Need Elaboration on 1996 3a). (1 Viewer)

[ProdigyInspired]

I need some clarification/explanation on why the outer radius is rather than .

I've been discussing this with my tutor and it still doesn't make sense to me.

The 4 is easily understood, as the graph shows that it goes from 0 to 4.

is what I would've thought would be added onto the radius of the outer annulus. I wouldn't think it is . Why would the radius get smaller than 4? The outer radius - what I'm looking for, covers the area of as well.

My tutor said subbing in as it is an end point of the segment, which of course gives you as needed. However if you sub in , which is the other end point of the segment, it gives you , giving you , and then .

 

[ProdigyInspired]

My tutor just clarified:

It is best to consider both 4-x and 4+x, since x values can be both positive or negative as shown by the segment.

 

[InteGrand]

I need some clarification/explanation on why the outer radius is rather than .

I've been discussing this with my tutor and it still doesn't make sense to me.

The 4 is easily understood, as the graph shows that it goes from 0 to 4.

is what I would've thought would be added onto the radius of the outer annulus. I wouldn't think it is . Why would the radius get smaller than 4? The outer radius - what I'm looking for, covers the area of as well.

My tutor said subbing in as it is an end point of the segment, which of course gives you as needed. However if you sub in , which is the other end point of the segment, it gives you , giving you , and then .

Note that x is negative (except at the origin, where it's 0) for the outer radius (since the curve is in the left half of the plane), so by doing 4 – x, we are adding on a positive number to 4 (since -x is positive, so it's 4 + (-x) > 4).

In general, if x < X, then the distance along the real line from x to X is X – x. So since here x < 4, the distance from x to 4 is 4 – x. (This'd be true even if x were positive, the only important thing is that x < 4.)

 

[InteGrand]

My tutor just clarified:

It is best to consider both 4-x and 4+x, since x values can be both positive or negative as shown by the segment.

It's always 4 – x (in this question). Firstly, for the outer radius, the x is always on the left side of the plane, so x is negative. Secondly, it doesn't actually matter, even if x were positive, it'd still be 4 – x. As long as x < 4 (which it always is for this Q.) the distance from x to 4 is just 4 – x. (If x were greater than 4 though, the distance would be x – 4. In general, distance between two points a and b on the real line is |a – b|. So if a > b, this becomes a – b. If a < b, it becomes -(a – b) = b – a.)

 

[ProdigyInspired]

Thanks Integrand, another question from the 1996 HSC.

Easy enough, but the answers show that not every term is accompanied by the factor of 1/2 (part iv)

And only has the factor.

Why is that so? I understand the last few lines of the working out, as all the terms end up having the factor of , but not so much of the first line, as it isn't factorized. Is it shown in a way that it's factorized?

 

[Trebla]

I think that's a typo. Each term should be accompanied by a factor of 1/2 when using the trapezoidal rule.

 

[ProdigyInspired]

I think that's a typo. Each term should be accompanied by a factor of 1/2 when using the trapezoidal rule.

Is the question a typo as well? Because it similarly only shows ln k to have a factor of 1/2.

However I understand that they're the same as it's just multiplied.

 

[InteGrand]

Is the question a typo as well? Because it similarly only shows ln k to have a factor of 1/2.

However I understand that they're the same as it's just multiplied.

The middle terms add up to 1 times a log. E.g. with k = 6, it would be:

(1/2)*(ln(1) + ln(2)) + (1/2)*(ln(2) + ln(3)) + (1/2)*(ln(3) + ln(4)) + (1/2)*(ln(4) + ln(5)) + (1/2)*(ln(5) + ln(6))

= ln(2) + ln(3) + ln(4) + ln(5) + (1/2)*ln(6)

= ln(5!) + (1/2)*ln(6) = ln((k-1)!) + (1/2)*ln(k).

The same idea applies for general k. So the answers just skipped a few steps maybe. I guess what they wrote in the first line is a bit sloppy but the rest is definitely right.

 

[ProdigyInspired]

I guess so.

Sorry mods if this violates anything, I'll just use this thread so it doesn't fill the forum up.
I'm getting confused on resolving forces in general.

1. What is the best way to find expressions for other forces, particularly if they aren't specified an angle? This includes questions that have like a slope with angle theta, and a box with several forces, then you have to find expressions of N, gravity etc using theta.

2. HSC 2015 Q14c)
I'm not exactly sure how frictional force is considered.
The horizontal forces what I would've thought: .
I'm assuming Ncostheta points positively i.e. right, and microN sintheta points to the left, as it is friction i.e. slows the object down.
But in the answers, microN i.e. the frictional force is considered positive. Why?

 

[ProdigyInspired]

Is it related to micro > 0 ?

 

[InteGrand]

I guess so.

Sorry mods if this violates anything, I'll just use this thread so it doesn't fill the forum up.
I'm getting confused on resolving forces in general.

1. What is the best way to find expressions for other forces, particularly if they aren't specified an angle? This includes questions that have like a slope with angle theta, and a box with several forces, then you have to find expressions of N, gravity etc using theta.

2. HSC 2015 Q14c)
I'm not exactly sure how frictional force is considered.
The horizontal forces what I would've thought: .
I'm assuming Ncostheta points positively i.e. right, and microN sintheta points to the left, as it is friction i.e. slows the object down.
But in the answers, microN i.e. the frictional force is considered positive. Why?

For convenience of anyone who wants to see the answers from BOSTES, here they are (see page 23): https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-maths-ext-2.pdf .

 

[ProdigyInspired]

For convenience of anyone who wants to see the answers from BOSTES, here they are (see page 23): https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-maths-ext-2.pdf .

Ah, I mixed them up as well as assuming right as positive. Thanks. I'd made this mistake since I had incorrectly placed theta's position. How do you suggest to move theta to create the triangles BOSTES had made? Sometimes I try to use trigonometry and do it but get confused.

 

[InteGrand]

Ah, I mixed them up as well as assuming right as positive. Thanks. I'd made this mistake since I had incorrectly placed theta's position. How do you suggest to move theta to create the triangles BOSTES had made?

Well the mu*N arrow is parallel to the large triangle's (the track's) hypotenuse, so if we draw in a horizontal there, the angle between the horizontal and the mu*N arrow is just theta. For the other one (angle for the normal force vector), since that comes up basically on every banked track question, you should probably just remember that the theta (the track's banking angle) is the angle between the normal vector and the vertical. You can prove this using parallel lines / similar triangles etc., but it's a waste of time to since it appears in every banked track Q., so you should just remember it. (It should stick in your mind if you do enough practice Q's, since it's something present in most banked track Q's.)

 

[ProdigyInspired]

Do you also suggest taking the direction of movement as positive? (as in this question, where the car tends to slide up. This is probably more obvious when all forces {normal, tension, friction, gravity} are shown.)

Or taking the direction of the given forces to make it easier? (in this question, left is taken positive as friction is the only force shown. Thus for ease, it is selected).

 

[InteGrand]

Do you also suggest taking the direction of movement as positive? (as in this question, where the car tends to slide up. This is probably more obvious when all forces {normal, tension, friction, gravity} are shown.)

Or taking the direction of the given forces to make it easier? (in this question, left is taken positive as friction is the only force shown. Thus for ease, it is selected).

In this one, left is positive because the circle centre is to left (like the centre of the turn). In general, for banked track Q's, it's good to make the positive horizontal direction the direction pointing towards the centre of the turn.

 

[ProdigyInspired]

In 2015's q16b iv), how is determined as a substitution?

By inspection? itute's answers display using then letting it be , which is pretty random.

 

[InteGrand]

In 2015's q16b iv), how is determined as a substitution?

By inspection? itute's answers display using then letting it be , which is pretty random.

Since we needed to prove something equalling (2^{2n} times) cos(n*pi/2), it's natural to make arccos(x) equal pi/4, so that the 2n*arccos(x) becomes n*pi/2.