4U Maths Question Thread

[amandali]

for Part (iv)  how do you know p is greater than q

[jakesilove]

for Part (iv)  how do you know p is greater than q

Hey Amandali!

To be completely honest, I'm not sure! I think that probably it doesn't really matter which one is which: Once you get the answer out, just say "therefore p=.... q=....".

I could be wrong; there could be a specific reason that I'm not seeing, and if so I'd love the community to correct me!

Jake

[Spencerr]

Hey I think by drawing out a unit circle and plotting the roots of z to the power of 5 equals 1 will let you see which p or q is bigger.
Once you've plotted it out the pots you can identify the real vales of alpha, alpha squared and so on.
Comparing the sizes of the real values using the diagram will let you determine which one p or q is bigger. Sorry I'm on my phone at the moment but hopefully jake gets my drift and explains better if needed

[amandali]

the answer is -inverse cos x - root (1-x^2)  but mine is   -1/2inversecosx -1/2 root (1-x^2)

[jakesilove]

the answer is -inverse cos x - root (1-x^2)  but mine is   -1/2inversecosx -1/2 root (1-x^2)

Hey!

Have you made sure that you used the double angle rules correctly? It is likely that when you converted from cos^2(theta), you forgot to multiply everything by a half! Go through, and just make sure your substitutions are absolutely right. If you're still unsure, I can post an answer!

Jake

[jamonwindeyer]

the answer is -inverse cos x - root (1-x^2)  but mine is   -1/2inversecosx -1/2 root (1-x^2)

Have a crack again using Jake's tip, and check your working against mine if you are still struggling! 

Spoiler

That last line is from using the original substitution, and the Pythagorean Identities! Let me know if it makes sense. Note that I forgot the proper notation in most of those integrals, from the moment I substitute I am integrating with respect to Theta.

[xXCandyDXx]

Hey guys ... Having some trouble starting this question... I don't know how to start it correctly...

[Happy Physics Land]

Hey guys ... Having some trouble starting this question... I don't know how to start it correctly...

(Image removed from quote.)

Hey Candy!

This is quite a typical question that you might see in your half yearlies or topic tests, perhaps not so much in the HSC. Sorry I cant give you a working solution on paper because of some technical issues and I will just type out my working procedure here.

Step1: apply the distance formula

Let P be the point (x,y)
sqrt((x-3)^2 + y^2) = 4/5 (2-x)

Step 2: square both sides:

(x-3)^2+y^2 = 16/25 (2-x)^2

Step 3: expand and simplify

25((x-3)^2 + y^2) = 16(2-x)^2
25(x^2 - 6x + 9 + y^2) = 16(4 - 4x + x^2)
25x^2 - 150x + 225 + 25y^2 = 64 - 64x + 16x^2
9x^2 - 86x + 161 + 25y^2 = 0
9x^2 - 86x + 25y^2 = -161
(9x^2 - 86x + 1849/9) + 25y^2 = -161 + 1849/9
9(x^2 - 86x/9 + 1849/81) + 25y^2 = 400/9
9(x-43/9)^2 + 25y^2 = 400/9
[81(x-43/9)^2]/400 + 9y^2/16 = 1

Ok I got some pretty dank answer here because I might have made a mistake somewhere in my calculations due to my late night practise. But you get my point, start with distance formula and then square both sides, and you will end up getting an ellipse. Hope my answer helped!

Best Regards
Happy Physics Land

[xXCandyDXx]

Hey Happy Physics Land,
Thanks so much! Sorry, I forgot to also post the solution!!! You got it btw! 9x^2 - 86x + 25y^2 + 161 = 0
The wording of the question is confusing to me ... When I saw the fraction I thought of the definition of an ellipse so like the "PS/PN =e" thing .. I also don't really understand why you put 4/5 with (2-X) ... 

[xXCandyDXx]

Oh no wait never mind ... I get it LOL thanks again !!!

[Happy Physics Land]

Oh no wait never mind ... I get it LOL thanks again !!!

Haha no worries, thanks for asking that question! Sometimes intimidating questions can actually be solved using the most basic principles, but ofc, as extension 2 students we are often too ignorant towards those basic formulae haha

[amandali]

I dont how its progressed from last 3rd line to last 2nd line  :/

[jakesilove]

I dont how its progressed from last 3rd line to last 2nd line  :/

Hey Amandali!

So, the only change from the third last line to the second last line is, originally, we have



Now, if we take out a factor of two, we have



And so, we have 2 multiplied (n+1) times! Or, in other words,



Which is what we needed in the second last line!

Hope that this helped

Jake

[amandali]

im lost :/

[Happy Physics Land]

im lost :/

Hey Amanda:

This is quite an interesting approach adopted by whoever came up with the answer. The first four steps were quite self explanatory, and quite smooth running. Now, when we get up to step 5, there isnt a clear indication of what the persona has exactly done to figure out why the expression 2u^2/(u^2+1)^2 equals to u(derivative of 1/u^2+1). But we do understand that step, because if we do the differentiation in step 5, we can get the same expression as in the fourth step. Step 6 is even more ambiguous, they simply just skipped from the integral to another integral without even telling you what has exactly happened! So what happened between step 5 and step 6, which is the step you underlined, is that they used integration by parts. I have attached a photo below to show how the transformation of the integral has occurred. Hope it helps!



If you are still confused dont hesitate to ask!

Best Regards
Happy Physics Land