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March 28, 2024, 06:59:45 pm

Author Topic: 3U Maths Question Thread  (Read 1230130 times)  Share 

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goodluck

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Re: 3U Maths Question Thread
« Reply #3945 on: February 19, 2019, 09:44:23 pm »
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It's 135 degrees because as the cruise ship is going 15 km/h SE, the relative motion of the fishing vessel is 25km/h N. The angle between N and SE is 135 degrees.

Also, your question is kinda unclear. Bridges can't move, and what is it? (ie. it steps off the bridge). A bit confused here :) Just verify those things and I'll be glad to help you out.

Sorry the question just said it was a moveable bridge/walkway, I'm not really sure what it means by that but I just went with it. and the "it" was a typo for I (I'll fix that right now :) )

fun_jirachi

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Re: 3U Maths Question Thread
« Reply #3946 on: February 19, 2019, 10:16:20 pm »
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Basically, you would've crossed the bridge in 8 seconds as you say. Also you're right in that you've moved '40cm and 96cm' due to the speed of both movements. Because the bridge moves vertically and you move horizontally, you form a right angled triangle with sides 40 and 96. The hypotenuse is how far you've moved (calculated through Pythag.) Because it's displacement, you need an angle (or more precisely a bearing.) The bearing is equal 180-tan^-1(40/96) degrees true, or Stan^-1(40/96)W (this makes sense if you draw the triangle properly.)
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david.wang28

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Re: 3U Maths Question Thread
« Reply #3947 on: February 21, 2019, 12:58:19 pm »
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Hello,
I'm stuck on this question: integrate sec(secx+tanx). (I'm wondering if this is a typo). Can anyone please help me with this question? Thanks :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3948 on: February 21, 2019, 01:24:47 pm »
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Hello,
I'm stuck on this question: integrate sec(secx+tanx). (I'm wondering if this is a typo). Can anyone please help me with this question? Thanks :)
That certainly does not look like it can be integrated, even if we used MX2 techniques.

david.wang28

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Re: 3U Maths Question Thread
« Reply #3949 on: February 21, 2019, 01:39:50 pm »
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That certainly does not look like it can be integrated, even if we used MX2 techniques.
I think the question has a typo: it is secx(secx+tanx). I think this can be integrated, right?
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3950 on: February 21, 2019, 01:52:23 pm »
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I think the question has a typo: it is secx(secx+tanx). I think this can be integrated, right?

Yep! \(\sec^2{x}\) is integrable by itself, and looking at \(\sec{x}\tan{x}\), check out what happens if we differentiate sec:



So the \(\tan{x}\sec{x}\) in the expansion of that bracket will integrate back to \(\sec{x}\), and you have all you need ;D

david.wang28

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Re: 3U Maths Question Thread
« Reply #3951 on: February 21, 2019, 02:23:12 pm »
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Yep! \(\sec^2{x}\) is integrable by itself, and looking at \(\sec{x}\tan{x}\), check out what happens if we differentiate sec:



So the \(\tan{x}\sec{x}\) in the expansion of that bracket will integrate back to \(\sec{x}\), and you have all you need ;D
Thank you! :)
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shaynec19

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Re: 3U Maths Question Thread
« Reply #3952 on: February 27, 2019, 05:47:26 am »
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Could someone please help me integrate this without using substitution, and express the answer in terms of sin2x, not sin^2x

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Re: 3U Maths Question Thread
« Reply #3953 on: February 27, 2019, 06:34:04 am »
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Not sure what you mean by express the answer in terms of sin2x.

The other result involves sin2x as you correctly say, but still not sure what you're asking.
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3954 on: February 27, 2019, 12:05:22 pm »
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Not sure what you mean by express the answer in terms of sin2x.

The other result involves sin2x as you correctly say, but still not sure what you're asking.

You could take this and use \(\sin^2{2x}+\cos^2{2x}=1\) to express in terms of \(\sin{2x}\) ;D

shaynec19

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Re: 3U Maths Question Thread
« Reply #3955 on: February 27, 2019, 01:01:41 pm »
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Thankyou!

shaynec19

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Re: 3U Maths Question Thread
« Reply #3956 on: February 27, 2019, 01:08:00 pm »
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Could someone please help me with the following two;


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Re: 3U Maths Question Thread
« Reply #3957 on: February 27, 2019, 01:21:17 pm »
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Could someone please help me with the following two;


Is there a typo in the first one? There's nothing fancy about it. It's just \( \int 2\sin^2 x\,dx\) which becomes \( \int (1-\cos 2x)\,dx \) following the usual double-angles approach.

The second integral is not a part of the MX1 course without any substitution being provided.
« Last Edit: February 27, 2019, 01:24:15 pm by RuiAce »

shaynec19

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Re: 3U Maths Question Thread
« Reply #3958 on: February 27, 2019, 01:33:54 pm »
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Haha sorry, terrible today.
Yes there was a typo,

RuiAce

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Re: 3U Maths Question Thread
« Reply #3959 on: February 27, 2019, 01:42:18 pm »
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That one would be unlikely to appear in a 3U exam (more likely in a 4U exam) but 3U techniques are certainly sufficient for it.
\begin{align*} \int \sin^2x\cos^2x\,dx &= \int \frac14 \sin^2 2x\,dx\\ &= \int \frac18 (1-\cos 4x)\,dx \end{align*}
Basically use both the sine and the cosine double angle formula.