Author Topic: Mathematics Question Thread (Read 1625939 times) Tweet Share

RuiAce · « **Reply #2295 on:** July 27, 2017, 07:05:43 pm »

Quote from: winstondarmawan on July 27, 2017, 05:10:31 pm

Hello, would appreciate help with the following, TIA.
Mainly not sure on how the n from n-sided polygon gets incorporated into the angle.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20427880_1282466078548868_1386597183_n.png?oh=a54f49ae1c9642fa30bdf54ebbf9cd6d&oe=597C71D1

This one was asked a while back

katnisschung · « **Reply #2296 on:** July 27, 2017, 07:10:08 pm »

Hi need help with this thanks ins advance!
Find the area between the curve y=2ln(x+1), the x axis and the line x=2
How do I rearrange this for x.

RuiAce · « **Reply #2297 on:** July 27, 2017, 07:12:58 pm »

Quote from: katnisschung on July 27, 2017, 07:10:08 pm

Hi need help with this thanks ins advance!
Find the area between the curve y=2ln(x+1), the x axis and the line x=2
How do I rearrange this for x.

$\begin{align*}\frac{y}2&=\ln (x+1)\\ e^{\frac{y}2}&=x+1\\ e^{\frac{y}{2}}-1&=x\end{align*}$
I'll let you have a go at the remainder first. Comment again if you need further help

katnisschung · « **Reply #2298 on:** July 27, 2017, 08:27:39 pm »

How do I find the points of intersections between the 2 trig graphs
Sin x and 1/2 tanx I tried it solving it graphically but that didn't work out so well

RuiAce · « **Reply #2299 on:** July 27, 2017, 08:29:41 pm »

Quote from: katnisschung on July 27, 2017, 08:27:39 pm

How do I find the points of intersections between the 2 trig graphs
Sin x and 1/2 tanx I tried it solving it graphically but that didn't work out so well

$\begin{align*}\sin x &= \frac12\tan x\\ {\sin x }&=\frac12 \times \frac{\sin x}{\cos x}\\ \sin x\left(1-\frac{1}{2\cos x}\right)&=0\end{align*}$
$\text{i.e. }\sin x = 0\text{ or }\cos x = \frac12$
The relevant related angle for whatever purpose is $\frac\pi3$

VanillaRice · « **Reply #2300 on:** July 27, 2017, 08:47:39 pm »

Quote from: katnisschung on July 27, 2017, 08:27:39 pm

How do I find the points of intersections between the 2 trig graphs
Sin x and 1/2 tanx I tried it solving it graphically but that didn't work out so well

We can use an algebraic method to solve for x, and then substitute those values into either equation to obtain the y value:
$\sin x = \frac {1}{2} \tan x$

Spoiler

$\sin x - \frac {1}{2} \tan x = 0$
$\sin x - \frac {1}{2} \frac{\sin x}{\cos x} = 0$
$\sin x (1 - \frac {1}{2}\frac {1}{\cos x}) = 0$

By the null factor law, we now need to solve for the two cases:
$\sin x = 0$
and

Spoiler

$1-\frac {1}{2} \frac {1}{\cos x} = 0$
$1=\frac {1}{2} \frac {1}{\cos x}$
$1 = 2\cos x$

$\cos x = \frac{1}{2}$
After finding a general solution for both cases, substitute a specific x-value from each general solution (should be four different cases) to obtain the y-values.

EDIT: Oops got beaten to it

EDIT2:

Quote from: RuiAce on July 27, 2017, 08:29:41 pm

$\begin{align*}\sin x &= \frac12\tan x\\ {\sin x }&=\frac12 \times \frac{\sin x}{\cos x}\\ \cos x &=\frac12\end{align*}$
The relevant related angle for whatever purpose is $\frac\pi3$

I think you've accidentally deleted a solution by dividing both sides by sin(x)

RuiAce · « **Reply #2301 on:** July 27, 2017, 09:02:36 pm »

Quote from: VanillaRice on July 27, 2017, 08:47:39 pm

EDIT2: I think you've accidentally deleted a solution by dividing both sides by sin(x)

Hmm. Walked into my own trap it seems; fixing.

kiiaaa · « **Reply #2302 on:** July 27, 2017, 10:17:38 pm »

hey
could someones please help me how to tackle this question please? I couldnt find the answers and nor do i know where to start from?
Thanks soo much guys!!:)))

RuiAce · « **Reply #2303 on:** July 27, 2017, 10:22:00 pm »

Quote from: kiiaaa on July 27, 2017, 10:17:38 pm

hey
could someones please help me how to tackle this question please? I couldnt find the answers and nor do i know where to start from?
Thanks soo much guys!!:)))

$\frac{dM}{dt}=-kM\implies M=M_0e^{-kt}$
$\text{Since }M_0\text{ is the initial mass, we are saying that}\\ \text{because 29yr is the half-life, when }t=29\text{ we have }M=\frac{M_0}{2}$
$\text{Hence}\\ \begin{align*}\frac{M_0}{2}&=M_0e^{-29k}\\ \frac12 &= e^{-29k}\\ k&=-\frac1{29}\ln \frac12\\ &= \text{whatever the calculator says}\end{align*}$
I will let you have a go at ii).

Mymy409 · « **Reply #2304 on:** July 28, 2017, 08:43:43 am »

Quote from: Shadowxo on July 27, 2017, 01:20:51 pm

The velocity is negative so it's moving to the left
The acceleration is positive so the 'velocity' is increasing but it's just becoming less negative. The 'speed' however is decreasing (as the initial speed is 3 m/s, speed doesn't have a direction). So it's slowing down (speed decreasing). It's just a terminology thing, does this help?

Thanks for your reply! Why is the speed decreasing, though? I still don't quite get it.

RuiAce · « **Reply #2305 on:** July 28, 2017, 09:25:19 am »

Quote from: Mymy409 on July 28, 2017, 08:43:43 am

Thanks for your reply! Why is the speed decreasing, though? I still don't quite get it.

We have stated that the particle is moving to the left (because v is negative).

We have also stated that the particle is accelerating to the right (a is positive). Consider what this means.

If the particle is moving to the left, AND accelerating to the left, it will move to the left more quickly. I.e. When the acceleration and velocity work in the SAME direction, the particle speeds up.

But we have the opposite. If the particle is moving to the left, BUT it's being accelerated towards the other direction (the right), something is causing the particle to actually WANT to move to the right, instead of to the left. This means that the particle slows down when the velocity and acceleration work against each other.

Mymy409 · « **Reply #2306 on:** July 28, 2017, 10:25:23 am »

Quote from: RuiAce on July 28, 2017, 09:25:19 am

We have stated that the particle is moving to the left (because v is negative).

We have also stated that the particle is accelerating to the right (a is positive). Consider what this means.

If the particle is moving to the left, AND accelerating to the left, it will move to the left more quickly. I.e. When the acceleration and velocity work in the SAME direction, the particle speeds up.

But we have the opposite. If the particle is moving to the left, BUT it's being accelerated towards the other direction (the right), something is causing the particle to actually WANT to move to the right, instead of to the left. This means that the particle slows down when the velocity and acceleration work against each other.

That makes sense, thank you!

Mounica · « **Reply #2307 on:** July 29, 2017, 06:58:49 pm »

Hey Guys,
Can someone pls help me with this question.
thanks

RuiAce · « **Reply #2308 on:** July 29, 2017, 07:07:05 pm »

Quote from: Mounica on July 29, 2017, 06:58:49 pm

Hey Guys,
Can someone pls help me with this question.
thanks

$\text{The diameter is just }2r\\ \text{so if the diameter and height sum to 60 we're just saying}\\ 2r+h=60\implies \boxed{h=60-2r}$
____________________________________
$\text{Plugging this into the volume formula:}\\ \begin{align*}V&=\frac13 \pi r^2 h\\ &= \frac13 \pi r^2 (60-2r)\\ &= 20\pi r^2 - \frac23 \pi r^3\end{align*}$
____________________________________
$\text{This is now a standard max-min problem}\\ \begin{align*}\frac{dV}{dr}&=40\pi r-2\pi r^2\\ &= 2\pi r(20-r)\\&=0\end{align*}\\ \text{when }r=0,r=20$
$\text{Rejecting }r=0\text{ as else we have no radius we gather }\boxed{r=20}\\ \text{Follow the usual process to check that it's a maximum}\\ \text{and then substitute back into }V\text{ to obtain the maximum volume }\frac{8000\pi}3$

Mounica · « **Reply #2309 on:** July 29, 2017, 07:15:34 pm »

Quote from: RuiAce on July 29, 2017, 07:07:05 pm

$\text{The diameter is just }2r\\ \text{so if the diameter and height sum to 60 we're just saying}\\ 2r+h=60\implies \boxed{h=60-2r}$
____________________________________
$\text{Plugging this into the volume formula:}\\ \begin{align*}V&=\frac13 \pi r^2 h\\ &= \frac13 \pi r^2 (60-2r)\\ &= 20\pi r^2 - \frac23 \pi r^3\end{align*}$
____________________________________
$\text{This is now a standard max-min problem}\\ \begin{align*}\frac{dV}{dr}&=40\pi r-2\pi r^2\\ &= 2\pi r(20-r)\\&=0\end{align*}\\ \text{when }r=0,r=20$
$\text{Rejecting }r=0\text{ as else we have no radius we gather }\boxed{r=20}\\ \text{Follow the usual process to check that it's a maximum}\\ \text{and then substitute back into }V\text{ to obtain the maximum volume }\frac{8000\pi}3$

Thank You

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