MAY13 Maths HL TZ2 Paper 2 Discussion

[Zuler]

@Zuler: Did that as well, but I think they asked for a value for d0/dt at a certain value of t (which I forgot) and I got a really small answer (like something to the power of 10^-4) so I'm not sure if I did it right

As for the binomial question, yeah I hope examiners will realize the ambiguity and credit both.

Ahhh looking at my calculator, I got -3.88 * 10^-5 rad s-1, which is small indeed O_O Who knows, hopefully we're right

[Garmr]

From reading all your replies, I already knew I got that total income question wrong

I guess the downside of discussing these papers is that your expected mark is at risk of decreasing over time >_>

[race2013]

For the binomial one, I simply did P (X=1) + P(X=3) with n=16 and p=0.25, why would it be anything else?

Also for the volume, wasn't it simply pi* S f(x)^2 - g(x)^2 dx, where g(x) = 1? So my integral was:

pi * 0->pi/2 S (sin2x + 3) - 1^2 dx (the square over f(x) cancelled the root)

= pi * 0-> pi/2 S (sin2x +2) dx = 13.0 units^3

I'm kinda confused. For the binomial one, I did what u did but I also subtracted P(X=1)* P(X=3) after adding the two.

I see your point. But what I dont get is that if they were mutually exclusive, then P(A intersect B) would have been 0 but when you multiplied the individual P(A) and P (B), the value was not 0.

[Christina perera]

@Zuler: Did that as well, but I think they asked for a value for d0/dt at a certain value of t (which I forgot) and I got a really small answer (like something to the power of 10^-4) so I'm not sure if I did it right

As for the binomial question, yeah I hope examiners will realize the ambiguity and credit both.

Ahhh looking at my calculator, I got -3.88 * 10^-5 rad s-1, which is small indeed O_O Who knows, hopefully we're right

Even I got the value as 2.26 * 10^-4. Pretty sure I am right. Had to substitute 10 in the equation and multiply by 0.5

[Garmr]

@Zuler: Did that as well, but I think they asked for a value for d0/dt at a certain value of t (which I forgot) and I got a really small answer (like something to the power of 10^-4) so I'm not sure if I did it right

As for the binomial question, yeah I hope examiners will realize the ambiguity and credit both.

Ahhh looking at my calculator, I got -3.88 * 10^-5 rad s-1, which is small indeed O_O Who knows, hopefully we're right

Even I got the value as 2.26 * 10^-4. Pretty sure I am right. Had to substitute 10 in the equation and multiply by 0.5

Alright then got the same answer. I worked in degrees throughout though since they weren't being specific, although in retrospect radians makes more sense since the value is that small. Hope they don't mark down for the unit lol

[Giveme45]

Im assuming we're allowed to discuss, when the moderator allows this thread.

Let's discuss paper 2.

I found the paper quite nice tbh, I did really well even though some questions were difficult. I might even score a full 120 if I'm lucky!

Though, I have two confusions, which bother me quite a lot.

For the question 10, finding the velocity, it was given that acceleration was (60-v)/40, so integrating both sides in respect to t gave the expression:

-ln|60-v|=t/40+c

Now we knew that initially the car was at rest so at t=0, v=0 giving c=-ln60.

We have: -ln|60-v|=t/40-ln60

Now at t=30 we have

-ln|60-v|=3/4-ln60

giving

|60-v|=e^(ln60-3/4)

This gave two results v=31.7 and v=88.3. The two results are there because of absolute value sign. I had no idea which one of the results too chose from the information given, and i assumed that the question was bad and they forgot about the absolute value sign so i went with 31.7 Can someone explain?

For question 12 I think it was, we had something like ydy/dx=cos2x or something like that and when we had to find the general function in form of y=f(x), i was quite baffled, because integrating gave: y²=sin2x+c, so i didnt know how to decide whether the function should be y=sqrt(sin2x+c) or y=-sqrt(sin2x+c). This is quite annoying me, and once again it seemed like not information was given?

I hope I won't lose a lot of marks because of these.

Shouldn't the integration of cos2x give (1/2)sin(2x) + c ? And I think IB wanted us to just leave it in the square root form, because when you think about it, y = underroot x has both the positive and negative values in it's range. So when you write sqrt(1/2 sin2x + C), you're not limiting your range, unless you went and specifically wrote (y>=0) which I'm assuming you didn't. But that's simply my opinion, I could be wrong.

Yes thats the integration cos2x, and that ultimateley gives y²=sin2x+c since the integration of y is also y²/2. What do u mean leave it just in the square root form?

y=sqrt(sin2x+c) and y=-sqrt(sin2x+c) are two different equations. But IB stated they only wanted one of them since they wanted us to give the general form y=f(x).

And I'm confused about how we were supposed to know which one of these they wanted.

What im trying to say is that both the negative and positive values of the function since when you write y = underroot x, both the positive values of negative values of y are included in the range. So in y = underroot 4, you will have both 2 and -2 as the answers

Well no, y=underroot 4 would only give the answer y=2. So yeah there is a difference between the two expressions.

@Christina: I get what you mean about the binomial, I guess we'll have to see Also for the volume, I remember the question said something along the lines of "The region R under the x-axis and above the line y=1", so isn't that simply two separate functions?

@Garmr: For the last one, I used chain rule so do/dt = do/dx * dx/dt = (the big fraction function) * 0.5?

@Rohan: Ahh right right I didn't remember that part, I hope they realise people found it ambiguous and award credit for both!

Also, I did laugh a bit when I got 47^2 - 14^2 = 2013 as the answer in the first part of that section B question, I bet the examiners did that on purpose

Also, for the weekly income, I got $165 and not $137, and here's what I did:

They said it was given that 3 requests happened in 2 days (mean =2.4), which had a certain probability (can't remember the exact value). Let's call this x.

Then, this could have happened in 4 ways: either 3 requests on one day then 0 the next (which would earn an income of 120), 0 on one day then 3 ($120), 1 on first day then 2 the next ($180) and 2 then 1 ($180)

Using P(A|B) = P(AUB)/P(B)

P (3 requests then 0| 3 requests) = the value we found in a) / x. I got this value as 0.125

P(0 then 3 | 3 total requests) = the value from a) / x

P(1 then 2| 3 total) = P(X=1) * P(X=2) (mean of 1.4)/x I got this value as 0.375

P(2 then 1| 3 total) = the same as above

Then, I did 2*0.125*$120 + 2*0.375*180 = $165. Idk if I'm right, but that's how I went about it

Don't worry, you're right on this. I can't believe a simple arithmetic mistake or even mistype on my calculator costed me my 120/120.... man.. I did exactly the same and got a different result, probably losing 1 mark. blunder.

[Giveme45]

Can anyone else elaborate on question 12 when you integrated the differential equation getting

y²=sin2x+k

did you choose

y=sqrt(sin2x+k)

or y=-sqrt(sin2x+k)

and why?

[Christina perera]

Can anyone else elaborate on question 12 when you integrated the differential equation getting

y²=sin2x+k

did you choose

y=sqrt(sin2x+k)

or y=-sqrt(sin2x+k)

and why?

y equals +- sqrt of the thing. What did you do for the 1 experiement...took n=1 or 16?

[Giveme45]

Can anyone else elaborate on question 12 when you integrated the differential equation getting

y²=sin2x+k

did you choose

y=sqrt(sin2x+k)

or y=-sqrt(sin2x+k)

and why?

y equals +- sqrt of the thing. What did you do for the 1 experiement...took n=1 or 16?

Yeah but they only wanted one specific expression in form of y=f(x) and y=+-thing is not a function... so that's weird.

As far as i remember we were given mean = 4 variance = 3

So np=4, np(1-p)=3

1-p=3/4

p=1/4

n=16.

Giving

X~B(16,1/4).

P(X=3)+P(X=1)=binompdf(16,0.25,3)+binompdf(16,0.25,1).

I think this is the only correct answer, since it wouldnt make sense with n=1, as X=3 would not be possible.

[Zuler]

Can anyone else elaborate on question 12 when you integrated the differential equation getting

y²=sin2x+k

did you choose

y=sqrt(sin2x+k)

or y=-sqrt(sin2x+k)

and why?

I chose plus, and I scribbled a note on the side saying "To be a function, it cannot be +-: it must be either +ve or -ve". So I think that should be fine Also on the income question, it's quite possible I made the calculator mistake and not you Just check if you want, the mean was 1.4 for one day and 2.8 for two.

[Giveme45]

Can anyone else elaborate on question 12 when you integrated the differential equation getting

y²=sin2x+k

did you choose

y=sqrt(sin2x+k)

or y=-sqrt(sin2x+k)

and why?

I chose plus, and I scribbled a note on the side saying "To be a function, it cannot be +-: it must be either +ve or -ve". So I think that should be fine Also on the income question, it's quite possible I made the calculator mistake and not you Just check if you want, the mean was 1.4 for one day and 2.8 for two.

I shoudlve probably scrillbed that note as well haha. Wasn'T the mean 1.2 for one day and 2.4 for two? Also I checked, and be reassured, its 165 I made a miistake and for the probability space i added the probabilities of the different outcomes, but forgot to double them. Jesus, i so wanted 120

[Zuler]

Can anyone else elaborate on question 12 when you integrated the differential equation getting

y²=sin2x+k

did you choose

y=sqrt(sin2x+k)

or y=-sqrt(sin2x+k)

and why?

I chose plus, and I scribbled a note on the side saying "To be a function, it cannot be +-: it must be either +ve or -ve". So I think that should be fine Also on the income question, it's quite possible I made the calculator mistake and not you Just check if you want, the mean was 1.4 for one day and 2.8 for two.

I shoudlve probably scrillbed that note as well haha. Wasn'T the mean 1.2 for one day and 2.4 for two? Also I checked, and be reassured, its 165 I made a miistake and for the probability space i added the probabilities of the different outcomes, but forgot to double them. Jesus, i so wanted 120

Yeah you're right about the means, I still have it in my calculator Btw for the range of cats, was it simply = mean + s.d. - (mean - s.d.) = 2 s.d.s? I think it was like 14 cats or something, I can't remember the standard deviation exactly. Though I have a feeling you have to find the range some other way

[Giveme45]

Can anyone else elaborate on question 12 when you integrated the differential equation getting

y²=sin2x+k

did you choose

y=sqrt(sin2x+k)

or y=-sqrt(sin2x+k)

and why?

I chose plus, and I scribbled a note on the side saying "To be a function, it cannot be +-: it must be either +ve or -ve". So I think that should be fine Also on the income question, it's quite possible I made the calculator mistake and not you Just check if you want, the mean was 1.4 for one day and 2.8 for two.

I shoudlve probably scrillbed that note as well haha. Wasn'T the mean 1.2 for one day and 2.4 for two? Also I checked, and be reassured, its 165 I made a miistake and for the probability space i added the probabilities of the different outcomes, but forgot to double them. Jesus, i so wanted 120

Yeah you're right about the means, I still have it in my calculator Btw for the range of cats, was it simply = mean + s.d. - (mean - s.d.) = 2 s.d.s? I think it was like 14 cats or something, I can't remember the standard deviation exactly. Though I have a feeling you have to find the range some other way

Haha initially when I saw this question I though we had to know that one standard deviation of the mean was 34.12% of the data space (i even knew this value but wasnt sure).

Then i realised that its simply what you did, so dont worry . I forgot the number but i remember that for the second part of the question i got 28 cats.

[Zuler]

Sweet! I can't remember what the second part asked but I think it was something standard/basic so no problems there. Also, I noticed you were asking about the velocity question, maybe I can explain why the only value is indeed 30.8 or whatever it was.

So, let's take an example of the differential equation dy/dx = y. Let's say when x =0, y =3.

dy/y = 1dx

ln |y| = x + c

|y| = e^(x+c)

y = +-e^c * e^x

y = Ae^x (where A = +-e^c)

So substituting our points,

3 = A(1)
A=3

So y = 3e^x

The point I'm trying to make is, the +- gets 'absorbed' by the constant, and the initial values given will determine whether the function is the + variant or the - variant. If I said when x=0, y = -3, our function would be -3e^x. Our initial condition leads to one function as desired.

In the case of our velocity function, the values made our equation the positive variant, so indeed we're all right and there's only one answer. Hope that makes sense

Edited May 11, 2013 by Zuler

[JohnSmith]

Sweet! I can't remember what the second part asked but I think it was something standard/basic so no problems there. Also, I noticed you were asking about the velocity question, maybe I can explain why the only value is indeed 30.8 or whatever it was.

So, let's take an example of the differential equation dy/dx = y. Let's say when x =0, y =3.

dy/y = 1dx

ln |y| = x + c

|y| = e^(x+c)

y = +-e^c * e^x

y = Ae^x (where A = +-e^c)

So substituting our points,

3 = A(1)

A=3

So y = 3e^x

The point I'm trying to make is, the +- gets 'absorbed' by the constant, and the initial values given will determine whether the function is the + variant or the - variant. If I said when x=0, y = -3, our function would be -3e^x. Our initial condition leads to one function as desired.

In the case of our velocity function, the values made our equation the positive variant, so indeed we're all right and there's only one answer. Hope that makes sense

I don't think you're right because here the problem was

y^2 = sin2x + c

so y= +- sqrt(sin2x + c)

but no matter what your value for c, it wont change the sign in front of the sqrt... I agree with your previous example, but multiplication and addition do not work in the same way. besides c has to be smaller than sin2x here. So both functions are still valid. This question had to be more clear, they'll definitely award marks for whatever you wrote so I souldn't worry. The IB makes mistakes quite often.

I just continued the question with the positive one since it didn't make much sense to continue with both.

[Giveme45]

Sweet! I can't remember what the second part asked but I think it was something standard/basic so no problems there. Also, I noticed you were asking about the velocity question, maybe I can explain why the only value is indeed 30.8 or whatever it was.

So, let's take an example of the differential equation dy/dx = y. Let's say when x =0, y =3.

dy/y = 1dx

ln |y| = x + c

|y| = e^(x+c)

y = +-e^c * e^x

y = Ae^x (where A = +-e^c)

So substituting our points,

3 = A(1)

A=3

So y = 3e^x

The point I'm trying to make is, the +- gets 'absorbed' by the constant, and the initial values given will determine whether the function is the + variant or the - variant. If I said when x=0, y = -3, our function would be -3e^x. Our initial condition leads to one function as desired.

In the case of our velocity function, the values made our equation the positive variant, so indeed we're all right and there's only one answer. Hope that makes sense

Hey, thanks for trying to explain it, i really appreciate you taking your time. It makes quite some sense, but still seems kind of vague, since I still didn't really understand how my method gave me two different results and how to decide between them ^^ I spent some serious in depth thoughts about it and if you're interested then have a read:

The differential equation is called "autonomous", because it does not explicitly contain the independent variable t (time).  Theinstantaneous time rate of change of v, acceleration, depends only upon the value of v.  If we have two objects moving according to this law of motion, one having velocity 0 at t=0, and the second having v=0 at t=10, then whatever the velocity of the first object is at time t, the second object will have the same velocity at  time t+10. The time-velocity graph of the second object will just be the graph of the first object shifted 10 units to the right (i.e. 10 seconds later). Note that v=60 is an "equilibrium solution".  If v = 0, then the acceleration is 0, so the object will neither speed up nor slow down, so it will continue moving at exactly velocity v.  If v < 60, then dv/dt > 0 so the object will speed up, and if dv/dt > 0, it will slow down, so in both situations, the velocity will move toward the equilibrium velocity, v = 60.  This is often represented by a phase space (or for a single dependent variable, a phase line):      |     |     v   v > 60     |     |     +----------- v = 60     |     |          ^   v < 60     | which shows that if v < 60, then the DE says that velocity will increase toward 60, while if v > 60, velocity will decrease toward 60. In solving the DE, when you reach     ln |60-v| = -(t/40) - c and exponentiate to get    |60-v| = e^((-t/40)-c)          = e^(-t/40) * e^(-c)          = C*e^(-t/40)          (where C = e^(-c)) The left side is either 60-v, if v < 60, or it is v-60, if v > 60, so we can write    60-v = C1 * e^(-t/40)     where C1 = -C if v > 60, and C1 = C if v < 60.  In this form, setting t=40 gives C1 = 60, and    60 - v = 60*e^(-t/40) ,   or  v = 60(1-e^(-t/40))  . The last formula shows that v < 60 for all t, so 31.7 must be the answer. We don't ignore or absorb the absolute value, we have to use what the DE tells you about the problem, which is this caseis the qualitative behavior of the object we see in the phase plane.

Edited May 11, 2013 by Giveme45

[hrach]

Guys, About the Probabilty problem with binomial distribution, I got n=16 p= 0,25, and P(X=1)+p(X=3)=0,261 if i'm not mistaken.

Just want your opinion about this, who got the same answer?

Edited May 11, 2013 by hrach

[kinguncaged1]

^ I did too!

[IB-Adam]

^ I did too!

Me as well.

[Garmr]

Guys, About the Probabilty problem with binomial distribution, I got n=16 p= 0,25, and P(X=1)+p(X=3)=0,261 if i'm not mistaken.

Just want your opinion about this, who got the same answer?

I did too I was worried it was some kind of trick question, to be honest.