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March 29, 2024, 04:52:58 am

Author Topic: 3U Maths Question Thread  (Read 1230339 times)  Share 

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jamonwindeyer

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3U Maths Question Thread
« Reply #2055 on: May 23, 2017, 08:02:11 am »
+1
Hi I was wondering if someone would be able to help me with 20a? This is my solution but the answer is -1cm/s?
(Image removed from quote.)

(Image removed from quote.)

Hey! Try again, but use the double angle formula:



I reckon you'll get it this way! I think there was a sign error somewhere in that integral is all, but converting the expression with the double angle formula is way easier

Edit, actually no I think that gives the same answer, and your integral looks fine - I reckon the answer is wrong!
« Last Edit: May 23, 2017, 02:22:21 pm by jamonwindeyer »

RuiAce

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Re: 3U Maths Question Thread
« Reply #2056 on: May 23, 2017, 08:11:11 am »
+1



I have a feeling that the textbook confused acceleration/velocity and velocity/DISTANCE.
« Last Edit: May 23, 2017, 08:15:51 am by RuiAce »

J.B

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Re: 3U Maths Question Thread
« Reply #2057 on: May 23, 2017, 02:12:45 pm »
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Thank you,
I ended up getting the -1cm/s using the double angle method.
:)

K9810

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Re: 3U Maths Question Thread
« Reply #2058 on: May 23, 2017, 08:30:25 pm »
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Hey!

Can you please help me with this question?
A particle moves according to x = t2 −8t+7, in units of metres and seconds.  What is the maximum distance from the origin, and when does it occur: (i) during the first 2 seconds, (ii) during the first 6 seconds, (iii) during the first 10 seconds?


jakesilove

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Re: 3U Maths Question Thread
« Reply #2059 on: May 24, 2017, 09:58:14 am »
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Hey!

Can you please help me with this question?
A particle moves according to x = t2 −8t+7, in units of metres and seconds.  What is the maximum distance from the origin, and when does it occur: (i) during the first 2 seconds, (ii) during the first 6 seconds, (iii) during the first 10 seconds?



Hey!

So, we have

and we want to find maximum distance (ie. x). It's helpful to have a sketch in mind before attempting questions like this. Clearly it's a positive parabola. So, it doesn't actually have a 'maximum' turning point; rather it has a minimum, which we're not interested in. In fact, if you sketch the graph itself, you can get the answer seriously quickly.


Based on the above (or just intuition), it's clear that the maximum displacement will either occur at the beginning (ie. t=0) or end (ie t=2). We can just sub our values in to see which is correct!


It doesn't matter if the value is positive or negative, as long as it has the greatest magnitude. However, clearly the greatest displacement in the first section (first two seconds) occurs at t=0, x=7m.

Now, onto the first six seconds. This contains the global minimum, which may have a greater magnitude than x=7. So, we need to find it! We can use calculus for this part.


Subbing this value into our formula

So, clearly the point of greatest displacement is at t=4, x=-9.

Finally, we look to the first 10 seconds. The only way this any value could be greater than the global minimum is if it is far enough to the right that it keeps rising, and rising, and rising. So, we only need to check t=10.


Which is way bigger than -9m. So, the greatest displacement occurs at t=10, x=27m.









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legorgo18

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Re: 3U Maths Question Thread
« Reply #2060 on: May 24, 2017, 04:05:27 pm »
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Hi, whats happening in this question:

For all real values of x for which cot(1/4)x- cotx = sinkx/[(sin1/4(x))(sinx)] is defined, find the value(s) of k.
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scienceislife

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Re: 3U Maths Question Thread
« Reply #2061 on: May 24, 2017, 06:19:59 pm »
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hey just wondering if I could please have some help with this questions; I've changed it into different letters by using PT^2 = TB x TA but cannot seem to get much further than that. Thank you!

scienceislife

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Re: 3U Maths Question Thread
« Reply #2062 on: May 24, 2017, 06:23:43 pm »
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not sure how to do this with the information given?

RuiAce

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Re: 3U Maths Question Thread
« Reply #2063 on: May 24, 2017, 07:04:17 pm »
+1

hey just wondering if I could please have some help with this questions; I've changed it into different letters by using PT^2 = TB x TA but cannot seem to get much further than that. Thank you!

RuiAce

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Re: 3U Maths Question Thread
« Reply #2064 on: May 24, 2017, 07:08:34 pm »
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Hi, whats happening in this question:

For all real values of x for which cot(1/4)x- cotx = sinkx/[(sin1/4(x))(sinx)] is defined, find the value(s) of k.
Since the expression is true for ALL relevant values of x, we may equate coefficients as appropriate. This is the bonus of having an identity instead of just some equation.

Fix up the LHS and things come out nicely.

Solution

RuiAce

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Re: 3U Maths Question Thread
« Reply #2066 on: May 24, 2017, 07:34:21 pm »
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help please i still cant see what's actually in the topic test book for ext 1 can someone please tell me i kinda wanna know what im buying?

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #2067 on: May 24, 2017, 09:34:35 pm »
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help please i still cant see what's actually in the topic test book for ext 1 can someone please tell me i kinda wanna know what im buying?


Hey! I just replied to your PM, but I'll pop it here too in case anyone else is curious:

I've not seen the Extension 1 Topic Tests, didn't have anything to do with this particular set actually, so can't comment on their style or content. I know the developer is working on fixing up the images on the site - But as it is a set of HSC Topic Tests, it will have (in some arrangement) topic tests covering all HSC topics. If your test is on HSC content, the topic tests will cover it in some fashion :)

Sorry for the issues with the images - It's a server issue that our developer's been working on fixing for a while!

legorgo18

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Re: 3U Maths Question Thread
« Reply #2068 on: May 24, 2017, 10:46:19 pm »
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Hi um stuck on integrating 2^lnx pls help :)
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Re: 3U Maths Question Thread
« Reply #2069 on: May 25, 2017, 12:03:04 am »
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Hi um stuck on integrating 2^lnx pls help :)

Hey! Tricky one this, we have to take an exponential/logarithm at the same time and then use the Log Laws to manipulate. See if this makes sense:



And that is just a power of \(x\) - Integrate that as normal! The power of \(x\) will be \(1+\ln{2}\) ;D