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March 28, 2024, 08:37:10 pm

Author Topic: 3U Maths Question Thread  (Read 1230185 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #60 on: March 20, 2016, 07:42:55 am »
+1
Hi, just wondering,
we have only just done inverse trigonometry but there was a question in our test that we had not learnt how to attempt. We had to find the integral of the inverse of sin. How would you attempt this question?

As this is a 3U forum, if you were asked such a question in the 3U exam there is a probability of 100% that the integral came with BOUNDARIES and was a DEFINITE integral

You need to draw out the graph of the inverse sine function, and then find
A = area of rectangle - integral of sin(y)dy for certain boundaries.

Integration by parts is NOT in the 3U course, as already mentioned.

Happy Physics Land

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Re: 3U Maths Question Thread
« Reply #61 on: March 20, 2016, 11:30:45 am »
+1
As this is a 3U forum, if you were asked such a question in the 3U exam there is a probability of 100% that the integral came with BOUNDARIES and was a DEFINITE integral

You need to draw out the graph of the inverse sine function, and then find
A = area of rectangle - integral of sin(y)dy for certain boundaries.

Integration by parts is NOT in the 3U course, as already mentioned.

OMG yes.... after all inverse sine x comes from the original curve sine y..... of course... this way the question is actually a lot easier. We are essentially just finding the area of sin y within the domain 0 <= x <= pi/2 and then using a rectangle to minus that area. And then because arcsin x is defined from -pi/2 to pi/2, just times the resultant area by 2. Am I right in saying this rui?
« Last Edit: March 20, 2016, 11:32:39 am by Happy Physics Land »
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starkidshani

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Re: 3U Maths Question Thread
« Reply #62 on: March 20, 2016, 12:13:47 pm »
0
Hi,
I'm having trouble with an Integration By Substitution question:

∫(x-2)/(√x+2) dx 
using u^2 =x+2

Can you help me out? Thanks very much :)

 

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Re: 3U Maths Question Thread
« Reply #63 on: March 20, 2016, 01:40:06 pm »
+1
Hi,
I'm having trouble with an Integration By Substitution question:

∫(x-2)/(√x+2) dx 
using u^2 =x+2

Can you help me out? Thanks very much :)

Hey shani!

I have posted the solution below, l guess the important thing in this question is to realise that x-2 = u-4 which is kind of tricky to realise I guess (Btw Im just assuming here that you meant sqrt(x+2) not sqrt(x) + 2. If the denominator is sqrt(x) + 2 then the outcome would be different). If you have any further questions please don't hesitate to ask! :)



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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #64 on: March 20, 2016, 01:53:24 pm »
+1
OMG yes.... after all inverse sine x comes from the original curve sine y..... of course... this way the question is actually a lot easier. We are essentially just finding the area of sin y within the domain 0 <= x <= pi/2 and then using a rectangle to minus that area. And then because arcsin x is defined from -pi/2 to pi/2, just times the resultant area by 2. Am I right in saying this rui?

It would depend on the exact boundaries! The boundaries would determine how much area of the siny curve you need, and then if we have boundaries either side of x=0, then yep we can multiply by two to take advantage of symmetry!  For example, if the boundaries are from x=1 to -1, you would find the area under the sine curve from y=0 to pi/2, subtract from the rectangle, and then multiply by two, if that makes sense :)

RuiAce

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Re: 3U Maths Question Thread
« Reply #65 on: March 20, 2016, 06:47:45 pm »
+1
Hi,
I'm having trouble with an Integration By Substitution question:

∫(x-2)/(√x+2) dx 
using u^2 =x+2

Can you help me out? Thanks very much :)

Not too sure why HPL used that substitution when you were given that specific one.
« Last Edit: March 20, 2016, 08:41:38 pm by jamonwindeyer »

Happy Physics Land

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Re: 3U Maths Question Thread
« Reply #66 on: March 20, 2016, 07:43:11 pm »
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Not too sure why HPL used that substitution when you were given that specific one.


Wait what I thought I did use that substitution? Its just with the numerator I subtracted 4 from u to make it the same as the numerator.
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #67 on: March 20, 2016, 08:39:28 pm »
+1
Wait what I thought I did use that substitution? Its just with the numerator I subtracted 4 from u to make it the same as the numerator.

You used:



RuiAce used:



Both work quite well actually!

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Re: 3U Maths Question Thread
« Reply #68 on: March 20, 2016, 09:25:42 pm »
+1
You used:



RuiAce used:



Both work quite well actually!

Yeah I agree Jamon, I realised that only shortly after I posted my comment. I guess at the end of the day it doesnt matter what you substitute it with. After all, substitution is only a technique, and if that technique helps you to solve the question then I guess regardless which mechanism you adopt you would get the same outcome.
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RuiAce

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Re: 3U Maths Question Thread
« Reply #69 on: March 20, 2016, 09:31:59 pm »
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Yeah I agree Jamon, I realised that only shortly after I posted my comment. I guess at the end of the day it doesnt matter what you substitute it with. After all, substitution is only a technique, and if that technique helps you to solve the question then I guess regardless which mechanism you adopt you would get the same outcome.

You mean in 4U and beyond it doesn't matter.

You should, of course, use the substitution given to you in 3U.

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Re: 3U Maths Question Thread
« Reply #70 on: March 24, 2016, 10:12:49 pm »
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Hey guys!

I was wondering whether you could show me how to sketch the function y=tan-1(tanx) over the domain -π≤x≤π <3 thank youu

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Re: 3U Maths Question Thread
« Reply #71 on: March 24, 2016, 10:27:48 pm »
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Hey guys!

I was wondering whether you could show me how to sketch the function y=tan-1(tanx) over the domain -π≤x≤π <3 thank youu

Neutron

Hey Neutron!

Tan-1(tanx) will cancel each other out, leaving us simply with y = x. So all you need to do is plot the points (pi, pi) and
(-pi, -pi), and then connect these two points (essentially just the linear graph y=x).

If you really need me to show you the sketch just tell me, but Im sure you are capable of doing it! :)

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« Last Edit: March 24, 2016, 10:29:31 pm by Happy Physics Land »
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Re: 3U Maths Question Thread
« Reply #72 on: March 25, 2016, 11:20:41 am »
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hey, i've been stuck on this question. i keep getting the answer 30.

In a tennis club, there are five married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In haw many ways can a group of four persons be chosen for this match if:

(b) a man and his wife may not play in the match either as partners or as opponents.

Answer is 60


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Re: 3U Maths Question Thread
« Reply #73 on: March 25, 2016, 11:59:49 am »
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Thanks Happy Physics land, I think the markers wanted us to work it out by finding the area under the graph in relation to the y-axis. Thank you for the integration by parts though, it was very helpful.  :)

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Re: 3U Maths Question Thread
« Reply #74 on: March 25, 2016, 12:24:57 pm »
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hey, i've been stuck on this question. i keep getting the answer 30.

In a tennis club, there are five married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In haw many ways can a group of four persons be chosen for this match if:

(b) a man and his wife may not play in the match either as partners or as opponents.

Answer is 60

Hey Chloe!

Okay, so we have here a combinations question! I find the best way to do these is to separate into any separate categories, in this case, men and women. Doing that, let's proceed!

So let's say we choose the two males first. This would be a combination of 2 males from a possible 5, and so:



Now, we can pick their partners. Two of the women are not allowed to be chosen, because their husbands are in the game already. Therefore, we are selecting 2 women from 3 possible choices:



Now at this point, the answer of 30 becomes apparent:



Now, I was a little confused, and I think the wording of the question is poor here. We haven't considered which team the women choose to join in the second step. That is, when we choose the women, they can join the teams in one way, or they can swap sides and join in another way. So, we multiply our final answer by 2:



Now, the question only said how many ways the groups can be chosen, which I suppose implicitly suggests that we consider different teams as well, but I'd totally forgive someone for not doing it, because technically, changing the teams does not change the group.

But yep, that's where the 60 comes from. Hope this helps Chloe!  ;D