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August 20, 2017, 12:02:43 pm

Author Topic: 3U Maths Question Thread  (Read 139625 times)  Share 

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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 »
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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:



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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
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winstondarmawan

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Opengangs

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Re: 3U Maths Question Thread
« Reply #2659 on: August 10, 2017, 07:21:41 pm »
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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




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.








(I'm not sure about this; you might have to wait for Rui to answer).
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RuiAce

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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 »
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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:


Using bounds, we get:
« Last Edit: August 10, 2017, 09:11:52 pm by Opengangs »
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Shadowxo

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Re: 3U Maths Question Thread
« Reply #2663 on: August 10, 2017, 09:18:31 pm »
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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

You should get the same answer as Opengangs :)
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Sukakadonkadonk

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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?











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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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Re: 3U Maths Question Thread
« Reply #2667 on: August 12, 2017, 12:41:35 am »
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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.

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).



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



So we rearrange this to get P:



And then just substitute what you need to find your answer ;D

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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