VCE Methods Question Thread!

[lzxnl]

The equation you gave me looks more like the solution to a cubic of the form x^3 = ax + b

sin(pi/8) using the half angle formula? Yeah, not the greatest fun.
sin(pi/8) = sqrt((1-cos(pi/4))/2) = sqrt((1-sqrt2/2)/2) = sqrt(2-sqrt2)/2
That's not quite as bad

[Orb]

1. Although we solved for different things, we should still end up with the same answer (or equation for the line really) so I don't think it's that.

2. I used the rule which is finding an equation of a tangent. I subbed in the point (1,-2) as it is known that it lies on the tangent and I'm just solving for the unknown gradient of the line which I can then put back and work out the equation of the tangent since the rule I originally used is finding the equation of the tangent.
It wouldn't really work out if I subbed (1,-2) into y=mx+c but yeah that's just how I thought it through :S

(Dunno if any of that made sense LOL but hopefully it did)

I don't see anything wrong with your working out, but when I put my equations into the CAS it does get me a tangent line and the line does pass through (1,-2)...

[nhmn0301]

It was from an ancient (and unrelated) math paper. Mind you, the only reason I posted it is because I've come across similar instances in spesh that I didn't know how to deal with. For example, if we were asked to find sin(pi/8) using half angle formula while keeping our answer in the simplest form. However, apparently VCAA lets you use the CAS for these questions.

Just a bit add up to lzxnl's post, you can also find sin (pi/8) by using cos (pi/4) = 1 - 2sin^2 (pi/8), problem solved from there. At the end, only choose the positive value since the angle is in quadrant 1.
Hope this helps

[lzxnl]

Just a bit add up to lzxnl's post, you can also find sin (pi/8) by using cos (pi/4) = 1 - 2sin^2 (pi/8), problem solved from there. At the end, only choose the positive value since the angle is in quadrant 1.
Hope this helps

You'll find that my formula is just the rearranged form of your formula already taking the positive root xP

[alchemy]

I don't see anything wrong with your working out, but when I put my equations into the CAS it does get me a tangent line and the line does pass through (1,-2)...

I think IndefatigableLover is correct. What is your complete equation btw hamo94?

[Cort]

Just for some clarity- unique solutions only exist if it does not equal zero right? That's the idea I'm getting from doing these exams. Beyond the fact that unique solutions m1=/=m2, is there any other reason why it cannot be =0 and will normally be ignored (eg: R\{-2,2}) ?

[IndefatigableLover]

I don't see anything wrong with your working out, but when I put my equations into the CAS it does get me a tangent line and the line does pass through (1,-2)...

Same here... I sub 'm' back into the equation and I get an answer that is a tangent as well as it passes through (1,-2)...
https://www.desmos.com/calculator/qvjclkggsl

[Zealous]

Hamo's working is correct or rather you are both correct.

Continuing on from Hamo's working out:



Sub that into (let's just do the positive solution)



Now we have two points:
and

Using those two points to form a linear equation:


The same line that IndefatigableLover got. (There's also another line if you consider the negative root 3)

[Orb]

Hamo's working is correct or rather you are both correct.

Continuing on from Hamo's working out:



Sub that into (let's just do the positive solution)



Now we have two points:
and

Using those two points to form a linear equation:


The same line that IndefatigableLover got. (There's also another line if you consider the negative root 3)

AHH cheers for the clarification!

I was so puzzled there for a second.

[IndefatigableLover]

Hamo's working is correct or rather you are both correct.

Continuing on from Hamo's working out:



Sub that into (let's just do the positive solution)



Now we have two points:
and

Using those two points to form a linear equation:


The same line that IndefatigableLover got. (There's also another line if you consider the negative root 3)

Oh nice! Looks like I got some skills that I need to run-over during the holidays again but thanks for that Zealous and you too hamo94 for your alternative solution

[Yacoubb]

Do we need to know about linear approximation in calculus?

[IndefatigableLover]

Do we need to know about linear approximation in calculus?

Had a check of the Methods Study Design and on Page 141 (if you just use 'Ctrl + F' and type in 'Linear Approximation' it'll say):

Polynomials and power functions including differential calculus

Polynomial functions and their graphs, average and instantaneous rates of change, use of ƒ´ to
define increasing and decreasing, use of linear approximation relationship.

I think it's a yes going by what they've said here but maybe others who have done previous Methods exams can give better insight (I think there was a question in a VCAA exam about Linear Approximation from memory..)

EDIT: VCAA 2009 Exam 1 (Last Question) uses Calculus with Linear Approximation
Exam
Assessor's Report

[Rishi97]

Is it possible to do addition of ordinates on CAS?

[lzxnl]

Is it possible to do addition of ordinates on CAS?

Put in f(x), put in g(x) and then put in f+g (x). Graph all three

[lzxnl]

Perhaps wrong thread but can you show me now you implicitly differentiated that? I forgot how to do it

So... r^2 = A/pi
Differentiating the right hand side with respect to A gives 1/pi
Differentiating the left hand side with respect to A? d (r^2)/dA = d (r^2)/dr × dr/dA by chain rule. The rest follows