Author Topic: Mathematics Question Thread (Read 1626435 times) Tweet Share

georgiia · « **Reply #2235 on:** July 20, 2017, 04:25:01 pm »

Thanks!!

jaskirat · « **Reply #2236 on:** July 20, 2017, 04:59:28 pm »

Need help with this please! Forgot this topic 😐

RuiAce · « **Reply #2237 on:** July 20, 2017, 05:03:29 pm »

Quote from: jaskirat on July 20, 2017, 04:59:28 pm

Need help with this please! Forgot this topic 😐

$\begin{align*}x&=3\sin 2t\\v&=6\cos 2t\\ a&= -12\sin 2t\end{align*}$
Literally sub in t=0 to do part a).
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$\text{Looking at the range of the function }x=3\sin 2t\\ \text{The maximum displacement is 3.}$
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$\text{At rest: }v=0\\ \text{Solving }6\cos 2t=0\text{ gives }\\ 2t=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\\ \text{i.e. }t=\frac{\pi}4, \frac{3\pi}4, \frac{5\pi}4, \dots$
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$\text{Compare the equations:}\\ \begin{align*}x&=3\sin 2t\\ a&=-12\sin 2t\end{align*}$
$\text{Literally sub the first into the second to obtain }a=-4x$

georgiia · « **Reply #2238 on:** July 21, 2017, 12:29:30 pm »

I ve solved the integral and get that K needs to be 0??

Can someone please explain this?

RuiAce · « **Reply #2239 on:** July 21, 2017, 12:39:32 pm »

Quote from: georgiia on July 21, 2017, 12:29:30 pm

I ve solved the integral and get that K needs to be 0??

Can someone please explain this?

$\text{When }k=0\text{ this is trivially true as}\\ \int_0^\pi \sin (0x)\,dx= \int_0^\pi 0\, dx = 0$
$\text{In the situations where }k\neq 0\\ \text{Evaluating the integral gives}\\ \begin{align*}\int_0^\pi \sin kx\, dx&=0\\ \left[\frac{1}{k}\cos kx\right]_0^\pi &=0\\ \frac{1}{k}(\cos k\pi - \cos 0)&=0\\ \cos k\pi = 1\end{align*}$
$\text{Of course, we don't have say }0\le x \le 2\pi\\ \text{So we want to write down }\textbf{ALL}\text{ the solutions to the above equation}$
$\begin{align*}\cos k\pi &=1\\ k\pi &= \pm 2\pi, \pm 4\pi, \pm 6\pi, \dots\\ k&= \pm2, \pm4, \pm6, \dots\end{align*}$
$\text{So the answer is all the even integers}$

georgiia · « **Reply #2240 on:** July 21, 2017, 01:45:29 pm »

Quote from: RuiAce on July 21, 2017, 12:39:32 pm

$\text{When }k=0\text{ this is trivially true as}\\ \int_0^\pi \sin (0x)\,dx= \int_0^\pi 0\, dx = 0$
$\text{In the situations where }k\neq 0\\ \text{Evaluating the integral gives}\\ \begin{align*}\int_0^\pi \sin kx\, dx&=0\\ \left[\frac{1}{k}\cos x\right]_0^\pi &=0\\ \frac{1}{k}(\cos k\pi - \cos 0)&=0\\ \cos k\pi = 1\end{align*}$
$\text{Of course, we don't have say }0\le x \le 2\pi\\ \text{So we want to write down }\textbf{ALL}\text{ the solutions to the above equation}$
$\begin{align*}\cos k\pi &=0\\ k\pi &= \pm 2\pi, \pm 4\pi, \pm 6\pi, \dots\\ k&= \pm2, \pm4, \pm6, \dots\end{align*}$
$\text{So the answer is all the even integers}$

Thanks!!

jaskirat · « **Reply #2241 on:** July 21, 2017, 04:52:14 pm »

Need help with these two questions, thanks

RuiAce · « **Reply #2242 on:** July 21, 2017, 04:57:14 pm »

Quote from: jaskirat on July 21, 2017, 04:52:14 pm

Need help with these two questions, thanks

$\begin{align*}\sec^2x+\sec x\tan x &= \frac{1+\sin x}{\cos^2 x}\\ &=\frac{1+\sin x}{1-\sin^2x}\\ &= \frac{1+\sin x}{(1+\sin x)(1-\sin x)}\\ & = \frac{1}{1-\sin x}\end{align*}$
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$\frac{d}{dx}\log_e(\cos x)=\frac{-\sin x}{\cos x}=-\tan x$
This second one involved nothing challenging. What was causing the struggle?

jaskirat · « **Reply #2243 on:** July 21, 2017, 05:04:39 pm »

Quote from: RuiAce on July 21, 2017, 04:57:14 pm

$\begin{align*}\sec^2x+\sec x\tan x &= \frac{1+\sin x}{\cos^2 x}\\ &=\frac{1+\sin x}{1-\sin^2x}\\ &= \frac{1+\sin x}{(1+\sin x)(1-\sin x)}\\ & = \frac{1}{1-\sin x}\end{align*}$
_____
$\frac{d}{dx}\log_e(\cos x)=\frac{-\sin x}{\cos x}=-\tan x$
This second one involved nothing challenging. What was causing the struggle?

Right after i posted about needing help with the second question, i remembered it lol! Thanks bro!

12070 · « **Reply #2244 on:** July 22, 2017, 02:05:42 pm »

Hey, was just wondering how if 3e^t=12/(e^2t), e^3t=4

RuiAce · « **Reply #2245 on:** July 22, 2017, 02:06:52 pm »

Quote from: 12070 on July 22, 2017, 02:05:42 pm

Hey, was just wondering how if 3e^t=12/(e^2t), e^3t=4

$\begin{align*}3e^t &= \frac{12}{e^{2t}}\\ e^t&=4e^{-2t}\end{align*}$
$\text{So multiplying }e^{2t}\text{ to both sides, from our index laws we have}\\ e^{3t}=4$

winstondarmawan · « **Reply #2246 on:** July 23, 2017, 02:29:51 pm »

Hello! Would appreciate help with Q3:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20270047_1538514876198918_1314381077_n.jpg?oh=3ce907b341fda76bc69f71d6d59fe730&oe=5975F432
TIA

RuiAce · « **Reply #2247 on:** July 23, 2017, 02:34:33 pm »

$\begin{align*}PA&=3PB\\ PA^2&=9PB^2\\ (x+2)^2 + (y-0)^2&=9\left[(x-2)^2+(y-0)^2\right]\\ x^2+4x+4+9y^2&=9x^2-36x+36+9y^2\\ 8x^2-40x+y^2+32=0\\ x^2+y^2-5x+4&=0\end{align*}$
$\text{You should be able to rearrange it to being the equation of a circle.}\\ \text{Completing the square gives}\\ \left(x-\frac52\right)^2+y^2&=-4+\frac{25}{4}$ Subject to computational inaccuracy

winstondarmawan · « **Reply #2248 on:** July 23, 2017, 02:57:22 pm »

Quote from: RuiAce on July 23, 2017, 02:34:33 pm

$\begin{align*}PA&=3PB\\ PA^2&=9PB^2\\ (x+2)^2 + (y-0)^2&=9\left[(x-2)^2+(y-0)^2\right]\\ x^2+4x+4+9y^2&=9x^2-36x+36+9y^2\\ 8x^2-40x+y^2+32=0\\ x^2+y^2-5x+4&=0\end{align*}$
$\text{You should be able to rearrange it to being the equation of a circle.}\\ \text{Completing the square gives}\\ \left(x-\frac52\right)^2+y^2&=-4+\frac{25}{4}$ Subject to computational inaccuracy

I was able to do that part, but theres a 2nd part to the Q

RuiAce · « **Reply #2249 on:** July 23, 2017, 03:11:14 pm »

Oh. I take the blame for not noticing it.

I won't be able to get to it for a while though; I can't do that without a diagram

That being said, please make it clearer what you required for future benefit

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