VCE Specialist 3/4 Question Thread!

[brightsky]

Yeah sorry, ignore that. Draw the triangle with vertices (-2,2sqrt(3)), (0,2sqrt(3)) and (0,0). It is a right angled triangle. Let theta be the angle which the line connecting (0,0) and (-2,2sqrt(3)) makes with the positive y-axis. theta = tan^-1 (2/(2sqrt(3)). Double that and add pi/2 to the result. That should be your argument.

[barydos]

Yeah sorry, ignore that. Draw the triangle with vertices (-2,2sqrt(3)), (0,2sqrt(3)) and (0,0). It is a right angled triangle. Let theta be the angle which the line connecting (0,0) and (-2,2sqrt(3)) makes with the positive y-axis. theta = tan^-1 (2/(2sqrt(3)). Double that and add pi/2 to the result. That should be your argument.

Can I ask why we have to double that answer before adding pi/2?

[brightsky]

Can I ask why we have to double that answer before adding pi/2?

Property of tangents. The origin is our external point. There are two tangents that can be drawn from the origin to the circle. One lies on the y-axis. The other, we must find. But we can circumvent this problem by considering only the angles. If we draw a line from the centre of the circle to the origin, we bisect the angle made by the two aforementioned tangents. I have found only one half of the angle in my working above. So I must double before adding pi/2.

[barydos]

Property of tangents. The origin is our external point. There are two tangents that can be drawn from the origin to the circle. One lies on the y-axis. The other, we must find. But we can circumvent this problem by considering only the angles. If we draw a line from the centre of the circle to the origin, we bisect the angle made by the two aforementioned tangents. I have found only one half of the angle in my working above. So I must double before adding pi/2.

OH MAN you're a genius! (or im just silly idk) thanks a lot,
BUT ONE LAST QUESTION sorry for troubling you!

Um, how come this wasn't solvable via the first method you suggested? I mean, it sounds logical.. :/

[brightsky]

Nah, the first method was based on the assumption that (-2, 2sqrt(3) - 2) is indeed the required z. I'm almost certain it isn't. But as I've explained in my latest post, it's not really necessary to find z at all. Exclusive consideration of angles will suffice.

[barydos]

Nah, the first method was based on the assumption that (-2, 2sqrt(3) - 2) is indeed the required z. I'm almost certain it isn't. But as I've explained in my latest post, it's not really necessary to find z at all. Exclusive consideration of angles will suffice.

Thanks a lot brightsky, you're the best!

[Stick]

[brightsky]

S_(10) = 1 + x/2 + x^2/2^2 + ... + x^9/2^9
x/2*S_(10) = x/2 + x^2/2^2 + x^3/2^3 ...+ x^10/2^10
(x/2-1) S_(10) = x^10/2^10 - 1
S_(10) = (x^10/2^10 - 1)/(x/2 - 1)
When x = 3/2
S_(10) = 3.7747...

b) i) -1 < x/2 < 1 which means -2 < x < 2
ii) S = 1/(1-x/2)
We require:
1/(1-x/2) =2*(x^10/2^10 - 1)/(x/2-1)
x = +- 2^(9/10)

[Stick]

Thanks. I didn't learn that the common ratio had to be between -1 and 1 for the infinite sum to exist.

[BubbleWrapMan]

Remember that all geometric series have a closed form expression of a(1-r^n)/(1-r). Note what happens when n approaches infinity. r^n becomes infinitely large if |r|>1, but infinitely small if |r|<1, so it will only approach a limiting value of a/(1-r) if r^n becomes 0, which is the case when |r|<1, or equivalently -1<r<1.

[507]

When sketching parametric equations of circles and ellipses, do you need to indicate which direction the graph is travelling in (clockwise/anticlockwise)?

[Jaswinder]

how would i find the volume generated when the region is rotated about the y axis. some how i keep getting 8pi/3 

[polar]

how would i find the volume generated when the region is rotated about the y axis. some how i keep getting 8pi/3 

isn't that the answer?
by integrating:


by using :

[BubbleWrapMan]

When sketching parametric equations of circles and ellipses, do you need to indicate which direction the graph is travelling in (clockwise/anticlockwise)?

Nope, unless they specifically ask you to

[Homer]

isn't that the answer?
by integrating:


by using :

the book gets 28pi/6 :/

Also with y = root(x), x=1, x=4 the volume generated when region is rotated about the y -axis. I keep getting 31pi/5 but the book says 124pi/5