Author Topic: Uni Maths Questions (Read 34708 times) Tweet Share

brightsky · « **Reply #120 on:** June 14, 2013, 07:25:32 pm »

let the denominator of the derivative = 0. that is, let x + 4y^3 = 0 => x = 4y^3. sub this into the original equation and solve to get the values of x for which tangent is vertical.

M-D · « **Reply #121 on:** June 15, 2013, 06:51:46 pm »

could someone please help with these two questions:

1) Use the complex exponential to express $sin^3\theta$ in terms of $sin\theta$ and $sin3\theta$

2) Find all the roots of the polynomial: $P(z)=z^6-9z^3+8$

thanks. i really appreciate your help

BubbleWrapMan · « **Reply #122 on:** June 15, 2013, 07:21:39 pm »

1) Expand out $\left(\frac{1}{2i}\left(\exp(\mathrm{i}\theta)-\exp(-\mathrm{i}\theta)\right)\right)^3$ , try and get it in the form $\frac{a}{2i}\left(\exp(\mathrm{i}\theta)-\exp(-\mathrm{i}\theta)\right)+\frac{b}{2i}\left(\exp(3\mathrm{i}\theta)-\exp(-3\mathrm{i}\theta)\right)+c$ , which is equal to $a\sin(\theta)+b\sin(3\theta)+c$ .

2) Let $a=z^3$ , solve resulting quadratic.

b^3 · « **Reply #123 on:** June 15, 2013, 07:29:07 pm »

EDIT: Technically used the complex exponential but this might not be the way you were meant to do it.

Quote from: M-D on June 15, 2013, 06:51:46 pm

1) Use the complex exponential to express $sin^3\theta$ in terms of $sin\theta$ and $sin3\theta$

Firstly we can use the complex exponential to obtain the result below, then we can find an equivalent term by expanding using the binomial theorem.

Spoiler

$\begin{alignedat}{1}e^{i\theta} & =\cos\left(\theta\right)+i\sin\left(\theta\right) \\ \left(\cos\left(\theta\right)+i\sin\left(\theta\right)\right)^{3} & =\left(e^{i\theta}\right)^{3} \\ & =e^{3i\theta} \\ & =\cos\left(3\theta\right)+i\sin\left(3\theta\right) \\ \left(\cos\left(\theta\right)+i\sin\left(\theta\right)\right)^{3} & =\dbinom{3}{0}\left(\cos\theta\right)^{3}\left(i\sin\theta\right)^{0}+\dbinom{3}{1}\left(\cos\theta\right)^{2}\left(i\sin\theta\right)^{1}+\dbinom{3}{2}\left(\cos\theta\right)^{1}\left(i\sin\theta\right)^{2}+\dbinom{3}{3}\left(\cos\theta\right)^{0}\left(i\sin\theta\right)^{3} \\ & =\cos^{3}\theta+3i\cos^{2}\theta\sin\theta+3i^{2}\cos\theta\sin^{2}\theta+i^{3}\sin^{3}\theta \\ & =\cos^{3}\theta+3i\cos^{2}\theta\sin\theta-3\cos\theta\sin^{2}\theta-i\sin^{3}\theta \\ & =\left(\cos^{3}\theta-3\cos\theta\sin^{2}\theta\right)+i\left(3\cos^{2}\theta\sin\theta-\sin^{3}\theta\right) \end{alignedat}$

Then we can equate this to what we found previously, since we want to be dealing with $\sin^{3}\theta$ we can just equate the imaginary part.

Spoiler

$\begin{alignedat}{1}\cos\left(3\theta\right)+i\sin\left(3\theta\right) & =\left(\cos^{3}\theta-3\cos\theta\sin^{2}\theta\right)+i\left(3\cos^{2}\theta\sin\theta-\sin^{3}\theta\right) \\ \sin\left(3\theta\right) & =3\cos^{2}\left(\theta\right)\sin\left(\theta\right)-\sin^{3}\left(\theta\right) \\ \sin\left(3\theta\right) & =3\left(1-\sin^{2}\left(\theta\right)\right)\sin\left(\theta\right) \\ \sin\left(3\theta\right) & =3\sin\left(\theta\right)-3\sin^{3}\left(\theta\right)-\sin^{3}\left(\theta\right) \\ 4\sin^{3}\left(\theta\right) & =3\sin\left(\theta\right)-\sin\left(3\theta\right) \\ \sin^{3}\left(\theta\right) & =\frac{3}{4}\sin\left(\theta\right)-\frac{1}{4}\sin\left(3\theta\right) \end{alignedat}$

Hope that helps

Deleted User · « **Reply #124 on:** June 15, 2013, 08:56:50 pm »

Hi guys. Could someone explain to me how the quadric x^2+8xy+7y^2=9 is equivalent to the hyperbola u^2 - v^2/9 =1? Thanks

M-D · « **Reply #125 on:** June 15, 2013, 10:01:22 pm »

Quote from: b^3 on June 15, 2013, 07:29:07 pm

Hope that helps

thanks b^3.

BubbleWrapMan · « **Reply #126 on:** June 15, 2013, 10:58:03 pm »

Quote from: Deleted User on June 15, 2013, 08:56:50 pm

Hi guys. Could someone explain to me how the quadric x^2+8xy+7y^2=9 is equivalent to the hyperbola u^2 - v^2/9 =1? Thanks

How much do you know about rotations and stuff (for getting rid of the xy term)? I don't wanna explain stuff you already know.

Deleted User · « **Reply #127 on:** June 15, 2013, 11:09:11 pm »

I have no idea. I probably should considering I have an exam on it in 4 days but yeah, do you mind explaining it from the basics? Thanks!

BubbleWrapMan · « **Reply #128 on:** June 15, 2013, 11:34:18 pm »

Do you know how to rotate a curve about the origin using the 2D rotation matrix?

Deleted User · « **Reply #129 on:** June 17, 2013, 04:22:48 pm »

Is it this matrix?
|cosθ -sinθ |
|sinθ cosθ|

BubbleWrapMan · « **Reply #130 on:** June 17, 2013, 07:17:38 pm »

Yes. Is there anything in your notes about removing the $xy$ term by rotation?

Deleted User · « **Reply #131 on:** June 17, 2013, 09:34:52 pm »

BubbleWrapMan · « **Reply #132 on:** June 17, 2013, 10:16:49 pm »

I don't know why you'd be expected to be able to do that question then. But just to give you a decent hint, try to rotate the graph of $x^2+8xy+7y^2=9$ clockwise about the origin by an angle of $\arctan(2)$ .

Deleted User · « **Reply #133 on:** June 17, 2013, 10:31:04 pm »

Wow thanks! So why do I use the angle arctan(2)? And why do I have to use u and v instead of x and y?

BubbleWrapMan · « **Reply #134 on:** June 18, 2013, 12:16:42 am »

This has a decent explanation of rotations (courtesy of $b^{3}$ ): http://math.sci.ccny.cuny.edu/document/show/2685

This method gives you something like $-\frac{1}{2}\arctan\left(\frac{4}{3}\right)$ for your rotation angle, which I think is equal to $-\arctan\left(\frac{1}{2}\right)$ by some double angle formula. If you use this angle you get a hyperbola with a top half and a bottom half (as opposed to left and right halves) so you can add $\frac{\pi}{2}$ to this angle and rotate it further to get the type of hyperbola you want. You can do this because $\cot\left(2\left(\theta+\frac{\pi}{2}\right)\right)=\cot(2\theta)$ , so this angle still works and gives you a better result.

$u$ and $v$ are really different coordinates to $x$ and $y$ , but you can express them both in terms of $x$ and $y$ . I think that's what the question would be asking you for: something like $u=ax+by$ and $v=cx+dy$ .

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