August 20, 2017, 12:02:43 pm

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#### bronwoolbank

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##### Re: 3U Maths Question Thread
« Reply #2655 on: August 10, 2017, 01:53:39 pm »
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Hallo, I am having trouble with this

#### Opengangs

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##### Re: 3U Maths Question Thread
« Reply #2656 on: August 10, 2017, 02:20:45 pm »
+5
$\frac{d}{dx}[e^{secx}] = secxtanx \cdot e^{secx}$

We can note that this simply uses the chain rule (or substitution), where we let secx = u.
From here, we can see that, by simple chain rule, we get the derivative.

Since the derivative of e^secx = secxtanx * e^secx, then we can note:
$4\frac{d}{dx}[e^{secx}] = 4secxtanx \cdot e^{secx}$
$4\int {\frac{d}{dx}[e^{secx}]}\,dx = \int{4secxtanx \cdot e^{secx}}\,dx$

$\int{4secxtanx \cdot e^{secx}}\,dx = 4e^{secx} + C$
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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2657 on: August 10, 2017, 02:22:27 pm »
+5
Hallo, I am having trouble with this
$\text{Note that the standard integral}\\ \int \sec ax \tan ax \, dx = \frac1a \sec ax \\ \text{has since been removed, and the result}\\ \frac{d}{dx}\sec x = \sec x \tan x\text{ does not need to be memorised.}$
$\text{However, to convince yourself that }\frac{d}{dx}\sec x = \sec x \tan x\\ \text{you should differentiate it manually, using the chain rule}\\ \text{i.e. consider }\frac{d}{dx}(\cos x)^{-1}$
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#### winstondarmawan

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##### Re: 3U Maths Question Thread
« Reply #2658 on: August 10, 2017, 06:17:42 pm »
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#### Opengangs

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##### Re: 3U Maths Question Thread
« Reply #2659 on: August 10, 2017, 07:21:41 pm »
+5
Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20771709_1295975427194644_112379427_o.png?oh=4e056ee112237fc0488eb6bc368903d0&oe=598ED61C
$\text{Recall:} a= \frac{dv}{dt} = \frac{d\frac{1}{2}v^{2}}{dx}$
$a = \frac{1}{2}\frac{d[\frac{2}{1 + 3x}]^{2}}{dx} = \frac{1}{2} \cdot \frac{d[\frac{4}{(1 + 3x)^{2}}]}{dx}$
$= \frac{1}{2}\frac{-24}{(1 + 3x)^{3}} = \frac{-12}{(1 + 3x)^{3}}$

Varies directly with v^3 means that a = kv^3 (I think), so all we need to show is that a can be written in a form: K v^3, where K is an arbitrary constant.

$\frac{-12}{(3x + 1)^{3}} = Kv^{3}\, K \in \mathbb{R}$
$LHS = \frac{-3(4)}{(3x + 1)^{2}(3x + 1)}$
$= \frac{-3v^{2}}{(3x + 1)}$
$= \frac{-\frac{3}{2} \cdot 2 \cdot v^{2}}{(3x + 1)}$
$= -\frac{3}{2}v^{3}$
$= Kv^{3}\, K = -\frac{3}{2}$

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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2660 on: August 10, 2017, 08:45:57 pm »
+7
$\text{Recall:} a= \frac{dv}{dt} = \frac{d\frac{1}{2}v^{2}}{dx}$
$a = \frac{1}{2}\frac{d[\frac{2}{1 + 3x}]^{2}}{dx} = \frac{1}{2} \cdot \frac{d[\frac{4}{(1 + 3x)^{2}}]}{dx}$
$= \frac{1}{2}\frac{-24}{(1 + 3x)^{3}} = \frac{-12}{(1 + 3x)^{3}}$
$\text{Dumping the entire fraction in the numerator and building a nested fraction}\\ \text{ really isn't good practice when it involves the derivative, and is also hard to read.}\\\text{The only thing I'd say is to vouch for something like}\\ \text{this instead: }\frac{d}{dx} \left(\frac{4}{1+3x^2}\right)$
Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20771709_1295975427194644_112379427_o.png?oh=4e056ee112237fc0488eb6bc368903d0&oe=598ED61C
$\text{Clearly, if when }t=0, x=0\\ \text{ we have}\\ v=2, a=-12\\ \text{You should be able to describe the motion easily.}$
$\text{Now, for the other part, think about it intuitively.}\\ \text{If you look at your equations, you will see that for any }x\\ v > 0\text{ and }a < 0$
$\text{This is the easiest way to observe that the particle always travels to the right}\\ \text{but also always accelerates towards the left}\\ \text{i.e. the particle is slowing down.}$
$\text{Now, the particle starts from the origin.}\\ \text{Because }v > 0\text{, it will always be moving to the right (and never left)}\\ \text{so it will always be moving towards the right of the origin}$
$\text{This will keep happening, as despite the fact that }a < 0\\ \text{i.e. it accelerates to the left}\\ \text{it never accelerates fast enough so that the particle turns around!}$
$\text{Because this keeps happening, we should anticipate that}\\ \text{as }t\to \infty, x\to \infty\text{ as well.}$
$\text{This also allows us to determine the limiting velocity and acceleration}\\ \lim_{t\to \infty} v = \lim_{x\to \infty}\frac{2}{1+3x}=\lim_{x\to \infty}\frac{\frac{2}{x}}{\frac{1}{x}+3}=0\\ \text{and similarly }\lim_{t\to \infty}a = 0$
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#### austv99

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##### Re: 3U Maths Question Thread
« Reply #2661 on: August 10, 2017, 09:00:05 pm »
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Hi,
Would appreciate some help with this question.
Thanks.
p is meant to be pi

#### Opengangs

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##### Re: 3U Maths Question Thread
« Reply #2662 on: August 10, 2017, 09:09:08 pm »
+4
Hi,
Would appreciate some help with this question.
Thanks.
p is meant to be pi
Recall that the derivative of 1/a sec(ax) is sec(ax)tan(ax).
Using this standard integral:
$\int sec(2x)tan(2x)\,dx = \frac{1}{2}sec(2x) + C$

Using bounds, we get: $\int_0^{\frac{\pi}{6}} sec(2x)tan(2x)\,dx = \frac{1}{2}$
« Last Edit: August 10, 2017, 09:11:52 pm by Opengangs »
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##### Re: 3U Maths Question Thread
« Reply #2663 on: August 10, 2017, 09:18:31 pm »
+4
Hi,
Would appreciate some help with this question.
Thanks.
p is meant to be pi
From what I know, that derivative has since been removed from the course, so I expect they want to use substitution
$\sec 2x \tan 2x = \frac{1}{\cos 2x}* \frac{\sin 2x}{\cos 2x} = \frac{\sin 2x}{\cos ^2 2x}\\ \text{Substitute u= cos(2x) and go from there}$
You should get the same answer as Opengangs
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##### Re: 3U Maths Question Thread
« Reply #2664 on: August 11, 2017, 10:54:56 pm »
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Hey, could I get some help part ii please?

#### Opengangs

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##### Re: 3U Maths Question Thread
« Reply #2665 on: August 11, 2017, 11:12:03 pm »
+4
Hey, could I get some help part ii please?
$\frac{dx}{dt} = sinxcosx$
$\frac{dx}{sinxcosx} = dt$
$\int \frac{dx}{sinxcosx} = \int dt$
$ln(tanx) = t + C$

$ln(1) = 0 + C$
$C = 0$
$\therefore ln(tanx) = t$

$\text{Solving for x:}$
$tanx = e^{t}$
$\therefore x = tan^{-1}(e^{t})$
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#### legorgo18

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##### Re: 3U Maths Question Thread
« Reply #2666 on: August 11, 2017, 11:37:14 pm »
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Help! Got super lost in this wall of text...

Brine containing 2g of salt/L flows into a container initially filled with 50L of water containing 10g of salt. If the brine enters the tank at 5L/min, the concentration is kept uniform by stirring and the mixture flows out at the same rate. P(t) is the amount of salt (g) in the tank after t mins,. Find the amount of salt in the tank after 10 mins.
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#### jamonwindeyer

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##### Re: 3U Maths Question Thread
« Reply #2667 on: August 12, 2017, 12:41:35 am »
+4
Help! Got super lost in this wall of text...

Brine containing 2g of salt/L flows into a container initially filled with 50L of water containing 10g of salt. If the brine enters the tank at 5L/min, the concentration is kept uniform by stirring and the mixture flows out at the same rate. P(t) is the amount of salt (g) in the tank after t mins,. Find the amount of salt in the tank after 10 mins.

Hey! So let's start by trying to focus only on the flow of the quantity we care about, that's salt. So, we know:

- 10 grams of salt flows into the tank per minute (5 litres, 2 grams per litre)
- Initially, there are 10 grams of salt in the tank

Let's look at the rate of change of salt. It is 10 grams going in per minute, and then $\frac{P}{10}$ is flowing out per minute (5 litres out of 50 litres, one tenth of the quantity of salt in the tank at that time).

$\frac{dP}{dt}=10-\frac{P}{10}\\\frac{dt}{dP}=\frac{1}{10-\frac{P}{10}}\\t=-10\ln{\left(10-\frac{P}{10}\right)}+C$

Now to find the constant C, we use the initial condition of 10 grams of salt in the tank.

$t=0, P=10\implies -10\ln{\left(10-\frac{10}{10}\right)}+C=0\\C=10\ln{9}\\\therefore t=-10\ln{\left(10-\frac{P}{10}\right)}+10\ln{9}$

So we rearrange this to get P:

$0.1t=\ln{9}-\ln{\left(10-\frac{P}{10}\right)}\\0.1t=\ln{\frac{9}{10-\frac{P}{10}}}\\e^{0.1t}=\frac{9}{10-\frac{P}{10}}\\10-\frac{P}{10}=9e^{-0.1t}\\100-P=90e^{-0.1t}\\P=100-90e^{-0.1t}$

#### sssona09

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##### Re: 3U Maths Question Thread
« Reply #2668 on: August 15, 2017, 07:28:29 am »
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heey <br>
so i know how to do this problem using yhe fact that the points are collinear thus they have same gradient.<br>
but is there a shorter method?? or how do we do this using x^2=4ay?<br>
<br>
the q is:<br>
the focal cord that cuts tge parabola at x^2= -6y at (6,-6) cuts the parabola again at X. find the coordinates of X.

thank you!!
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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2669 on: August 15, 2017, 12:07:37 pm »
+4
heey <br>
so i know how to do this problem using yhe fact that the points are collinear thus they have same gradient.<br>
but is there a shorter method?? or how do we do this using x^2=4ay?<br>
<br>
the q is:<br>
the focal cord that cuts tge parabola at x^2= -6y at (6,-6) cuts the parabola again at X. find the coordinates of X.

thank you!!
To be honest, the only "other way" to do it is to explicitly find the equation of the chord (i.e. the equation of the line passing through (6,-6) and (0,-1.5)), then also explicitly find the point of intersection between this line and the parabola.

This approach would make more intuitive sense, but it's bound to be slower. If you want a faster way then I don't really think there is any.
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