3U Maths Question Thread

[RuiAce]

Totally agree: I can prove that it equals e, but only using university level maths. I'm sure you can prove the relationship somehow using iii) but honestly it doesn't fall out as easily as I would have expected.

The proof I got given demonstrated how by the monotone sequence theorem the binomial expansion as a sum is greater than 2 upon inspection. It also assumes n!/(n-k)! < n^k.to prove the < 3 component.

[jakesilove]

The proof I got given demonstrated how by the monotone sequence theorem the binomial expansion as a sum is greater than 2 upon inspection. It also assumes n!/(n-k)! < n^k.to prove the < 3 component.

Actually, I think that that's fine. Once you get to the Taylor serious of e, you can easily calculate that the number will be between 2 and 3 by summing up some terms. Maybe that's the best way?

[RuiAce]

Actually, I think that that's fine. Once you get to the Taylor serious of e, you can easily calculate that the number will be between 2 and 3 by summing up some terms. Maybe that's the best way?

Well I mean, in 3U we can't assume the Taylor series either. But I mean, we can't assume that the sequence will be monotone without proof right?

Whilst we're here I want to learn them already. Slight motivation to want to go back to uni to learn them next sem.

[zoe_rammie]

Hey Zoe,

First of all, your description of a potential method of answering this question is absolutely gold. Keep at it, love your work.

Your method, and the method in the answers, is actually the same. If you do what you were planning on doing, that's great, but you would have to equate coefficients as there isn't another way to compare the RHS to the LHS and prove that they are equal. So, you would equate coefficients, and get exactly the same answer as they did. They're answer probably isn't comprehensive enough anyway; if your method gets you the right relationship, you're absolutely doing the right thing.

If you wanted a more comprehensive proof, though, let us know, however as far as I can tell your "do shit get marks" methodology is spot on.

Jake

So would this cut it?

[jakesilove]

Well I mean, in 3U we can't assume the Taylor series either. But I mean, we can't assume that the sequence will be monotone without proof right?

Whilst we're here I want to learn them already. Slight motivation to want to go back to uni to learn them next sem.

Using the proof below, and letting x equal one, we can actually directly find the Taylor series (although obviously a 3U student won't know that; they'll just do the maths). Then, instead of recognising the final step, by typing in the first few terms into a calculator, a student could find that the limit is between 2 and 3!

[jakesilove]

So would this cut it?

I reckon so! I don't remember if the question asked you to "use" a specific part of the question prior to the proof, and if it did you may not get all the marks using your method. However, I think that your method is the quickest way to get to an answer.

[RuiAce]

So would this cut it?

I'm with Jake; it's fine.

Nice Pokemon Go reference

[conic curve]

Can someone here please help me with this

cos^-1[cos1050].

This is what I did:

cos1050=cos330
cos^-1[cos-30] (answer can't be -30 because it is not in the restriction)

Thanks

[jakesilove]

Can someone here please help me with this

cos^-1[cos1050].

This is what I did:

cos1050=cos330
cos^-1[cos-30] (answer can't be -30 because it is not in the restriction)

Thanks

You were pretty much there! Type



into your calculator, and you'll see the answer is 30. This comes about due to the even/odd nature of the graph etc. but really, if you're ever struggling, see if your calculator can give you the answer!

[jakesilove]

Can someone here please help me with this

cos^-1[cos1050].

This is what I did:

cos1050=cos330
cos^-1[cos-30] (answer can't be -30 because it is not in the restriction)

Thanks

In fact, you could have done this right from the start. To do the maths properly, you needed to recognise that

[RuiAce]

[conic curve]

Thanks Jake and Rui

How do I do this: if a=tan^-1 (-1/3) find cosa and sina

The reason why I am confused is because you can't have negative dimensions

Does this mean I'll need o draw it on a coordinate geometry graph?

[RuiAce]

Thanks Jake and Rui

How do I do this: if a=tan^-1 (-1/3) find cosa and sina

The reason why I am confused is because you can't have negative dimensions

Does this mean I'll need o draw it on a coordinate geometry graph?

[vamshimadas]

Quick Question on Parametrics:
Find the equation of the tangent to the parabola x=4t, y=2t^2 at the point where t=3.

Thank you

[jakesilove]

Quick Question on Parametrics:
Find the equation of the tangent to the parabola x=4t, y=2t^2 at the point where t=3.

Thank you

You can do this a whole bunch of ways. I think it's easiest to find the parabola first, differentiate, and find the tangent the usual way etc.





Then, we can find the tangent the normal way! First, you will need to x and y coordinates at t=3. We can sub this t value straight into the equations we are given to find them.




I'll leave you to find the tangent from there: just use normal calculus!

You can also use the chain rule method (ie find the first derivative of x with respect to t, y with respect to t, multiply them etc.). That may or may not be quicker, just depends on how you visualise the maths.

Jake