Author Topic: TT's Maths Thread (Read 143992 times) Tweet Share

kamil9876 · « **Reply #930 on:** February 12, 2010, 11:22:27 pm »

$=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} * (\frac{\pi}{4})^{2n+1}$

so it really does tell you that it's a series evaluated at $x=\frac{\pi}{4}$ And the $2n$ tells you that every second derivative is 0. More precisely every odd derivative is 0. This already hints at something trigonometric.

TrueTears · « **Reply #931 on:** February 15, 2010, 06:23:51 pm »

Let P be a point on the line L that passes through the points Q and R. Show that the distance d from the point P to the line L is $d = \frac{ \left | \mathbf{a} \times \mathbf{b} \right |}{\left |\mathbf{a} \right |}$ where $\mathbf{a} = \overrightarrow{QR}$ and $\mathbf{b} = \overrightarrow{QP}$

Thanks

kamil9876 · « **Reply #932 on:** February 15, 2010, 07:07:54 pm »

let $<CQP=\theta$

$d=|QP|\sin\theta=|b|sin\theta$

$\implies |a|d=|a||b|\sin\theta$

$d=\frac{|a||b|\sin\theta}{|a|}$

and the result follows.

TrueTears · « **Reply #933 on:** February 15, 2010, 07:36:37 pm »

Oh yeah that makes sense, thanks.

TrueTears · « **Reply #934 on:** February 15, 2010, 09:06:45 pm »

Find the distance between the skewed lines:

Line 1. $x= 1+t, y = 1+6t, z = 2t$ where $t \in \mathbb{R}$

Line 2. $x=1+2s, y = 5+15s, z=-2+6s$ where $s \in \mathbb{R}$

Here's my working, can someone check if it's correct, cause there are no answers for this question.

The line which is parallel to line 1 is $v_1 = \mathbf{i} + 6\mathbf{j} + 2\mathbf{k}$

The line which is parallel to line 2 is $v_2 = 2\mathbf{i} +15\mathbf{j} +6\mathbf{k}$

Now let line 1 and line 2 can be considered to be on two parallel planes $P_1$ and $P_2$ .

Thus the normal vector for $P_1$ and $P_2$ must be the same.

This normal vector is given by:

$v_1 \times v_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 6 & 2 \\ 2 & 15 & 6 \end{vmatrix}$

$= \mathbf{i} \begin{vmatrix} 6 & 2 \\ 15 & 6 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2 \\ 2 & 6 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 6 \\ 2 & 15 \end{vmatrix}$

$= 6\mathbf{i} - 2\mathbf{j} +3 \mathbf{k}$

When $t = 0$ we have the point $(1,1,2)$ lying on the plane $P_1$ .

Thus the plane is given by: $6(x-1) -2(y-1) +3(z-2) = 0 \implies 6x-2y+3z-10=0$

Now when $s = 0$ we have the point $(1,5,-2)$ lying on line 2.

The distance that this point is from $P_1$ is given by:

$D = \frac{\left | 6(1) - 2(5) +3 (-2) - 10 \right |}{\sqrt{6^2+(-2)^2+3^2}} = \frac{20}{7}$

kamil9876 · « **Reply #935 on:** February 15, 2010, 10:44:31 pm »

Quote

When t = 0 we have the point (1,1,2) lying on the plane P_1.

(1,1,0)

TrueTears · « **Reply #936 on:** February 17, 2010, 11:24:23 am »

Reparametrize the curve $\mathbf{r}(t) = \left( \frac{2}{t^2+1}-1\right)\mathbf{i} + \frac{2t}{t^2+1}\mathbf{j}$ with respect to arc length measured from the point $(1,0)$ in the direction of increasing $t$ .

This is my working, can someone check if it's right because there are no answers =(

Define $\mathbf{r}(u) = \left( \frac{2}{u^2+1}-1\right)\mathbf{i} + \frac{2u}{u^2+1}\mathbf{j}$

$\implies \mathbf{r'}(u) = \frac{-4u}{(u^2+1)^2}\mathbf{i} - \frac{2(u^2-1)}{(u^2+1)^2}\mathbf{j}$

$\implies \left | \mathbf{r'}(u) \right | = \frac{2}{u^2+1}$

$(1,0)$ corresponds to when $t = 0$

$\implies s(t) = \int_0^t \frac{2}{u^2+1} du = 2\tan^{-1}(t)$

$\implies s = 2\tan^{-1}(t)$

$\implies \tan\left(\frac{s}{2}\right)=t$

$\implies \mathbf{r}(s) = \left( \frac{2}{\tan^2\left(\frac{s}{2}\right)+1}-1\right)\mathbf{i} + \frac{2\tan\left(\frac{s}{2}\right)}{\tan^2\left(\frac{s}{2}\right)+1}\mathbf{j}$

$\implies \mathbf{r}(s) =\left(2\sec^2\left(\frac{s}{2}\right)-1\right)\mathbf{i} + \sin(s) \mathbf{j}$

Damo17 · « **Reply #937 on:** February 17, 2010, 12:03:40 pm »

Quote from: TrueTears on February 17, 2010, 11:24:23 am

$\implies \mathbf{r}(s) = \left( \frac{2}{\tan^2\left(\frac{s}{2}\right)+1}-1\right)\mathbf{i} + \frac{2\tan\left(\frac{s}{2}\right)}{\tan^2\left(\frac{s}{2}\right)+1}\mathbf{j}$

$\implies \mathbf{r}(s) =\left(2\sec^2\left(\frac{s}{2}\right)-1\right)\mathbf{i} + \sin(s) \mathbf{j}$

$<br />\left( \frac{2}{\tan^2\left(\frac{s}{2}\right)+1}-1\right)\mathbf{i}=\left( \frac{1-\tan^2\left(\frac{s}{2}\right)}{\sec^2\left(\frac{s}{2}\right)}\right)\mathbf{i}=\left( {\cos^2\left(\frac{s}{2}\right)\ - sin^2\left(\frac{s}{2}\right)}\right)\mathbf{i}=cos(s)\mathbf{i}$

Everything else is correct.

TrueTears · « **Reply #938 on:** February 17, 2010, 12:09:07 pm »

!!! Thanks Damo!!

TrueTears · « **Reply #939 on:** February 18, 2010, 01:43:16 am »

Use polar coordinates to find the limit: $\lim_{(x,y) \to (0,0)} \frac{e^{-x^2-y^2}-1}{x^2+y^2}$

oh man i have no idea on this one wdf

Mao · « **Reply #940 on:** February 18, 2010, 01:46:35 am »

Any point in the x-y plane can be represented as a modulus and an angle. Modulus is r, distance from origin, angle is theta, angle from positive x axis in the ccw direction.

Hence, $x = r \cos \theta,\; y = r \sin \theta$

$L = \lim_{r \to 0} \frac{e^{-r^2} - 1}{r^2}$

TrueTears · « **Reply #941 on:** February 18, 2010, 01:48:11 am »

OH

MY

FKN

GOD

HOW DID I NOT REMEMBER THAT WDF, MUST BE LACK OF SLEEP

thanks mao btw

TrueTears · « **Reply #942 on:** February 18, 2010, 01:45:05 pm »

Say we have the function $f(x_1,x_2,x_3 \cdots x_n) = c_1x_1 + c_2x_2 + \cdots c_nx_n$

Then this can be represented by a single vector variable $\mathbf{x} = <x_1,x_2,x_3 \cdots x_n>$ and $\mathbf{c} = <c_1,c_2,c_3 \cdots c_n>$ so that we get $f(\mathbf{x}) = \mathbf{x}.\mathbf{c}$

I understand where the dot product comes from but the $f(\mathbf{x})$ doesn't make sense, isn't that saying we have $f(\mathbf{x}) = f(x_1\mathbf{i},x_2\mathbf{j},x_3\mathbf{k} \cdots x_n\mathbf{n})$ rather than $f(x_1,x_2,x_3 \cdots x_n)$ ?

In other words inside the function we now have a vector instead of a set of coordinates?

Mao · « **Reply #943 on:** February 19, 2010, 12:58:35 am »

Yes, in this case it's the same thing.

TrueTears · « **Reply #944 on:** February 19, 2010, 08:07:02 pm »

Thanks Mao!

Can someone show me how to use an $\epsilon - \delta$ proof to show $\lim_{(x,y,z) \to (3,0,1)} e^{-xy} \sin\left(\frac{\pi z}{2}\right) = 1$

I've done some stuff, but it doesn't really lead anywhere... so I won't be bothered posting it.

Search the forums now!

We have moved!

Author Topic: TT's Maths Thread (Read 143992 times) Tweet Share

kamil9876

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

kamil9876

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

kamil9876

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Damo17

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Mao

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Mao

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Recent Posts

User:
Password: