My Suggested Solutions to SP2008 Exam 2 [Confirmed]

[Neobeo]

b) Graph: a figure 8 on its side, symmetrical about the axes; (see Neobeo's post#6 in this thread for visualisation)
x-intercepts (-1,0), (1,0), four stationary points in each quadrant at (±1/sqrt(2),±0.5)

The stationary points have y-coordinate ±0.25.

The fourth line of working for a)ii)

y^2 = sin^2t/3cos^2t/3, not y !!!!

I miswrote that when copying from my other draft. If you see the graph it is clearly the y^2 interpretation. Also, the cartesian equation doesn't carry to the other questions other than the graph.

[hamtarofreak]

You missed the ^2 on the cartesian.

y = 0 FOUR times in each circuit, look at the graph.. You need to use x=0, and find the period of it, which is 6pi.

I have no idea what you were trying to do with 3d.. You integrated it correctly the first time, then changed it into some weird-ass thing. v(0) = 1/3cos(0)i + 1/3cos(0)j |v(0)|= sqrt(1/9+1/9)=sqrt(2/9)=sqrt(2)/3.

Consequential error with 3e, the integral should be from 6pi to 0.

[Nelle91]

For MC Istnt 2e because (x+1)2 + y2 =0 isn't a circle right

[hamtarofreak]

For MC Istnt 2e because (x+1)2 + y2 =0 isn't a circle right

It's a circle translated 1 unit to the left, but still a circle.

[Nelle91]

Whats the radius? wouldnt e be right

[Synesthetic]

b) Graph: a figure 8 on its side, symmetrical about the axes; (see Neobeo's post#6 in this thread for visualisation)
x-intercepts (-1,0), (1,0), four stationary points in each quadrant at (±1/sqrt(2),±0.5)

The stationary points have y-coordinate ±0.25.

The fourth line of working for a)ii)

y^2 = sin^2t/3cos^2t/3, not y !!!!

I miswrote that when copying from my other draft. If you see the graph it is clearly the y^2 interpretation. Also, the cartesian equation doesn't carry to the other questions other than the graph.

Yes, they would have y-coordinate 0.25 without the square root, which is what you missed.

I have just constructed the curve using my graphics calculator and I am finding y-maximum = 0.5

You missed the ^2 on the cartesian.

y = 0 FOUR times in each circuit, look at the graph.. You need to use x=0, and find the period of it, which is 6pi.

I have no idea what you were trying to do with 3d.. You integrated it correctly the first time, then changed it into some weird-ass thing. v(0) = 1/3cos(0)i + 1/3cos(0)j |v(0)|= sqrt(1/9+1/9)=sqrt(2/9)=sqrt(2)/3.

Consequential error with 3e, the integral should be from 6pi to 0.

Yes, this aligns with my interpretation.

[hamtarofreak]

Oh wow. I didn't even realise there was a 0 on the other side o.O

You would be right then, E.

[Nelle91]

and for 11 isnt it D as you forgot to add the initial 1

[Synesthetic]

and for 11 isnt it D as you forgot to add the initial 1

My answer for 11 is D.

[sb3700]

Many people were asking about question 3, the train one. So coblin and I took it upon ourselves to come with fully worked out solutions. Enjoy.

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I agree with Synesthetic and disagree with NeoBeo:

For NeoBeo:
in aii - lost a square root
in c) it only passes origin when both sin(t/3) and sin(2t/3) are zero, so time is 6 pi
d) not zero, cos(t) = sin(t) / tan(t) is dividing by zero which is why you came up with zero. Just sub in t = 0 or t = 3pi so v = +/- 1/3i + 1/3j, |v| = sqrt(2)/3
ei) cos(2t/3) <> sin(t/3), cos(t/3) = sin(pi/2 - 2t/3), actual answer just evaluate integral from fourth step to get 6.1 metres i think

[Nelle91]

Lol soz my bad

[Nelle91]

I got 7 D though

[Synesthetic]

I got 7 D though

Yeah you're right, I'll change the OP.

After expanding the factors and converting z to z+yi, you have 4x, which is used to construct (x+2)^2 => centre at (-2,0)

[hamtarofreak]

I'm fairly sure 7 is D and 12 is A =/

[Nelle91]

I think 16 is C too as arctan(sqr3)= pie/3