VCE Methods Question Thread!

[keltingmeith]

I feel really bad about this, e^1, sorry about all this faulting I'm going to throw at you... >.<;

The question's worth 4 marks - from that, I can instantly tell you that this isn't enough. In fact, to answer it I'm going to use knowledge of physics that is NOT required for methods (if you're also doing specialist, you should know this, though):

Velocity is a vector quantity, and so is acceleration. If you want to play with them, you have to take into account their directions. The methods study design only considers kinematics in straight-line motion, so you don't need to worry about the direction of motion beyond if it should be negative or positive. In this case, it's not straight-line motion - and so, you have to break it up into rectangular components. I've attached a picture of my working out of it, tell me if any part confuses you and I'll explain it. (the red numbers represent the order of my steps)

[Note the triangle symbol (uppercase delta) is used instead of the 'd' (lowercase delta). This is because we are not dealing with infinitely small quantities. Also, acceleration is the change in velocity with respect to time.]

Let's play "where calculus symbols come from"! Hahah, lowercase delta actually looks like this: . See, when we change from bigger numbers to infinitesmalls, someone decided we're going to change from greek letters to arabic (who knows why?). So, the "d" in dy/dx is, in fact, just a normal d that we use in everyday life (including the word everyday ).

For more fun, you may now that becomes when we make it infinitely small. The integral sign is actually an elongated S, which stands for "sum" (as the first statement is read "the sum from ecks equal ay to be...")

Just to add a tiny bit, when I did methods I wrote it like this, being aware (as Euler had said) that infinity is a concept. This however is not necessary.

I'm pretty that's what was trying to be avoided, but that way of writing is certainly more correct than what I did.

EDIT: Fixed image, with a weird white box to reduce the size.

[e^1]

Hm, I see. Thanks for the correction and sorry for the error.

(edit: oh just realise I read question wrong. Shouldn't do math with only few hours sleep and coffee :/ )

[Valyria]

Hey AN,

Is the minimum value of f(x) = k^2(x + k^2)(x + k^2 -6), where k is not equal to 0, a positive constant or negative constant? Surely there's a more efficient approach than finding the turning point.

[lzxnl]

Let's complete the square on this.



As you can see, there is a turning point when x = 3-k^2

Need more info to determine if the turning point has a positive or negative y value

Alternatively, use the symmetry of the parabola to say that the x value of the turning point is midway between the two x intercepts

[keltingmeith]

Let's complete the square on this.



As you can see, there is a turning point when x = 3-k^2

Need more info to determine if the turning point has a positive or negative y value

Alternatively, use the symmetry of the parabola to say that the x value of the turning point is midway between the two x intercepts

Actually, using your working we know that the turning point occurs at (3-k^2, 9k^2). This means that the turning point must have a positive y-value or lie on the x-axis, as k^2>=0. But, since k=/=0, it must be a positive constant. (in fact, k >0 for the turning point to be a minimum, but that's a different story)

[Rishi97]

How do I differentiate the following?

Thanks

[Valyria]

Let's complete the square on this.



As you can see, there is a turning point when x = 3-k^2

Need more info to determine if the turning point has a positive or negative y value

Alternatively, use the symmetry of the parabola to say that the x value of the turning point is midway between the two x intercepts

I initially went through this process but the suggested solution took an alternative path and reached a conclusive statement:

f(1-k^2) = k^2 x 1 x (-5)
              = -5k^2

Therefore, the Turning Point of the parabola is below the x-axis since k^2>0, so the minimum value of f(x) is a negative constant.
Could someone please explain this deduction?

[keltingmeith]

I initially went through this process but the suggested solution took an alternative path and reached a conclusive statement:

f(1-k^2) = k^2 x 1 x (-5)
              = -5k^2

Therefore, the Turning Point of the parabola is below the x-axis since k^2>0, so the minimum value of f(x) is a negative constant.
Could someone please explain this deduction?

Refer to my post, but notice I did make a mistake. The turning point isn't (3-k^2,9k^2) like I said it was, it's actually (k^2-3,-9k^2). This means that the turning point must be below the y-axis, as where k doesn't equal 0. (I typed it out on tapatalk, sorry, which does not work well with looking at previous posts >.<)

I gotta say, though, I don't know how the suggested solution got those values - |zxn|'s completing the square is flawless, which I did double check on the CAS.

[keltingmeith]

How do I differentiate the following?

Thanks

We have a product, so apply the product rule:



Then, here you'll notice that the first derivative is a composite function, so we'll need to apply the chain rule:

[speedy]

Does anyone know how to find the asymptotes on a casio classpad calculator and does anyone also know how to complete the square on the claculator

+1

[lzxnl]

Actually, using your working we know that the turning point occurs at (3-k^2, 9k^2). This means that the turning point must have a positive y-value or lie on the x-axis, as k^2>=0. But, since k=/=0, it must be a positive constant. (in fact, k >0 for the turning point to be a minimum, but that's a different story)

Yeah I think I didn't read the question. It wanted the y-value not the x value. Sigh. Doing stuff at 1 am in the morning really doesn't work.

[Anchy]

An isosceles triangle has its equal sides of length 10cm with an included angle (theta). If theta changes from 60degrees to 61 degrees, find correct to 2 decimal places:

1. The approx area of triangle when theta=61degrees
2. The approximate increase in the area, A of a triangle

Thanks

[ETTH96]

Hey guys, how do i do this?

let f(x)= 6sin(pi/6 x)
and
let h(x)= 6cos(pi/2 x)

State a sequence of 2 transformations which takes the graph of f(x) to h(x).

[Reus]

Someone pls

[soNasty]

Hey guys, how do i do this?

let f(x)= 6sin(pi/6 x)
and
let h(x)= 6cos(pi/2 x)

State a sequence of 2 transformations which takes the graph of f(x) to h(x).

Ok for this one what youd do is graph them both, and then you'd realise that the sin graph is much more wide than the cos graph. To overcome this, we would:
- Dilate it by a factor of 1/3 parallel to the x axis /from y
this basically turns the (pi/6 x) into (pi/2 x) like the cos

and now all were left with is a translation..
right now we have this image of the graph:


so now we must translate it to the left one unit to make it the same (as u can see from the image)
so it'd require: - a translation of one unit in the negative direction of the x axis.

and hence it'll become