VCE Specialist 3/4 Question Thread!

[random_person]

I think I may have missed something, as the method that I've used isn't exactly in the course... using trigonometric substitution, but depending on your school you may cover it. Where exactly is the question from?

Thank you very much!
My teacher gave us this question as a challenge, he didn't expect many of us to get it 

[lzxnl]

I recall what someone did, that yields the same answer, is to unravel the torus into a cylinder. The height of this cylinder is then given by the distance the center of the circle travels in the revolution, which is 2pi*a.
The radius of this cylinder is b, so the volume is simply pi*r^2*h=2pi^2*a*b^2, the same answer that you got. Not one bit of calculus needed.
Although honestly, I prefer the integral method xP
You don't NEED to actually use calculus to evaluate the integral. The integral of sqrt(b^2-x^2) over x=+-b can be interpreted as the area of a semicircle of radius b, which is obviously pi*b^2/2.

[b^3]

I recall what someone did, that yields the same answer, is to unravel the torus into a cylinder. The height of this cylinder is then given by the distance the center of the circle travels in the revolution, which is 2pi*a.
The radius of this cylinder is b, so the volume is simply pi*r^2*h=2pi^2*a*b^2, the same answer that you got. Not one bit of calculus needed.
Although honestly, I prefer the integral method xP
You don't NEED to actually use calculus to evaluate the integral. The integral of sqrt(b^2-x^2) over x=+-b can be interpreted as the area of a semicircle of radius b, which is obviously pi*b^2/2.

This is basically leading into the shell method for vols of rev (again not on the course). Which would be easier to use than the method I used above (I avoided using it as it isn't on the course, just to see that I was going to go off the course anyways).

But yeah, for others don't worry about it too much, as it's no on the course (unless you want to learn a bit extra).

[brightsky]

This is basically leading into the shell method for vols of rev (again not on the course). Which would be easier to use than the method I used above (I avoided using it as it isn't on the course, just to see that I was going to go off the course anyways).

But yeah, for others don't worry about it too much, as it's no on the course (unless you want to learn a bit extra).

nah shell method is slightly different to the method nliu has suggested above. nliu is proposing the following: rotate the circle around the x-axis, obtain the torus, cut the torus, 'unravel' it to obtain a long cylinder, and then find the volume of that cylinder using year 9 maths. the shell method involves rotating rectangles parallel to the axis of revolution (as opposed to rotating rectangles perpendicular to the axis of revolution, as is the case with the disk method, i.e. the method taught in spesh).

it is certainly possible to use the shell method here, but i don't think it'll make life any easier....

[stolenclay]

So it's part b), c) and d) that I don't know how to do.
If you could help out, I'd appreciate it

I have a solution to part b, and I think part d follows on quickly from part b.
The obvious thing to try is calculating , and see if we can somehow prove it equals 0.




Now for a little Pythagoras' Theorem...





Hence !

Because our dot product is 0, and are perpendicular!

For part d:
During our solution, we showed ! How convenient!

For part c (apologies for messed up order):
During our solution, we showed ! How convenient!
and (as well as , I guess) together imply that (congruent triangles, RHS).
and mean that is the perpendicular bisector of .

[b^3]

nah shell method is slightly different to the method nliu has suggested above. nliu is proposing the following: rotate the circle around the x-axis, obtain the torus, cut the torus, 'unravel' it to obtain a long cylinder, and then find the volume of that cylinder using year 9 maths. the shell method involves rotating rectangles parallel to the axis of revolution (as opposed to rotating rectangles perpendicular to the axis of revolution, as is the case with the disk method, i.e. the method taught in spesh).

it is certainly possible to use the shell method here, but i don't think it'll make life any easier....

Sorry I misread what he meant, although what I was getting at is you wouldn't have to use a trigonometric substitution in the integration, rather a simpler substitution would be applicable(the integration technique would be in spesh, but as a whole still not inside the course, unfortunately. I wish they'd beef the course up a little bit -.-)

EDIT: Actually yes, the terminals will get pretty ugly, doesn't really help in the end (I was using the wrong circle equation earlier on when I tried to do it).

[barydos]

I have a solution to part b, and I think part d follows on quickly from part b.
The obvious thing to try is calculating , and see if we can somehow prove it equals 0.




Now for a little Pythagoras' Theorem...





Hence !

Because our dot product is 0, and are perpendicular!

For part d:
During our solution, we showed ! How convenient!

For part c (apologies for messed up order):
During our solution, we showed ! How convenient!
and (as well as , I guess) together imply that (congruent triangles, RHS).
and mean that is the perpendicular bisector of .

Hey thanks a lot, that was excellent

[~T]

Exercise 9C of the Essential textbook... Differential equations of the form dy/dx = f(y)

I understand how to get the solutions, I'm having no trouble with that. However I'm a little confused with absolute value signs, and when to use them and when not to.

For example, question 1a): Find the general solution of













, after letting

... but the answer only has the positive, rather than the plus/minus. Can someone please explain why?

Conversely, question 1c) is and I come out with:



.. and the answers say:





Thank you to anyone who can help

[BubbleWrapMan]

The reason it was dropped in the first instance is because can be any real constant, which accounts for the negative sign.

In most instances, you need to use the initial condition in order to get rid of the modulus function. You basically substitute the initial condition in and decide which of the two 'options' it works with. For instance, if you had something with and your initial condition included you would take . The reason you need to drop the modulus sign is that the solution can't cross any singularities (e.g. you can't have so solutions can't cross this 'boundary').

[Alwin]

<-- THIS LINE IS IMPORTANT

... but the answer only has the positive, rather than the plus/minus. Can someone please explain why?

consider :



Hence,
becomes

This is because in the first like RHS is positive and LHS positive too. Hence, the absolute values can be discarded.
It's the same as saying

Look at Timmeh's explanation below. especially that:

The reason it was dropped in the first instance is because can be any real constant, which accounts for the negative sign.

Conversely, question 1c) is and I come out with:



.. and the answers say:

Basically, when integrating always add the absolute signs, unless the domain is stated in the question. We assume maximal domain which is R\{c}

If, the question had been: integrate and domain given
then you can disregard the absolute signs because hence the exists when

Hope that clears things up a bit!

Edit: messed up, deleted that part and referred to Timmeh's explanation.

[~T]

becomes


This is because in the first like RHS is positive and LHS positive too. Hence, the absolute values can be discarded.
It's the same as saying

But I still don't understand... in the simpler example, couldn't you still have and

would be true, and would not??? Before getting rid of the modulus, we could have negative y values, and you can't afterwards.


Though the second of my queries makes sense now, thanks for that.

[BubbleWrapMan]

consider :



Hence,
becomes


This is because in the first like RHS is positive and LHS positive too. Hence, the absolute values can be discarded.
It's the same as saying

That isn't true. , you can't get rid of the modulus sign on the basis that one side is positive. For example, the point satisfies but not .

[ikhalid]

how do you solve this question?

2cosec^2 (theta) + 5cot(theta) =5 , find tan(theta) [0,2pie]

[~T]

Also, just checking the answers by putting them back into the calculator:
For and the answer supposedly equalling

Substituting and should both work if it is an absolute value, right?

The first solution gives which seems true...

The second "solution" gives which seems not so true...



Can anyone else please help with my query? I really cannot see why there should be a modulus sign added. My first question, I'm guessing is just correct because A can be positive or negative, but I'm still stumped on the second...

[Alwin]

That isn't true. , you can't get rid of the modulus sign on the basis that one side is positive. For example, the point satisfies but not .

Yeah, i realised.  I just had no better way of explaining it and I took that example out of context of the log_e function and that the constant A can be positive or negative. sorry, i just sort of smudged over it. I'll go back and change it now

how do you solve this question?

2cosec^2 (theta) + 5cot(theta) =5 , find tan(theta) [0,2pie]

hint: cot^2(x) + 1 = cosec^2(x)

I think you can do it from here