Post by: Synesthetic on November 11, 2008, 06:07:37 pm
1)D 2)E 3)B 4)E 5)B 6)C 7)D 8)D 9)B 10)C 11)D 12)A 13)E 14)B 15)D 16)C 17)A 18)E 19)C 20)E 21)A 22)B
ANALYSIS SECTION
1)a) Maximum (1,1.5)
b)i) 9x^4-26x^2+1=0
b)ii) (0.2,0.5), (1.7,1.4)
c) Graph rising steeply to maximum at (1,1.5), tapering off to around (4,1)
Mark in the previously stated inflection points at (0.2,0.5), (1.7,1.4)
d)i) 2pi*integral[(18x^3)/(3x^2+1)^2 dx,0 to 1/sqrt(3)]
d)ii) V = 2pi * integral[1/u - 1/u^2 du,1 to 2]
d)iii) V = pi[2loge(2)-1] cubic units
2)a) a=4.5 m/s^2
b) v=12 m/s
c) a=(180-3v)/40 m/s^2
d) t=1.8 seconds
e)i) From top anticlockwise, the forces are T, 100, 80g, 390
e)ii) Tcos(theta)=195sqrt(3)-100
and Tsin(theta)=979
e)iii) tan(theta)=4.118 (3 d.p.)
e)iv) T=1007 N
3)a)i) y^2=sin^2(t/3)*cos^2(t/3), via sine double angle formula
a)ii) y=±sqrt[x^2(1-x^2)], by substituting the parametric equation for x into y
b) Graph: a figure 8 on its side, symmetrical about the axes; (see Neobeo's post#6 in this thread for visualisation...
with ±1/2 instead of ±1/4 on the y-axis)
x-intercepts (-1,0), (1,0), four stationary points in each quadrant at (±1/sqrt(2),±0.5)
c) 6pi seconds
d) sqrt(2)/3 m/s
e)i) (1/3)*integral[sqrt(cos^2(t/3)+cos^2(2t/3))dt,0 to 6pi]
e)ii) 6.1 metres
(Re. 'solution' in post#6: ;D )
4)a) [(x-10)^2]/9 + (y-5)^2 = 1
b) Graph of the ellipse, endpoints clockwise from y-maximum: (10,6),(13,5),(10,4),(7,5)
Centre (10,5), semi-major axis length=3, semi-minor axis length=1
c)i) t=6 months
c)ii) 500 foxes
d)i) dy/dx=[xy-10y]/[25x-5xy] by related rates
d)ii) Use implicit differentiation to find the derivative of the given expression, such that dy/dx=[xy-10y]/[25x-5xy] in d)i)
e) Minimum number of rabbits: 6590 (to the nearest ten rabbits)
and Maximum number of rabbits: 14430 (to the nearest ten rabbits)
5)a) By conversion to polar form, sqrt(3)/2+0.5i is a root of z^3=i
b) The three roots are z={-i, sqrt(3)/2+0.5i, -sqrt(3)/2+0.5i}
c) The intersection points between the two relations are (-sqrt(3)/2, 1.5), (0,0)
d) Circle of radius=1, centre (0,1), and linear y=-sqrt(3)x
e) Shade area to the right of the linear graph, contained within the circle
f) A~2.53 square units - hamtarofreak's working: http://vcenotes.com/forum/index.php/topic,7511.msg93089.html#msg93089
======
Final edit [2]: itute confirms all these answers :)
http://www.itute.com/wp-content/uploads/2008_vcaa_specialist_mathematics_exam_2_solutions.pdf
Exam 2 mark = 93% = SAC mark :P
Post by: 08jmy on November 11, 2008, 06:09:06 pm
Post by: csq7 on November 11, 2008, 06:12:37 pm
Post by: Nelle91 on November 11, 2008, 06:14:27 pm
Post by: nrisme on November 11, 2008, 06:14:43 pm
Post by: deezowashington on November 11, 2008, 06:20:35 pm
Post by: Neobeo on November 11, 2008, 06:21:17 pm
I left the graph, since that's probably the only useful thing from it.
(http://img385.imageshack.us/img385/6050/partbmh9.png)
Post by: darlok on November 11, 2008, 06:21:30 pm
Post by: nrisme on November 11, 2008, 06:21:51 pm
Post by: Synesthetic on November 11, 2008, 06:22:09 pm
I had Tsintheta=784I can't confirm which is correct :( We'll see when others post solutions.
I'm not very confident Tsintheta=979 is correct, either, not having had time to go over the paper...
Post by: Synesthetic on November 11, 2008, 06:22:33 pm
dude do u remember the question for 4c??I just posted it, 6 months and 500 foxes.
Post by: nrisme on November 11, 2008, 06:23:39 pm
Post by: Synesthetic on November 11, 2008, 06:26:33 pm
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.
[images]
Damn, the latter part of these answers is completely at odds with mine ;D
Post by: xers on November 11, 2008, 06:27:36 pm
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.I got the time as 3pi seconds too, but many people said its 6pi. And the speed, i dont think its 0.
Post by: Synesthetic on November 11, 2008, 06:31:07 pm
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.I got the time as 3pi seconds too, but many people said its 6pi. And the speed, i dont think its 0.
Yeah I was going to say that I was skeptical about speed=0, perhaps a factor strengthening the t=6pi argument.
Post by: xers on November 11, 2008, 06:32:49 pm
Post by: pastiche on November 11, 2008, 06:33:00 pm
Post by: Nelle91 on November 11, 2008, 06:35:00 pm
Post by: kurrymuncher on November 11, 2008, 06:37:02 pm
Post by: csq7 on November 11, 2008, 06:37:06 pm
i think you forgot to square root the y to get the cartesian equation..
Post by: Synesthetic on November 11, 2008, 06:37:15 pm
I think actually Tsintheta = 979 N my bad :(Okay thanks for confirmation, and bad luck.
Post by: jamestaR` on November 11, 2008, 06:37:45 pm
and speed = 0? dun think so
Post by: Synesthetic on November 11, 2008, 06:38:10 pm
Synesthetic:Do you mean this? a)ii) y=±sqrt[x^2(1-x^2)], by substituting the parametric equation for x into y
i think you forgot to square root the y to get the cartesian equation..
I included 'sqrt'.
As if your train one is right...you went from y^2 = whatever to say that y = the same thing?I didn't have speed=0 ...
and speed = 0? dun think so
Hey, was it harder than last years?YES.
Post by: jamestaR` on November 11, 2008, 06:40:12 pm
edit sorry meant for the guy who hand wrote his train solution
Post by: darlok on November 11, 2008, 06:42:02 pm
Post by: csq7 on November 11, 2008, 06:42:05 pm
Post by: xers on November 11, 2008, 06:42:19 pm
Post by: Ahmad on November 11, 2008, 06:47:45 pm
I have also verified Neobeo's other solutions and they are indeed correct.
Post by: Synesthetic on November 11, 2008, 06:54:04 pm
(http://img353.imageshack.us/img353/2411/page1kk3.png)
The fourth line of working for a)ii) :
y^2 = sin^2t/3cos^2t/3, not y !!!!
So does that make the rest of his solutions all incorrect? Is that why we have different answers to [my solutions:] t= 6pi, v=sqrt(2)/3, and s=6.1m?
Post by: xers on November 11, 2008, 06:54:37 pm
Post by: Neobeo on November 11, 2008, 06:56:53 pm
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.
Post by: hamtarofreak on November 11, 2008, 06:59:00 pm
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.
Post by: Nelle91 on November 11, 2008, 07:00:47 pm
Post by: hamtarofreak on November 11, 2008, 07:01:49 pm
For MC Istnt 2e because (x+1)2 + y2 =0 isn't a circle rightIt's a circle translated 1 unit to the left, but still a circle.
Post by: Nelle91 on November 11, 2008, 07:02:36 pm
Post by: Synesthetic on November 11, 2008, 07:03:34 pm
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.
Post by: hamtarofreak on November 11, 2008, 07:04:04 pm
You would be right then, E.
Post by: Nelle91 on November 11, 2008, 07:05:26 pm
Post by: Synesthetic on November 11, 2008, 07:06:24 pm
and for 11 isnt it D as you forgot to add the initial 1
My answer for 11 is D.
Post by: sb3700 on November 11, 2008, 07:07:54 pm
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.
(http://img353.imageshack.us/img353/2411/page1kk3.png)
(http://img353.imageshack.us/img353/4959/page2tv8.png)
(http://img353.imageshack.us/img353/7898/page3yr5.png)
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
Post by: Nelle91 on November 11, 2008, 07:08:23 pm
Post by: Nelle91 on November 11, 2008, 07:09:16 pm
Post by: Synesthetic on November 11, 2008, 07:11:57 pm
I got 7 D thoughYeah 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)
Post by: hamtarofreak on November 11, 2008, 07:12:05 pm
Post by: Nelle91 on November 11, 2008, 07:14:05 pm
Post by: Synesthetic on November 11, 2008, 07:15:36 pm
I'm fairly sure 7 is D and 12 is A =/
Yes, you're right, I rushed those :( Changing
Post by: stanas on November 11, 2008, 07:17:00 pm
Post by: benwonglovesdong on November 11, 2008, 07:18:27 pm
I'm fairly sure 7 is D and 12 is A =/
yeah, 12 must be A. f is the quadratic and then F is the cubic... antidif from -3 to 0 must be negative.... =A
Post by: Synesthetic on November 11, 2008, 07:19:33 pm
;DI'm fairly sure 7 is D and 12 is A =/
yeah, 12 must be A. f is the quadratic and then F is the cubic... antidif from -3 to 0 must be negative.... =A
Post by: pfftlah on November 11, 2008, 07:25:37 pm
did anyone finish with time to spare?!
Post by: xers on November 11, 2008, 07:26:09 pm
Post by: eza on November 11, 2008, 07:27:34 pm
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
In reference to c)
3 pi is fine - it is when the train is travelling from Q4 to Q2
6 pi is when it is travelling from Q3 to Q1
for T= 3pi
x = sin (3pi / 3) = sin (pi) = 0
y = 0.5 sin (2* 3pi / 3) = 0.5 sin (2pi) = 0
You can also graph this on the calc to verify that x=0 y=0 when t=3pi
Post by: Synesthetic on November 11, 2008, 07:30:29 pm
But has the train completed one CIRCUIT of the track at t=3pi?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
In reference to c)
3 pi is fine - it is when the train is travelling from Q4 to Q2
6 pi is when it is travelling from Q3 to Q1
for T= 3pi
x = sin (3pi / 3) = sin (pi) = 0
y = 0.5 sin (2* 3pi / 3) = 0.5 sin (2pi) = 0
You can also graph this on the calc to verify that x=0 y=0 when t=3pi
Post by: chid on November 11, 2008, 07:30:46 pm
For multiple choice question 2 E? and question 4 C?
Hmmm
Post by: eza on November 11, 2008, 07:32:25 pm
Post by: Synesthetic on November 11, 2008, 07:34:32 pm
I had a few minutes to spare-it was close...
For multiple choice question 2 E? and question 4 C?
Hmmm
How can 2) be E? a needs to take an integer value for the circle to be defined => C
I'm not sure about 4 though.
Post by: pfftlah on November 11, 2008, 07:37:29 pm
Post by: eza on November 11, 2008, 07:38:35 pm
Post by: Synesthetic on November 11, 2008, 07:38:46 pm
I had a few minutes to spare-it was close...
For multiple choice question 2 E? and question 4 C?
Hmmm
Re. 4, consider the curve, it's hyperbolic along the x-axis. It would make sense that the gradient can take any value outside the domain [-2,2] because they are the gradients of the respective asymptotes. The curve's gradient approaches this value but never reaches it.
At the x-intercepts, the gradient approaches infinity (I think). My point is in simply visualising the graph, the gradient ranges from (-inf,2) u (2,inf) => E is correct.
Post by: pfftlah on November 11, 2008, 07:39:13 pm
Post by: Synesthetic on November 11, 2008, 07:40:08 pm
i got E for 2 as well..
If a = ±2, x^2 ±2x + y^2 = 1 => (x±1)^2 + y^2 = 1 => circle defined
That's why I chose C.
Post by: Synesthetic on November 11, 2008, 07:40:43 pm
and for MC, Q6, doesnt imaginary include i?
No that's a common misconception, Im(z) refers to the value preceding i.
e.g. Im(5i) = 5
Post by: pfftlah on November 11, 2008, 07:41:07 pm
Post by: eza on November 11, 2008, 07:43:18 pm
(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 = (a^2)/4 - 1
For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.
Post by: Synesthetic on November 11, 2008, 07:43:28 pm
+ can someone please expain why 12 is A?
Integral from -3 to 0 = F(0) - F(-3)
Observe the cubic (the antiderivative function), the corresponding points at -3 and 0 are -2 and 7 respectively.
Thus F(0) - F(-3) = -2-7 = -9 [A]
Post by: chid on November 11, 2008, 07:44:47 pm
this is how I did mc2) - not sure if it's rightYes that was my appraoch also.
(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 = (a^2)/4 - 1
For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.
If a=2 then circle is not defined (radius=0). Therefore C is incorrect.
Post by: csq7 on November 11, 2008, 07:45:27 pm
Post by: hamtarofreak on November 11, 2008, 07:45:52 pm
Area = Area under the top half of circle
Minus area under the bottom half of the circle
Minus the triangle of area under the point
Triangle area =
Therefore, area within bounds is:
Post by: Synesthetic on November 11, 2008, 07:46:45 pm
this is how I did mc2) - not sure if it's right
(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 = (a^2)/4 - 1
For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.
Yeah I see your logic, I was fixated on the circles with centre (±1,0). I'm still not entirely sure which is correct.
*edit*
this is how I did mc2) - not sure if it's rightYes that was my appraoch also.
(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 = (a^2)/4 - 1
For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.
If a=2 then circle is not defined (radius=0). Therefore C is incorrect.
Point taken I'll change it
Post by: csq7 on November 11, 2008, 07:46:55 pm
Post by: Synesthetic on November 11, 2008, 07:49:30 pm
I got E for MC 12.. Anyone else? maybe just me? but its a definite integral.. and the area is clearly positive.. so E??
It's the area bounded by the parabola and the axes, under the x-axis, so it IS negative.
Post by: negatron on November 11, 2008, 07:49:59 pm
isnt it jst tha top bit? not the bottom because t is greater than zero
also with tcostheta and tsintheta can you do it straight away n obtain tantheta straight away?
Post by: eza on November 11, 2008, 07:51:25 pm
Quote
"If a = ±2, x^2 ±2x + y^2 = 1 => (x±1)^2 + y^2 = 1 => circle defined"should be
x^2 ± 2x + y^2 + 1 = 0
for example, let us take a=2
x^2 + 2x + y^2 + 1 = 0
(x + 1)^2 + y^2 = 0
Post by: Captain on November 11, 2008, 07:51:51 pm
(http://img353.imageshack.us/img353/4959/page2tv8.png)
t=6
To verify this [the easy way]
Get your TI-89 Titanium
Mode --> Graph --> Parametric
y= editor
x(t) =
y(t) =
Window editor
tmin: 0
tmax: 3pi
xmin: -1.2
xmax: 1.2
ymin: -0.6
ymax: 0.6
Graph.
Now, try the following
tmin: 0
tmax: 6pi
xmin: -1.2
xmax: 1.2
ymin: -0.6
ymax: 0.6
QED :P
Post by: Synesthetic on November 11, 2008, 07:53:34 pm
Ah you have the 1 on the wrong side of the equation.Quote"If a = ±2, x^2 ±2x + y^2 = 1 => (x±1)^2 + y^2 = 1 => circle defined"should be
x^2 ± 2x + y^2 + 1 = 0
for example, let us take a=2
x^2 + 2x + y^2 + 1 = 0
(x + 1)^2 + y^2 = 0
Thanks for that. Definitely E then :)
Post by: negatron on November 11, 2008, 07:57:12 pm
Post by: Synesthetic on November 11, 2008, 07:58:52 pm
wtf..... hey how come you dont restrict the ellipse?
isnt it jst tha top bit? not the bottom because t is greater than zero
also with tcostheta and tsintheta can you do it straight away n obtain tantheta straight away?
Which ellipse
Tan(theta) is found in the question succeeding that, they required you to state two equations so that T and theta could be evaluated.
Post by: Captain on November 11, 2008, 08:04:14 pm
wtf..... hey how come you dont restrict the ellipse?
isnt it jst tha top bit? not the bottom because t is greater than zero
also with tcostheta and tsintheta can you do it straight away n obtain tantheta straight away?
Question 4?
It's made up of trig functions. As time goes on, the pattern repeats.
Hence, it's a full ellipse.
Post by: hanhanchampion on November 11, 2008, 08:09:17 pm
I got E for MC 12.. Anyone else? maybe just me? but its a definite integral.. and the area is clearly positive.. so E??same.. haha
Post by: Rayle on November 11, 2008, 08:16:55 pm
Post by: azn_kiwi91 on November 11, 2008, 08:18:16 pm
Post by: ed_saifa on November 11, 2008, 08:20:01 pm
did anyone think of using the segment formula we learnt last year for the area in the last Question?Yes, but i couldn't remember it.
Post by: sida on November 11, 2008, 08:23:36 pm
Post by: Synesthetic on November 11, 2008, 08:26:10 pm
is it possible if neoen posts the solutions to mc?..as in worked solutions ..i dun rememeber any of the mutiple choice questions?!!trinon uploaded the blank exam here:
http://trinon.info/exams/VCAA_2008_Specialist_Mathematics_Exam_2.pdf
Post by: azn_kiwi91 on November 11, 2008, 08:28:38 pm
Post by: Synesthetic on November 11, 2008, 08:29:36 pm
Cool, nice MC answers, I got the same *thumbs up*That's encouraging :P I got at least 3 wrong, though.
Post by: azn_kiwi91 on November 11, 2008, 08:32:14 pm
Cool, nice MC answers, I got the same *thumbs up*That's encouraging :P I got at least 3 wrong, though.
WHAT???!!! which ones?
Post by: Synesthetic on November 11, 2008, 08:38:00 pm
Cool, nice MC answers, I got the same *thumbs up*That's encouraging :P I got at least 3 wrong, though.
WHAT???!!! which ones?
I changed them from my answers to the correct ones in the OP, so you're looking at the correct answers.
In any case, the ones I got wrong were 2,7,12 - all careless...
Post by: chris on November 11, 2008, 08:40:24 pm
Post by: azn_kiwi91 on November 11, 2008, 08:41:40 pm
For those wondering:
1/2 r^2 (theta - sin theta), theta in our shaded circle was 2pi.
PS: btw, what do you guys use to input those mathematical symbols.
Post by: jamestaR` on November 11, 2008, 08:43:15 pm
And extended can get fucked, how are u supposed to do that last area question! I forgot the year 11 segment formula and could think of no other way.
Post by: azn_kiwi91 on November 11, 2008, 08:53:42 pm
Yeh i made exact same mistakes as you synesthetic for 2 and 12, and had no clue for 17 so guessed it and got it wrong.
And extended can get fucked, how are u supposed to do that last area question! I forgot the year 11 segment formula and could think of no other way.
Yeah i thought that it was rat of VCAA to put a Q related to the year 11 syllabus. I took a guess at the formula and before using it, tried it with random circles with random radii to with angle pi to make sure that it was right.
Post by: junaman on November 11, 2008, 09:01:32 pm
Yeh i made exact same mistakes as you synesthetic for 2 and 12, and had no clue for 17 so guessed it and got it wrong.
And extended can get fucked, how are u supposed to do that last area question! I forgot the year 11 segment formula and could think of no other way.
Yeah i thought that it was rat of VCAA to put a Q related to the year 11 syllabus. I took a guess at the formula and before using it, tried it with random circles with random radii to with angle pi to make sure that it was right.
You could have integrated...
Post by: azn_kiwi91 on November 11, 2008, 09:14:51 pm
Yeh i made exact same mistakes as you synesthetic for 2 and 12, and had no clue for 17 so guessed it and got it wrong.
And extended can get fucked, how are u supposed to do that last area question! I forgot the year 11 segment formula and could think of no other way.
Yeah i thought that it was rat of VCAA to put a Q related to the year 11 syllabus. I took a guess at the formula and before using it, tried it with random circles with random radii to with angle pi to make sure that it was right.
You could have integrated...
Oh yeah that's right. 2 decimal places, and we got our calcs. lol Stupid me
Post by: julz on November 11, 2008, 09:31:46 pm
so
dw
u guys will beat me
and can people who did shit please post coz ur making me feel liek crap with all these solutions
i didnt fkn do anythign!!!!
it was sooooooo harddddddddddddddddddddd
Post by: jamestaR` on November 11, 2008, 09:41:23 pm
And i dunno how to use calc to find 'top minus bottom' areas so..
Post by: Blizz on November 11, 2008, 09:47:00 pm
well i slept thro my exam and i got 12/22 multiple choice rightI did shit
so
dw
u guys will beat me
and can people who did shit please post coz ur making me feel liek crap with all these solutions
i didnt fkn do anythign!!!!
it was sooooooo harddddddddddddddddddddd
i answered 2.5 questions out of 5 in analysis and i did worse than you in multiple
those fucking cartesian equations......
Post by: ed_saifa on November 11, 2008, 09:48:11 pm
I couldn't get the cartesian equations as well. GRRwell i slept thro my exam and i got 12/22 multiple choice rightI did shit
so
dw
u guys will beat me
and can people who did shit please post coz ur making me feel liek crap with all these solutions
i didnt fkn do anythign!!!!
it was sooooooo harddddddddddddddddddddd
i answered 2.5 questions out of 5 in analysis and i did worse than you in multiple
those fucking cartesian equations......
Post by: Blizz on November 11, 2008, 09:48:52 pm
I couldn't get the cartesian equations as well. GRRwell i slept thro my exam and i got 12/22 multiple choice rightI did shit
so
dw
u guys will beat me
and can people who did shit please post coz ur making me feel liek crap with all these solutions
i didnt fkn do anythign!!!!
it was sooooooo harddddddddddddddddddddd
i answered 2.5 questions out of 5 in analysis and i did worse than you in multiple
those fucking cartesian equations......
With 10 mins to go i think i did find one of them
but i forgot the +-
so when i drew the graph it was half an '8'
Im so fucking dopey lol
will be heaps happy with 40/80 although i dunno if i'll get that
i would take 30 raw atm and run with it.......
Post by: polky on November 11, 2008, 09:53:27 pm
Don't know if that's technically correct? Might have gotten 2.53 as a fluke.
[IMG]http://img91.imageshack.us/img91/5045/1111082203vh5.th.jpg[/img][IMG]http://img91.imageshack.us/images/thpix.gif[/img]
EDIT: Linked to a easier-to-see picture
Post by: csq7 on November 11, 2008, 09:54:32 pm
Post by: Synesthetic on November 11, 2008, 09:58:10 pm
Can someone pleeease explain to me.. how do you do MC 18? the one with the magnitude of F in the direction of dThat's the scalar product.
F . d hat = [F . d] / magnitude of d
*edit* Answer is E ... as per the post below yours :)
Post by: csq7 on November 11, 2008, 09:59:35 pm
Post by: hamtarofreak on November 11, 2008, 10:00:19 pm
Can someone pleeease explain to me.. how do you do MC 18? the one with the magnitude of F in the direction of d
It's just asking you to find the component of F in the direction of d.
Thus, you're looking for
Which is also
Post by: csq7 on November 11, 2008, 10:02:13 pm
Post by: Blizz on November 11, 2008, 10:06:32 pm
Post by: riadnicolas on November 11, 2008, 10:21:03 pm
Post by: junaman on November 11, 2008, 10:40:05 pm
INtegrated how? you couldnt use the "top minus bottom" method because the circle went went outside the point of intersection.
And i dunno how to use calc to find 'top minus bottom' areas so..
The circle went outside the point of intersection? The circle _was_ the point of intersection? What you talkin' bout?
You integrate by finding the equation of the circle in terms of x (top half of the circle), which was rt(1-x^2) +1
And you integrated between -rt(3)/2 and 0 with the circle minus the line. The calculator does it for you if you subtract Y2 from Y1...
Then to that you add the pi/2 which was the semi circle...
Post by: riadnicolas on November 11, 2008, 10:41:38 pm
INtegrated how? you couldnt use the "top minus bottom" method because the circle went went outside the point of intersection.
And i dunno how to use calc to find 'top minus bottom' areas so..
i attempted that but lost time and ended up with 2.52 or sumthing funny
The circle went outside the point of intersection? The circle _was_ the point of intersection? What you talkin' bout?
You integrate by finding the equation of the circle in terms of x (top half of the circle), which was rt(1-x^2) +1
And you integrated between -rt(3)/2 and 0 with the circle minus the line. The calculator does it for you if you subtract Y2 from Y1...
Then to that you add the pi/2 which was the semi circle...
Post by: junaman on November 11, 2008, 10:45:08 pm
Post by: riadnicolas on November 11, 2008, 10:54:17 pm
2.52 is hardly funny...
nah im just guessing i did some weird things i remember getting two point something two and integrating about y
Post by: chris on November 11, 2008, 11:09:08 pm
Post by: Captain on November 11, 2008, 11:13:56 pm
I did the area of the circle thing this way (picture attached), which didn't require any integration.
Don't know if that's technically correct? Might have gotten 2.53 as a fluke.
[IMG]http://img91.imageshack.us/img91/5045/1111082203vh5.th.jpg[/img][IMG]http://img91.imageshack.us/images/thpix.gif[/img]
EDIT: Linked to a easier-to-see picture
That looks right. Good thinking :P
I've got a friend who by inspecting, decided to try and model a sin wave to fit the area he needed to find, so he could integrate :P HAHAHA. I think examiners will see lots of weird methods here.
Post by: junaman on November 11, 2008, 11:40:38 pm
That looks right. Good thinking :P
I've got a friend who by inspecting, decided to try and model a sin wave to fit the area he needed to find, so he could integrate :P HAHAHA. I think examiners will see lots of weird methods here.
Fail.
Post by: chris on November 11, 2008, 11:42:37 pm
-- anyoen im desperate to know i cant sleep otherwise,, and i cant figure it out.
Post by: Toothpaste on November 11, 2008, 11:44:20 pm
****** alrite guys what would g1 c+, g2 b+ and g3 C get me in spec raw, and then standardised,,,, --- these r following 2007's distribution?http://vcenotes.com/forum/index.php/topic,6930.msg93460/topicseen.html#msg93460
-- anyoen im desperate to know i cant sleep otherwise,, and i cant figure it out.
Post by: jamestaR` on November 11, 2008, 11:48:40 pm
INtegrated how? you couldnt use the "top minus bottom" method because the circle went went outside the point of intersection.
And i dunno how to use calc to find 'top minus bottom' areas so..
The circle went outside the point of intersection? The circle _was_ the point of intersection? What you talkin' bout?
You integrate by finding the equation of the circle in terms of x (top half of the circle), which was rt(1-x^2) +1
And you integrated between -rt(3)/2 and 0 with the circle minus the line. The calculator does it for you if you subtract Y2 from Y1...
Then to that you add the pi/2 which was the semi circle...
Yeah i realise the point of intersection is where you integrate from, but part of the circle stuck out to the left of that particular point. Additionally, you are dealing with 2 different equations of the circle, where the top semi circle has a different equation to the bottom. For these 2 reasons you cant simply integrate from -rt(3)/2 to 0 with say, the bottom half, as you are not takin into account that little bit of area to the left of -rt(3)/2, as well as the fact that theres 2 equations of the circle to consider.
So i had no clue.
Post by: junaman on November 11, 2008, 11:55:29 pm
INtegrated how? you couldnt use the "top minus bottom" method because the circle went went outside the point of intersection.
And i dunno how to use calc to find 'top minus bottom' areas so..
The circle went outside the point of intersection? The circle _was_ the point of intersection? What you talkin' bout?
You integrate by finding the equation of the circle in terms of x (top half of the circle), which was rt(1-x^2) +1
And you integrated between -rt(3)/2 and 0 with the circle minus the line. The calculator does it for you if you subtract Y2 from Y1...
Then to that you add the pi/2 which was the semi circle...
Yeah i realise the point of intersection is where you integrate from, but part of the circle stuck out to the left of that particular point. Additionally, you are dealing with 2 different equations of the circle, where the top semi circle has a different equation to the bottom. For these 2 reasons you cant simply integrate from -rt(3)/2 to 0 with say, the bottom half, as you are not takin into account that little bit of area to the left of -rt(3)/2, as well as the fact that theres 2 equations of the circle to consider.
So i had no clue.
You only needed the top half of the semi circle as that was the area under it. So you needed the positive sqrt, which was easily found by transposing to find y in terms of x. Then just integrate from -rt(3)/2 to 0. The circle is the circle you used to find the point of intersection of the line, so there can't be a bit "stuck out to the left of that particular point". The circle that was already drawn for you as part of the complex field had nothing to do with it.
Post by: eza on November 12, 2008, 12:28:23 am
Was it required for the max/min points to be labelled? I have a feeling I only chucked in the axis intercept points.
Q1)c - another graph drawing question, specifically asks for these to be labelled. However they both were worth 2 marks :( lol
Post by: Synesthetic on November 12, 2008, 12:34:19 am
With Q3)b - drawing the path of the train.
Was it required for the max/min points to be labelled? I have a feeling I only chucked in the axis intercept points.
Q1)c - another graph drawing question, specifically asks for these to be labelled. However they both were worth 2 marks :( lol
I'm quite sure 1)c) asked for the two inflection point coordinates, as well as the maximum turning point. You would probably need to label them for both marks.
I think the 'figure eight' sketch [3)b] only required the shape of the curve, though. But because it asked for the 'path' I think arrows on the curve, representing the direction of motion, might be necessary for the second of the two marks. However they may not care...
Post by: eza on November 12, 2008, 12:50:40 am
Post by: Nelle91 on November 12, 2008, 09:48:55 am
Then take area of circle which is pie - .5236, which gives 2.62
Post by: xers on November 12, 2008, 09:52:51 am
Post by: Nelle91 on November 12, 2008, 09:53:34 am
Post by: Synesthetic on November 12, 2008, 10:15:57 am
http://vcenotes.com/forum/index.php/topic,7511.msg93089.html#msg93089
Or wait for itute :P
Post by: xers on November 12, 2008, 10:17:47 am
Post by: Synesthetic on November 12, 2008, 10:24:51 am
Also, for the train question, i got 3pi seconds which made the answers that follow wrong too. You think i would get consequential mark if working out and reasoning is correct?In my opinion you will most likely get consequential marks for e)i) and ii) - perhaps 1/2 + 1/2 = 1 of the 2 available.
I'm not too sure about d) though, perhaps 1 / 2 marks again if the examiner is generous.
In short I would expect some marks to be awarded.
Post by: xers on November 12, 2008, 10:29:29 am
Post by: orsel on November 12, 2008, 02:16:20 pm
There's no such thing as consequential/half marks in maths. Only method marks.Also, for the train question, i got 3pi seconds which made the answers that follow wrong too. You think i would get consequential mark if working out and reasoning is correct?In my opinion you will most likely get consequential marks for e)i) and ii) - perhaps 1/2 + 1/2 = 1 of the 2 available.
I'm not too sure about d) though, perhaps 1 / 2 marks again if the examiner is generous.
In short I would expect some marks to be awarded.
Post by: riadnicolas on November 12, 2008, 02:50:46 pm
Yes i got 2.62 (converted from 5pi/6) too. But majority said 2.53.i think i finished on 2.62 as well
Post by: Mao on November 12, 2008, 04:34:48 pm
I think I agree with your answers there, never bothered to write mine down.
and a note to VCAA: no benefits from using CAS my ass. I relied on it the whole way.
Post by: eza on November 12, 2008, 06:23:43 pm
Post by: jamestaR` on November 12, 2008, 07:48:09 pm
INtegrated how? you couldnt use the "top minus bottom" method because the circle went went outside the point of intersection.
And i dunno how to use calc to find 'top minus bottom' areas so..
The circle went outside the point of intersection? The circle _was_ the point of intersection? What you talkin' bout?
You integrate by finding the equation of the circle in terms of x (top half of the circle), which was rt(1-x^2) +1
And you integrated between -rt(3)/2 and 0 with the circle minus the line. The calculator does it for you if you subtract Y2 from Y1...
Then to that you add the pi/2 which was the semi circle...
Yeah i realise the point of intersection is where you integrate from, but part of the circle stuck out to the left of that particular point. Additionally, you are dealing with 2 different equations of the circle, where the top semi circle has a different equation to the bottom. For these 2 reasons you cant simply integrate from -rt(3)/2 to 0 with say, the bottom half, as you are not takin into account that little bit of area to the left of -rt(3)/2, as well as the fact that theres 2 equations of the circle to consider.
So i had no clue.
You only needed the top half of the semi circle as that was the area under it. So you needed the positive sqrt, which was easily found by transposing to find y in terms of x. Then just integrate from -rt(3)/2 to 0. The circle is the circle you used to find the point of intersection of the line, so there can't be a bit "stuck out to the left of that particular point". The circle that was already drawn for you as part of the complex field had nothing to do with it.
No, you need both as the area encompasses both equations of the circle. The area of the whole circle was obviously just pi, but the little bit enclosed by the straight line and the circle which had to be subtracted I could not do by hand. That little bit that is enclosed is not a simple integration thing, it was fucked up coz, as I said, part of the circle sticks out to the left. Do i have to draw you a picture>
Post by: Mao on November 12, 2008, 08:08:33 pm
Post by: chid on November 12, 2008, 08:45:52 pm
Post by: jamestaR` on November 12, 2008, 09:18:27 pm
I don't know if you can assume that the line going through the circle is one side of an equilateral triangle in the circle...there's only one length that line can have for it to be true so it seems unlikely...
ASeriously someone show me a proper way to do this question, assuming you have a graphics calc like i did...get fucked vcaa putting questions that advantage ppl with cas.
Post by: polky on November 12, 2008, 09:34:59 pm
I did the area of the circle thing this way (picture attached), which didn't require any integration.
Don't know if that's technically correct? Might have gotten 2.53 as a fluke.
[IMG]http://img91.imageshack.us/img91/5045/1111082203vh5.th.jpg[/img][IMG]http://img91.imageshack.us/images/thpix.gif[/img]
EDIT: Linked to a easier-to-see picture
You can calculate the side lengths of each side. The lengths of each side is
Post by: Mao on November 12, 2008, 09:58:31 pm
Post by: jamestaR` on November 12, 2008, 10:20:06 pm
As you can see whoever I was talking to before, the intersection is above the halfway point of the circle, so if you were gonna integrate you would need to consider both functions. And, the intersection is to the RIGHT* of some of the enclosed area u have to subtract from pi...so how the fuck do you use a GRAPHIX calc to do that?
The questions should be equally difficult regardless of which calc u have.
Post by: Captain on November 13, 2008, 12:22:57 am
Yeah thats a good method, but is there any other method thats actually relevant to thecourse! LIke integration...
Im pretty sure it's relevant to the course. Area of circles and triangles = pretty basic.
Post by: Mao on November 13, 2008, 07:32:19 am
Yeah thats a good method, but is there any other method thats actually relevant to thecourse! LIke integration...
As you can see whoever I was talking to before, the intersection is above the halfway point of the circle, so if you were gonna integrate you would need to consider both functions. And, the intersection is to the RIGHT* of some of the enclosed area u have to subtract from pi...so how the fuck do you use a GRAPHIX calc to do that?
The questions should be equally difficult regardless of which calc u have.
being above the half-way point actually makes it easier, you'll just work out the area on the LHS separately to the RHS
Alternatively, you can take the inverse of everything, so now you have a circle centred at (1,0) with the the intersection below the x axis, in which case
Post by: jamestaR` on November 13, 2008, 01:08:59 pm
Since the fun is over, I'd like to mention that coblin, Ahmad and I conspired to come up with reasonable-looking yet incorrect answers. Hopefully none of you were bothered too much by it. Kudos to the various people who managed to pinpoint exactly where the errors were :P
I left the graph, since that's probably the only useful thing from it.
(http://img385.imageshack.us/img385/6050/partbmh9.png)
Even your graph is wrong. Look at itute its a figure 8 shape, theres no turning points at the origin...unless thats meant to be a figure 8.
And ye I get how to do that cirlce one now..I was just obsessed with finding that enclosed area and minusing from pi
Post by: chid on November 13, 2008, 01:36:00 pm
For the train question, i wrote the parametric equation as y^2=... instead of y=... (itute did this too so i assume its acceptable)-after exam people were saying that cartesian equation strictly mean y=...though :|
I also got 3pi as the time for one circuit instead of 6pi. I then used this as the value of the upper limit of the integral. Would this generally get a consequential mark (since i've already been penalised for making the mistake). Thing is it's only one mark-so i assume it's an answer mark? Then obviously the next 1 mark question involved evaluating this integral which would again provide the wrong answer!
It seems really harsh to lose 2 or 3 marks for a small oversight, but the marking scheme allows this to hapen.
Does anyone have any thoughts on what might happen/is in a similar situation?
Thanks guys.
Post by: Mao on November 13, 2008, 02:23:46 pm