f(a-x)dx \\ &= RHS \end{align*}$

Sorry

lzxnl · « **Reply #22 on:** July 27, 2016, 10:59:03 pm »

Lol it's fine
I asked a question hoping someone would try it.

Second one?

If you want, try $\int \sqrt{\tan{x}} dx$
Messy but doable. Wolfram is cheating here.

Here's a fun one. Simplify the integral $\int \frac{dx}{(1+x^n)^{1/n}}$ into an integral of a rational function, where n is integral.

RuiAce · « **Reply #23 on:** July 27, 2016, 11:26:41 pm »

I did do the √tan(x) integral by brute force once before; haven't attempted a clever way yet

Maybe give it more thought another time

RuiAce · « **Reply #24 on:** July 28, 2016, 04:33:56 pm »

lzxnl · « **Reply #25 on:** July 28, 2016, 11:58:22 pm »

Quote from: RuiAce on July 28, 2016, 04:33:56 pm

(Image removed from quote.)

I'm actually quite impressed with the way you did the rational integral. I always solved that by partial-fractioning it into quadratics.

The questions I can ask though can only be so hard to stay in the confines of VCE spesh. I'm already pushing the limits a bit.

RuiAce · « **Reply #26 on:** July 30, 2016, 06:48:07 pm »

zsteve · « **Reply #27 on:** July 30, 2016, 10:39:56 pm »

Question for Week 3 is up - hopefully this is more doable than the previous questions

Let me know if it's still too challenging

lzxnl · « **Reply #28 on:** July 31, 2016, 11:03:04 am »

For lols, I'll do one of the previous questions without a substitution.

$I = \int \sqrt{1-x^2} dx = \int \frac{1-x^2}{\sqrt{1-x^2}} dx = \arcsin{x} - \int \frac{x^2}{\sqrt{1-x^2}} dx = \arcsin{x} + \int x \frac{d}{dx}(\sqrt{1-x^2}) dx$

Integrate by parts (hence why I drew the x out).

$I = \arcsin{x} + \int x \frac{d}{dx}(\sqrt{1-x^2}) dx = \arcsin{x} + x \sqrt{1-x^2} - \int \sqrt{1-x^2} dx = \arcsin{x} + x \sqrt{1-x^2} - I$
$2I = \arcsin{x} + x \sqrt{1-x^2}$
$I = \frac{1}{2} (\arcsin{x} + x \sqrt{1-x^2})$

You can put your arbitrary constant somewhere if you want.

zsteve, q3 is a question I routinely explain to my spesh students

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Author Topic: [Specialist] - Maths questions of the week thread! (Read 9015 times) Tweet Share

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