VCE Methods Question Thread!

[Sine]

In VCAA 2015 Q4 (d)(i)
why have they integrated it from pi to 0? how do we know where we're supposed to integrate it from?

Try to include a copy of the question in your posts in the future.

They were able to choose from 0 to pi because the required integral is the "same" from 0 to pi as it is from pi to 2pi thus we can use 0 to pi and just multiply the integral by 2.

[minhalgill]

for (c)(i) i understand how to solve for d, and get 20/3, which is the correct answer, but the marking scheme states d can be anywhere from 20/3 till 10? i dont understand how?

[S_R_K]

for (c)(i) i understand how to solve for d, and get 20/3, which is the correct answer, but the marking scheme states d can be anywhere from 20/3 till 10? i dont understand how?

This is a very difficult question, requiring some big leaps of logic. Very tough to do under time pressure in an exam.

Here is one way you might arrive at the answer:

First, the earlier parts of the question suggest that the stationary point of V(t), if it exists in the interval 0 ≤ t ≤ 5, will be a local minimum.

Hence, we need to find all values of d for which the stationary point of V(t) does not occur in the interval 0 < t ≤ 5. That is, if the stationary point is not in the interval, then V(t) will be increasing on [0, 5], and so the minimum will occur at t = 0.

To do this, we express the t-coordinate of the stationary point as a function of d. This is:



If you sketch a graph of t(d), you'll notice that t(d) > 0 for 0 < d < 20/3, and t(d) = 0 when d = 20/3, and t(d) < 0 when d > 20/3. The interval we are looking for is when t(d) < 0, since this corresponds to t-coordinates that are outside the domain of V(t). Hence, the stationary point of V(t) does not occur in the interval 0 < t ≤ 5 when d > 20/3. But we also know that d is restricted to values between 0 and 10, so our final answer is 20/3 ≤ d < 10. We include d = 20/3 in the interval, because this is when the stationary point is at t = 0.

[minhalgill]

in part g, when finding the angle, i try using m=tan(x), and the gradient is negative, i get the wrong value of the angle, yet when i remove the negative sign of the gradient, and then try solving it, it gives me the correct value for the angle, does anyone know why?

[S_R_K]

in part g, when finding the angle, i try using m=tan(x), and the gradient is negative, i get the wrong value of the angle, yet when i remove the negative sign of the gradient, and then try solving it, it gives me the correct value for the angle, does anyone know why?

If you use m = tan(x), you are assuming that x is the angle measured as the amount of anti-clockwise rotation from the positive x-axis.

For this question, the angle is defined as the amount of clockwise rotation from the negative x-axis.

This is equivalent to the identity -tan(x) = tan(180° – x).

also, does anyone know how to do h?

What about the solution in the examiner's report do you not understand?

[minhalgill]

If you use m = tan(x), you are assuming that x is the angle measured as the amount of anti-clockwise rotation from the positive x-axis.

For this question, the angle is defined as the amount of clockwise rotation from the negative x-axis.

This is equivalent to the identity -tan(x) = tan(180° – x).

so if we're measuring from the clockwise direction, should we always reverse the sign of the gradient? so the gradient can be negative as long as the angle measured is from the anticlockwise direction?

[minhalgill]

where it says 2k=tan(30) should this not be 1/2k=tan(30)?

[Lear]

That is a very tough question. I’d recommend you look back a few pages in this very thread as there was some good discussion on it.

[minhalgill]

That is a very tough question. I’d recommend you look back a few pages in this very thread as there was some good discussion on it.

ive looked back and i understand all that. what ive done is , i found the gradients for the tangents, and i understand the tangent of the inverse will make an angle of 30 with the x axis, and i understnad the tangent of the function will make a angle of 60 from the x axis. ive then applied the formula m=tan(x). i get correct value for the value of k from the functins tangent. for the tangent of the inverse, i get the correct graident which is 1/2k, as in the marking scheme, but the marking scheme goes from 1/2k to just 2k, and i dont understnad why

[S_R_K]

ive looked back and i understand all that. what ive done is , i found the gradients for the tangents, and i understand the tangent of the inverse will make an angle of 30 with the x axis, and i understnad the tangent of the function will make a angle of 60 from the x axis. ive then applied the formula m=tan(x). i get correct value for the value of k from the functins tangent. for the tangent of the inverse, i get the correct graident which is 1/2k, as in the marking scheme, but the marking scheme goes from 1/2k to just 2k, and i dont understnad why

Because there are two possible gradients that the tangent to g at x = 0 can have, such that the angle between the tangents of g and g^–1 is 30°. Either the tangent to g has an angle of 30° with the positive x-axis, or it has an angle of 60° with the positive x-axis. Hence you need to solve two equations for k.

[sailinginwater]

Can someone please explain why mcq 17 of last year's exam 2 is C? I know why it's not A,B, D or E but I don't know why it's C

[GMT. -_-]

Last year's MCQ 17 is D, not C...

[sailinginwater]

Last year's MCQ 17 is D, not C...

My bad, can someone please explain why it's D?

[minhalgill]

Because there are two possible gradients that the tangent to g at x = 0 can have, such that the angle between the tangents of g and g^–1 is 30°. Either the tangent to g has an angle of 30° with the positive x-axis, or it has an angle of 60° with the positive x-axis. Hence you need to solve two equations for k.

dont we obtain 1 tangent from the inverse and 1 tangent from the graph of g? and use the gradient of the inverse with 30 and the gradient of g with 60?

can someone please explain how to do this? i dont understnad the effect of the square power on the period and amp

mod edit: merged double post

[sailinginwater]

2013 methods exam 1
Question 10c
Is it possible to integrate a log in the methods study design?