Author Topic: 4U Maths Question Thread (Read 665360 times) Tweet Share

RuiAce · « **Reply #960 on:** March 11, 2017, 08:15:36 pm »

Quote from: jakesilove on March 11, 2017, 08:14:17 pm

First, we need an equation. We know that

$x=4\cos\theta, y=2\sin\theta$

Clearly,

$x^2+2y^2=16\cos^2\theta+16\sin^2\theta=16$
$\frac{x^2}{16}+\frac{y^2}{8}=1$

We want to find the tangent at the point

$(4\cos(-\frac{\pi}{4}), 2\sin(-\frac{\pi}{4}))$

Honestly Rui, I thought I'd get this one.

But your way is 100% faster.

Far out.

Aha.

Pro tip: If things are given to you parametrically, try to avoid Cartesian stuff

jakesilove · « **Reply #961 on:** March 11, 2017, 08:17:51 pm »

Quote from: RuiAce on March 11, 2017, 08:15:36 pm

Aha.

Pro tip: If things are given to you parametrically, try to avoid Cartesian stuff

Wise words from a freakin' genius. Legend as always matey.

VydekiE · « **Reply #962 on:** March 11, 2017, 09:02:17 pm »

Quote from: RuiAce on March 11, 2017, 08:12:58 pm

$\text{From parametric differentiation, we have}\\ \begin{align*}\frac{dy}{dx}&=\frac{dy/d\theta}{dx/d\theta}\\ &= \frac{2\cos \theta}{-4\sin \theta}\\ &= -\frac12 \tan \theta\end{align*}$
$\text{When }\theta=-\frac{\pi}{4}\\ \begin{align*}x&=\frac{4}{\sqrt2}=2\sqrt{2}\\ y&=-\frac{2}{\sqrt{2}}=-\sqrt{2}\\ \frac{dy}{d\theta}&=1\end{align*}$
$\text{Substitute straight into }y-y_1=m(x-x_1)$

Thank you!!

Kle123 · « **Reply #963 on:** March 12, 2017, 12:19:27 pm »

Could someone please help me with this question

cutiepie30 · « **Reply #964 on:** March 12, 2017, 01:04:11 pm »

Quote from: RuiAce on March 11, 2017, 06:33:55 pm

Pretty much, +C is the cause of it. Because in indefinite integration, we always have to keep in mind that the statement is true to within a constant.

If you ask me though, "true to within a constant" is the key thing that needs to be stated. Of course, you can, by computation, find the value of C (which is -1 or 1, depending on if your C is on the LHS or RHS) to reunite the solutions if you wish.

Thanks RuiAce

jakesilove · « **Reply #965 on:** March 12, 2017, 01:40:08 pm »

Quote from: Kle123 on March 12, 2017, 12:19:27 pm

Could someone please help me with this question

Hey! So,

$P(x)=x^4-2Ax^3+B$
$\alpha, \beta, \gamma, \alpha+\beta+\gamma$

First, let's use the sum of roots. This will equal -b/a, thus

$\frac{2A}{1}=\alpha+\beta+\gamma+\alpha+\beta+\gamma=2(\alpha+\beta+\gamma)$
$A=\alpha+\beta+\gamma$

Next, let's use product of roots. This will be equal to e/a, thus

$\frac{B}{1}=\alpha \times \beta \times \gamma \times (\alpha + \beta + \gamma) = \alpha \times \beta \times \gamma \times A$

Oh. This was stupid. If A= alpha + beta + gamma, then A is a root of the equation! So,

$P(A)=0=A^4-2A(A^3)+B$
$B=A^4$

as required.

Now,

$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha \beta + \alpha \gamma + \beta \gamma)$

Hmmm... will keep working on this part inb4 Rui posts his answer before me

Okay, so $\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha \beta + \alpha \gamma + \beta \gamma)=(\alpha+\beta+\gamma)^2-2(\frac{\alpha \beta \gamma}{\gamma} + \frac{\alpha \beta \gamma}{\beta} + \frac{\alpha \beta \gamma}{\alpha})$

From the above step that I didn't really need to do (but turns out I did!), we know that

$\frac{B}{A}=\alpha \beta \gamma$

New tactic

$A^3=\alpha \beta \gamma$
$A=\alpha + \beta + \gamma$

$(\alpha + \beta + \gamma)^3=\alpha \beta \gamma$

I have to run; hopefully Rui can swoop in and save the day, otherwise I'll pick this up later.

RuiAce · « **Reply #966 on:** March 12, 2017, 06:45:14 pm »

$\text{Note that}\\ \begin{align*}\alpha^2+\beta^2+\gamma^2+A^2&=(\alpha+\beta+\gamma+A)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma+A\alpha+A\beta+A\gamma)\\ &= (2(\alpha+\beta+\gamma))^2-2(\alpha\beta+\alpha\gamma+\beta\gamma+A\alpha+A\beta+A\gamma)\\&= 4(2A)^2-2(0)\\ \implies \alpha^2+\beta^2+\gamma^2&=-A^2+16A^2\end{align*}$
Subject to inaccuracies

cutiepie30 · « **Reply #967 on:** March 12, 2017, 07:29:26 pm »

How would you do this question?

Mahan · « **Reply #968 on:** March 12, 2017, 07:45:31 pm »

Quote from: cutiepie30 on March 12, 2017, 07:29:26 pm

How would you do this question?

Note:
$Cosh^{3}(x) = \frac{1}{8}(e^{x} + e^{-x})^{3} = \frac{1}{8}(e^{3x} + 3e^{2x}e^{-x} + 3e^{x}e^{-2x} + e^{-3x})$
$= \frac{1}{8}(e^{3x}+ e^{-3x} + 3(e^{x} + e^{-x})) = \frac{1}{4}(cosh(3x) + 3cosh (x))$
$\therefore cosh(3x) = 4cosh^{3} (x) -3cosh(x)$

cutiepie30 · « **Reply #969 on:** March 12, 2017, 08:25:16 pm »

Quote from: Mahan on March 12, 2017, 07:45:31 pm

Note:
$Cosh^{3}(x) = \frac{1}{8}(e^{x} + e^{-x})^{3} = \frac{1}{8}(e^{3x} + 3e^{2x}e^{-x} + 3e^{x}e^{-2x} + e^{-3x})$
$= \frac{1}{8}(e^{3x}+ e^{-3x} + 3(e^{x} + e^{-x})) = \frac{1}{4}(cosh(3x) + 3cosh (x))$
$\therefore cosh(3x) = 4cosh^{3} (x) -3cosh(x)$

thanks Mahan

but the question says find cosh(3x) in terms of cosh(x) and sinh(x), your answer only has cosh(x) not sinh(x)?

RuiAce · « **Reply #970 on:** March 12, 2017, 08:30:06 pm »

Quote from: cutiepie30 on March 12, 2017, 08:25:16 pm

thanks Mahan

but the question says find cosh(3x) in terms of cosh(x) and sinh(x), your answer only has cosh(x) not sinh(x)?

Hmm.

What answer do they want you to get to? Because the convention is to stop at where he did.

jakesilove · « **Reply #971 on:** March 12, 2017, 10:15:42 pm »

Quote from: RuiAce on March 12, 2017, 08:30:06 pm

Hmm.

What answer do they want you to get to? Because the convention is to stop at where he did.

Is sinh and cosh even part of the HSC?

RuiAce · « **Reply #972 on:** March 12, 2017, 10:28:25 pm »

Quote from: jakesilove on March 12, 2017, 10:15:42 pm

Is sinh and cosh even part of the HSC?

Nope.

At least it got defined for them though.

cutiepie30 · « **Reply #973 on:** March 12, 2017, 11:46:15 pm »

Quote from: RuiAce on March 12, 2017, 08:30:06 pm

Hmm.

What answer do they want you to get to? Because the convention is to stop at where he did.

i am not sure myself Rui

, our teacher just gave us this question as an extension and said we should be able to do it from the info provided

RuiAce · « **Reply #974 on:** March 13, 2017, 06:56:15 am »

Quote from: cutiepie30 on March 12, 2017, 11:46:15 pm

i am not sure myself Rui , our teacher just gave us this question as an extension and said we should be able to do it from the info provided

But then again, if you get it in terms of only cosh it's still in terms of both cosh and sinh. This is because you're basically adding $0\sinh x$ anyhow.

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