HSC 2013 Maths Marathon (archive) (5 Viewers)

[mahmoudali]

Re: HSC 2013 2U Marathon

First 2U Q I made.

EDIT: The Q means Tangent at that x-value for any clarification issues.

realise put the solution up for this, i cant take it anymore =(

 

[SpiralFlex]

Re: HSC 2013 2U Marathon

I like this question. Seems like the kind of thing I would write.

Are you coldlipstick?

 

[SpiralFlex]

Re: HSC 2013 2U Marathon

realise put the solution up for this, i cant take it anymore =(

Clue: "The tangent of the parabola at this value passes through the origin".

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

realise put the solution up for this, i cant take it anymore =(

Idk if you do MX1, but it's very similar to the proof for Newton's Method.

 

[mahmoudali]

Re: HSC 2013 2U Marathon

Idk if you do MX1, but it's very similar to the proof for Newton's Method.

yeah FrostKruncher explained it to me and i forgot how to prove newtons method

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

 

[anthony_wheels]

Re: HSC 2013 2U Marathon

What do you want us to do here? Solve for theta?

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

What do you want us to do here? Solve for theta?

Yep.

 

[anthony_wheels]

Re: HSC 2013 2U Marathon

Hmmm, looks easier than it actually is (in my opinion anyway - most of you probably find it easy ahah)... The answers I originally got were not possible... Any hints?

Thanks

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

Hmmm, looks easier than it actually is (in my opinion anyway - most of your probably find it easy ahah)... The answers I originally got were not possible... Any hints?

Thanks

Nope I think you are right. You are meant to get answers, but when you test them they become undefined, hence no solutions.

 

[anthony_wheels]

Re: HSC 2013 2U Marathon

Yeah that's strange ey... I graphed y=tanxcosxsinx and it still had times where y=1. But they were at my original solutions (if that makes sense) but as we know they're not possible... wtf? haha

 

[mahmoudali]

Re: HSC 2013 2U Marathon

Nope I think you are right. You are meant to get answers, but when you test them they become undefined, hence no solutions.

if a question like this came in HSC i'd expect more than half the state not to test

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

if a question like this came in HSC i'd expect more than half the state not to test

That's why it's good, you should always test solutions if you have time. Though no solutions will come up only when you are dividing by something usually.

 

[mahmoudali]

Re: HSC 2013 2U Marathon

That's why it's good, you should always test solutions if you have time. Though no solutions will come up only when you are dividing by something usually.

and if you dont have time leave it?

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

and if you dont have time leave it?

Take the risk.

 

[girlworld_club]

Re: HSC 2013 2U Marathon

how do you find possible inflection points:

y= 3cos(2x + pi/4)

 

[HeroicPandas]

Re: HSC 2013 2U Marathon

how do you find possible inflection points:

y= 3cos(2x + pi/4)

Differentiate twice, equate to 0

 

[anthony_wheels]

Re: HSC 2013 2U Marathon

Find the second derivative and then let that = 0.

 

[girlworld_club]

Re: HSC 2013 2U Marathon

yes i did that i get to

-12 sin (2x + pi/4) = 0

And a take the basic angle as 0 degrees? So i end up with 0, 180, 360, etc then minus pi/4 and divide by 2 for each.

feel like i'm doing something wrong can't seem to get answer.

 

[Menomaths]

Re: HSC 2013 2U Marathon

yes i did that i get to

-12 sin (2x + pi/4) = 0

And a take the basic angle as 0 degrees? So i end up with 0, 180, 360, etc then minus pi/4 and divide by 2 for each.

feel like i'm doing something wrong can't seem to get answer.

Should be cos?