VCE Specialist 3/4 Question Thread!

[zvezda]

I got the same answer as you (either by using a program on cas calculator or by hand).
I think this question has the same procedure as part b, just changing the upper limit.
This is how I solved it:
Find equation of line AB which has m=H/(b-a)
So the equation is: y=Hx/(b-a) - Ha/(b-a)
Rearranging the equation gives x=(y+Ha/(b-a))*(b-a)/H (1)
Now we need the upper limit: you draw the side view of the bowl, it should look like a trapezium with the larger base on top, now draw two vertical lines starting from (b,H) going down and another one from (a,0) going up. Now you pick an arbitrary value of y, and use similar triangles: y/H=(r-a)/(b-a) => y=H*(r-a)/(b-a) (2)
Now just put this on your calculator: pi*integration of x^2 (given by (1)) dy from 0 to y=H*(r-a)/(b-a) (given by (2))
Sorry if my explanation is hard to understand, I don't have access to a computer atm.

Ahhh yeah i see. Thanks heaps

[zvezda]

Hey,
With q2ci in exam 2 of last yr, im not following the solutions given.
What i have is n=24k +/- 6, yet the solutions have 12k+6???
Help is much appreciated

[brightsky]

z1 = cis(pi/12) from previous section
so we require:
Re(cis(n*pi/12)) = 0
cos(n*pi/12) = 0
n*pi/12 = (2k+1)pi/2 (by considering the graph of cos(x))
n = 6(2k+1), which coincides with solution key

what was your working?

[lzxnl]

Hey,
With q2ci in exam 2 of last yr, im not following the solutions given.
What i have is n=24k +/- 6, yet the solutions have 12k+6???
Help is much appreciated

I'm not convinced that there's a difference. If we have n = 24k +- 6, then the first few solutions are (n>0) 6, 18, 30, 42 etc. This coincides perfectly with 12k + 6.

[zvezda]

z1 = cis(pi/12) from previous section
so we require:
Re(cis(n*pi/12)) = 0
cos(n*pi/12) = 0
n*pi/12 = (2k+1)pi/2 (by considering the graph of cos(x))
n = 6(2k+1), which coincides with solution key

what was your working?

Ah ok. Yeah i see that. What i did though was i used the 2npi +/- cos^-1.
Thanks. Appreciate it

[zvezda]

I'm not convinced that there's a difference. If we have n = 24k +- 6, then the first few solutions are (n>0) 6, 18, 30, 42 etc. This coincides perfectly with 12k + 6.

So technically, i should be receiving the full marks for that q shouldnt i?

[lzxnl]

Yeah your answer is perfectly fine; if you haven't left out any steps or done anything stupid you'll get full marks for the question.

[jono88]

How do i express 1/x^2(2x+4) in partial fractions form?

[Alwin]

How do i express 1/x^2(2x+4) in partial fractions form?

The trick is knowing how to do the first setup equation, since x² is considered as an irreducible factor ...

[Limista]

Need some help with the following antidifferentiation:

dP/dt = (k-mP)P     where m = 0.000004    and initial condition is such that Po=12000 when t = 0

& thanks 

[Alwin]

Need some help with the following antidifferentiation:

dP/dt = (k-mP)P     where m = 0.000004    and initial condition is such that Po=12000 when t = 0

& thanks 

I'd hazard a guess that you are using the Heinemann textbook? I don't recall Modified exponential growth being covered in the Essential book. Also, you aren't given enough info to find the specific solution . . .

Anyways,



Maybe in a next part of the q, or something it gives you more conditions so you can solve for k, but that's as far as I think I can go ... hopefully without any mistakes so far! =D

[Limista]

^ thanks for that ; but how do you go from t=... to P=... by "WolframAlpha"? It's just that I got

Yeah you are right, 3 different k values were supplied in the next part of the question. They were k=0.12, k=0 and
k = - 0.13.

There is another part to this question, which I am stuck on that asks me to give another k value that would yield a different type of graph (for this same P=... equation) and explain as to why I chose this k value/why would this k value be appropriate?

[Alwin]

^ thanks for that.

Yeah you are right, 3 different k values were supplied in the next part of the question. They were k=0.12, k=0 and
k = - 0.13.

There is another part to this question, which I am stuck on that asks me to give another k value that would yield a different type of graph (for this same P=... equation) and explain as to why I chose this k value/why would this k value be appropriate?

Hmm, I would just try plug in some random values. You have +ve, -ve and 0 covered, but they are all |k|<1.

I would guess (haven't tried sketching all of them yet) that a value of k>1 will give a different result, but no conclusive reason until I can graph them myself -W|A is being a pain and my cas is somewhere... haven't seen it in a while.

Is this q a textbook question, or.. ?

[SocialRhubarb]

Umm, this works for SugarMinted's solution.

[Limista]

Hmm, I would just try plug in some random values. You have +ve, -ve and 0 covered, but they are all |k|<1.

I would guess (haven't tried sketching all of them yet) that a value of k>1 will give a different result, but no conclusive reason until I can graph them myself -W|A is being a pain and my cas is somewhere... haven't seen it in a while.

Is this q a textbook question, or.. ?

Well, I tried a k value of 1, and then of 1.2 - what I found was that the graph shape of these two was the same as when k=0.12, however the graph was translated upward and stretched in the direction of the Yaxis a whole lot more when k=1 compared to k=0.12

However, when I made the k value = 5, my calculator was not able to show the graph ( it was taking an unreasonably long time to graph k=5).

I don't know what sort of conclusion to draw from this?

edit: practice SAC question

Umm, this works for SugarMinted's solution.

thanks for the clarification, but why did you sub. in k=6/125 for only some of the k's in the population equation and not all of them?