Author Topic: TT's Maths Thread (Read 158460 times) Tweet Share

TrueTears · « **Reply #825 on:** February 01, 2010, 12:12:21 am »

O yeah, cause what happened was I zoomed in on my calc, it didn't look like a vertical tangent all O.O but then I realised the calc was on shit settings lol.

And also how do you know what values of $\theta$ to solve for? Is it typically between 0 and $2\pi$ since that sweeps every angle for the polar graph?

TrueTears · « **Reply #826 on:** February 01, 2010, 12:25:37 am »

Quote from: Damo17 on January 31, 2010, 11:36:13 pm

Quote from: TrueTears on January 31, 2010, 10:58:53 pm
So for horizontal tangents: $\sin(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$

For vertical tangents: $\cos(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$

Values of $\theta$ do not exist in the 2nd and 4th quadrants for this graph. That should clear up many questions.

Yeah, I should have been more specific and eliminated those such as $\frac{5\pi}{6}$ etc, but still I'm a bit hazy on why you solve across the the domain 0 to $2\pi$ .

Damo17 · « **Reply #827 on:** February 01, 2010, 12:28:25 am »

Quote from: TrueTears on February 01, 2010, 12:12:21 am

O yeah, cause what happened was I zoomed in on my calc, it didn't look like a vertical tangent all O.O but then I realised the calc was on shit settings lol.

And also how do you know what values of $\theta$ to solve for? Is it typically between 0 and $2\pi$ since that sweeps every angle for the polar graph?

Quote from: TrueTears on February 01, 2010, 12:25:37 am

Yeah, I should have been more specific and eliminated those such as $\frac{5\pi}{6}$ etc, but still I'm a bit hazy on why you solve across the the domain 0 to $2\pi$ .

Yeah, usually $[0, 2 \pi]$ but remember any restrictions. e.g in your question $sin(2 \theta) \ge 0$ , so there is a restriction on $\theta$ to the 1st and 3rd quadrants.

TrueTears · « **Reply #828 on:** February 01, 2010, 12:29:12 am »

Yeah, so what's the reason for solving across that domain?

TrueTears · « **Reply #829 on:** February 01, 2010, 09:32:38 pm »

Thanks ^^

Weird question, I don't know why my way doesn't work...

Find the area bounded by the 2 curves $r^2 = \sin(2\theta)$ and $r^2 = \cos(2\theta)$

Now after some working, the 2 curves look like the following: (p means $\pi$ )

Red is the curve $r^2 = \cos(2\theta)$

Purple is the curve $r^2 = \sin(2\theta)$

Now we need to find the area defined by $4 \times \int_0^{\frac{\pi}{8}} \frac{1}{2}r^2d\theta = 4 \times \int_0^{\frac{\pi}{8}} \frac{1}{2}\sin(2\theta)d\theta$

But answers has $4 \times \int_0^{\frac{\pi}{8}} \frac{1}{2}\sin^2(2\theta)d\theta$

How is it $\sin^2(2\theta)$ ...?

kamil9876 · « **Reply #830 on:** February 01, 2010, 11:12:19 pm »

TrueTears · « **Reply #831 on:** February 02, 2010, 05:02:25 pm »

I have a few questions regarding conic sections.

Let F be a fixed point called the focus and L be a fixed line called the directrix in a plane. Let e be a fixed positive number called the eccentricity. The set of all points P in the plane such that $\frac{|PF|}{|PL|} = e$ is a conic section. The conic is an ellipse if $e<1$ , a parabola if $e=1$ or a hyperbola if $e>1$ .

I understand how a parabola works, since there is only 1 focus and 1 directrix and also the line joining the point P to the directrix is perpendicular to the directrix.

But what if you have an ellipse? They have 2 foci so does that mean we only pick 1 focus to deal with? Also does the line joining the point P to the directrix have to be perpendicular to the directrix?

Christiano · « **Reply #832 on:** February 02, 2010, 10:45:58 pm »

I need help on rearranging this formula to make q the subject

$2w^2q=2\sqrt{3}-3q$

TrueTears · « **Reply #833 on:** February 02, 2010, 10:48:55 pm »

You can bring the $3q$ to the LHS so you get

$2w^2q+3q = 2\sqrt{3}$

Then factorise the $q$

$q(2w^2+3) = 2\sqrt{3}$

Now divide both sides by $(2w^2+3)$

$q = \frac{2\sqrt{3}}{(2w^2+3)}$

Christiano · « **Reply #834 on:** February 02, 2010, 10:53:27 pm »

And .. to make i the subject

$f=\sqrt{\frac{h^3i^2-3g}{2}}$

Christiano · « **Reply #835 on:** February 02, 2010, 10:55:24 pm »

Quote from: TrueTears on February 02, 2010, 10:48:55 pm

You can bring the $3q$ to the LHS so you get

$2w^2q+3q = 2\sqrt{3}$

Then factorise the $q$

$q(2w^2+3) = 2\sqrt{3}$

Now divide both sides by $(2w^2+3)$

$q = \frac{2\sqrt{3}}{(2w^2+3)}$

+1 .. How do you bring karma up ? hehe

Edmund · « **Reply #836 on:** February 02, 2010, 11:00:19 pm »

$f=\sqrt{\frac{h^3i^2-3g}{2}}$

Square both sides:

$f^2=\frac{h^3i^2-3g}{2}$

Multiply both sides by 2:

$2f^2=h^3i^2-3g$

$2f^2+3g=h^3i^2$

Divide both sides by h³

$\frac{2f^2+3g}{h^3}=i^2$

Square root both sides

$i=\sqrt{\frac{2f^2+3g}{h^3}}$

Quote from: Christiano on February 02, 2010, 10:55:24 pm

+1 .. How do you bring karma up ? hehe

Once you reach 50 posts

Christiano · « **Reply #837 on:** February 02, 2010, 11:20:14 pm »

Thanks a lot!

Christiano · « **Reply #838 on:** February 03, 2010, 12:06:57 am »

More simple equations I'm stuck on.

$3-(x-2)^2$

$64y^4-16y^2+1$

$(3p-5)^2-(p+2)^2$

$a^3-3a^2+a-3$

TrueTears · « **Reply #839 on:** February 03, 2010, 02:57:49 pm »

$\lim_{x \to \infty} x^x$ (For $x >1$ )

Can we use the comparison tool here?

$x^x > x \ \forall \ x>1$

$\lim_{x \to \infty} x^x > \lim_{x \to \infty} x = \infty$

So $\lim_{x \to \infty} x^x = \infty$

Or can we just treat $x^x$ as a continuous function for $x>1$

So $\lim_{x \to \infty} x^x = x^{\lim_{x \to \infty} x} = x^{\infty} = \infty$

But $\infty$ is not a number so can you still substitute it in the exponent of $x$ ?

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