VCE Methods Question Thread!

[zhen]

Hi!
Does anyone know of any good methods tutors that give really good notes + guidance + practice sheets? Im in dire need of a tutor so if please me some!
Thanks!

You can check out the tutors advertising here. https://atarnotes.com/forum/index.php?board=310.0

[ImmaculateJeff]

You can check out the tutors advertising here. https://atarnotes.com/forum/index.php?board=310.0

I did haha, most are either full or don't offer methods

[wyzard]

Hi!
Does anyone know of any good methods tutors that give really good notes + guidance + practice sheets? Im in dire need of a tutor so if please me some!
Thanks!

Hey Jeff! I have availability for methods tutoring if you're up for it! PM me for details

[captkirk]

Need help with this question. A line perpendicular to the graph of y=-2x^2 has the equation y=2x+c, where c is a real constant. Determine the value of c.

[Syndicate]

Need help with this question. A line perpendicular to the graph of y=-2x^2 has the equation y=2x+c, where c is a real constant. Determine the value of c.

perpendicular gradient = -1/2 at the graph of -2x^2. Note, this is not a straight line, the gradient is not constant. So you'll need to calculate the instantaneous gradient.

dy/dx = -4x = -1/2
x = 1/8 (substitute this in 2x^2)
y = -1/32

Now you have coordinates of where the line intersects with the graph, so substitute them to calculate c.
-1/32 = 1/4 + c
c = -9/32

[clarke54321]

Hi everyone,

Could I please have some help with part C. The solutions say to use symmetry, but I don't understand how this works.

Thanks!

[Syndicate]

Hi everyone,

Could I please have some help with part C. The solutions say to use symmetry, but I don't understand how this works.

Thanks!

Since methods doesn't show us how to integrate natural logs, we have to integrate the log in terms of it's inverse. By symmetry, the area bounded between an inverse and the non-inverse is same respective to their axis.

So, ln(2x) = \(\frac{e^y}{2} \).
We now need to integrate this, but wait... this time we are integrating in respect to the y-axis. So first, well need to calculate the y-coordinates.

at x = 2, y = In(4)
at x = 0, y = In(1) = 0

Now integrate.


Looks familiar to the previous question?
= 3/2 units^2

However, we aren't done yet. We have found the area between the y-axis and \( \frac{e^y}{2} \) (or in other words, between the y-axis and ln(2x)).

So to calculate the area between the x-axis and In(2x), we'll need to calculate the area bounded between (0,0) and (2, In(4)) (which is just a square).

Area = 2 x In(4)  = 2 ln(2^2) = 4ln(2)

So now we subtract the area of the area with the area calculated between the y-axis and the equation to get the area between the x-axis and the equation.

[undefined]

hi,

can someone please work through solutions for these 2 questions for me please?

thanks.

btw, the family they are referring to is ax^2+bx (c=0)

[zhen]

hi,

can someone please work through solutions for these 2 questions for me please?

thanks.

btw, the family they are referring to is ax^2+bx (c=0)

So the equation is
This is my answer. It might be wrong, so probably check with your teacher.

For the second part, I would say something like a tangent to the curve only meets at one point, since there is only one turning point and no asymptote and it isn't a hybrid function. Since cubics and things with more than one turning point can have tangents that meet the curve at multiple points and so can hybrid functions. But I'm not too sure about this so someone else should answer it properly.

[geminii]

Ahhhh I'm so confused

If y = x^2, then is dilating y by a factor of 2 from the y axis (x/2)^2 and dilating y by a factor  of 2 from the x axis 2x^2??

[Sine]

Ahhhh I'm so confused

If y = x^2, then is dilating y by a factor of 2 from the y axis (x/2)^2 and dilating y by a factor  of 2 from the x axis 2x^2??

Yes! confusing I know.

So when you are dilating by factor 2 from the y-axis, what are you doing? You are replacing each x with x/2.

When you are dilating by factor 2 from the x-axis, you may say you are sticking a 2 in front of the equation but that is wrong. You are actually replacing y with y/2 which yields y/2 = x^2 this gets you to y = 2x^2

Hope this helps

[captkirk]

Ok.. it's not allowing me to attach picture so I need to write out the question .-.

Anyways this is what the question asks

After the rebellion was successful in overthrowing the  empire, the number of Jedi  that were discovered  and trained increased according to the rule  J=alog2(t)+b, where a and b are constants and t is the time in years after he battle of endor.
Two years after the battle of Endor,  The new Republic was formed, and Luke and leila were the only two known jedi.
Thirty years later, after Luke established a jedi school on yavin four, there were 82 known jedi in existence

A) determine the values of a and b
B) sketch the graph against J against t for the first 50 years of the new republics existance.

Ok so I need help with A)
So I'm confused. This is what I did. Made simultaneous equations of
2=alog2(2)+b
And
82=alog2(32)+b

Then I got a=20 and b =-18 then subbed in actual rule. But when I sketched this it was giving me wrong numbers when x=2.11 it is giving y=3.55. (That is wrong) y is supposed to be equal to 2. What am I doing wrong. Thanks

[humblepie]

Hi! Could someone please help me with the attached question? How would you find the asymptote for the graph? - I've tried making t approach infinity but it doesn't seem to work (CAS is allowed btw)

Thanks!

[zhen]

Hi! Could someone please help me with the attached question? How would you find the asymptote for the graph? - I've tried making t approach infinity but it doesn't seem to work (CAS is allowed btw)

Thanks!

You've got what I think the asymptote is already in your graph. If you let t approach infinity, e^-bt becomes a really small number since you get 1/(a really big number), so as t approaches infinity you get w/b+a infinitely small number. So, the asymptote is w/b. Hope that helps. 

[humblepie]

You've got what I think the asymptote is already in your graph. If you let t approach infinity, e^-bt becomes a really small number since you get 1/(a really big number), so as t approaches infinity you get w/b+a infinitely small number. So, the asymptote is w/b. Hope that helps. 

I copied the graph from the answers haha! Thanks zhen!