VCE Specialist 3/4 Question Thread!

[deStudent]

Q13d) does VCAA ever asks these types of questions?

Q13f) Really confused on their gradient tables. For the first one when x = -1/2, how do you determine if dy/dx is positive or negative because from part b - dy/dx = -3x^2 / 2y. What is the value of y or how do we determine if it's positive to negative, because subbing x = 1/2 in to the original equation gives plus and minus of the y value.

http://m.imgur.com/a/wqdpt

[lzxnl]

you probs can't solve it because there aren't any solutions 

Think about at what point must cos²x sin² x  = 1. Only when cos²x and sin²x are equal to 1. cos²x and sin² x will equal 1 only when cos(x)=1,-1 same goes with sin(x)=1,-1. However this never happens at the same time hence no solutions.

Another reason is that sin^2 x cos^2 x = (1/2 sin 2x)^2 = 1/4 sin^2 2x, which clearly isn't ever 1.

[Shadowxo]

Q13d) does VCAA ever asks these types of questions?

Q13f) Really confused on their gradient tables. For the first one when x = -1/2, how do you determine if dy/dx is positive or negative because from part b - dy/dx = -3x^2 / 2y. What is the value of y or how do we determine if it's positive to negative, because subbing x = 1/2 in to the original equation gives plus and minus of the y value.

http://m.imgur.com/a/wqdpt

d) Normally not, but you should know this anyway. VCAA wouldn't ask it in that format either
f) I don't think VCAA would ever ask a question like that
So if you rearrange the question, y=±√(1-x³+)
If you consider the first half, where y=√(1-x³+) then y>0 therefore the gradient is negative, as x²≥0, that's how they got the first table, and vice versa for the other half.

[excelsiorxlcr]

Hi please help
Find the Cartesian equation of the curve with parametric equations x = 2cos (3t) and y = 2sin(3t), and determine the domain and range of the corresponding relation.
I don't know how to deal with the 3t?

[Quantum44]

Hi please help
Find the Cartesian equation of the curve with parametric equations x = 2cos (3t) and y = 2sin(3t), and determine the domain and range of the corresponding relation.
I don't know how to deal with the 3t?

Sin2(3t) + cos2(3t) = 1 just like sin2(t) + cos2(t) = 1 so the 3t does not really change it from a basic parametric circle

[deStudent]

http://m.imgur.com/a/p5bIC
Q14a) More specifically for horizontal asymptotes, so I just found out/learnt that you can cross them.
So in general when dealing with horizontal asymptotes in spesh/this exercise, we need to simplify the function in to a form where there is a constant term by itself? This is our horizontal asymptote, right? After, to solve when the graph intersects it, you need to solve the numerator of the fraction = 0. Is this all I need to know?

Q16e) the book's answer just said the first line in blue writing. I wrote the second line trying to figure this out. I'm not sure if my working is correct, how would I correctly go about solving this question because I was totally confused?

Thanks

[Shadowxo]

http://m.imgur.com/a/p5bIC
Q14a) More specifically for horizontal asymptotes, so I just found out/learnt that you can cross them.
So in general when dealing with horizontal asymptotes in spesh/this exercise, we need to simplify the function in to a form where there is a constant term by itself? This is our horizontal asymptote, right? After, to solve when the graph intersects it, you need to solve the numerator of the fraction = 0. Is this all I need to know?

Q16e) the book's answer just said the first line in blue writing. I wrote the second line trying to figure this out. I'm not sure if my working is correct, how would I correctly go about solving this question because I was totally confused?

Thanks

14.a) The way I learnt to do it in these situations with xⁿ on top and bottom was divide all terms top and bottom by xⁿ.
That way as x-> infinity y-> (2+0+0)/(2+0+0) = 1. When the graph intersects the asymptote, horizontal asymptote = function so 1=(function) and you end up with x=1/2 as the point of intersection. The way to do it may have changed but this is the way I used and I'd probably recommend it over your method, although you can still find it using your method. Simpler this way though.
Also, the other asymptote would be denominator = 0, but in this case it's never zero, discriminant<0 so there is no asymptote there.

16. e) An asymptote can be a function. In this case the asymptote is y=√x. As x-> infinity, y->x/√x = √x. That's all you need to write, as in this case, the asymptote is that function.

[Joseph41]

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[deStudent]

 I'm having difficulties with integration by substitution. I think I understand the method of solving these but I don't actually understand what I'm doing.

For example http://m.imgur.com/a/dFVTS
For l and m) can you check my working out? I'm not sure if I set out each line correctly, since I'm not entirely sure what I'm doing in each line. I'll try explan it but I'm not sure if it'll be clear though.

My thinking:
- let some difficult function = u (preferably if it's derivative becomes = or similar to another function in the equation).
- Find du/dx
- My problems usually arise from finding f(u). I literally cannot explain what my thinking is when I find this, I just make it something simple like 1/u..
- I set up the integrand so that it resembles the original equation, so I take the reciprocal of the additional factors caused by finding du/dx and factor out other constants so we don't lose them. For example in (l), I factor out 3/2, so we still get 3 at the top and 2 cancels the 2 from du/dx

[Shadowxo]

I'm having difficulties with integration by substitution. I think I understand the method of solving these but I don't actually understand what I'm doing.

For example http://m.imgur.com/a/dFVTS
For l and m) can you check my working out? I'm not sure if I set out each line correctly, since I'm not entirely sure what I'm doing in each line. I'll try explan it but I'm not sure if it'll be clear though.

My thinking:
- let some difficult function = u (preferably if it's derivative becomes = or similar to another function in the equation).
- Find du/dx
- My problems usually arise from finding f(u). I literally cannot explain what my thinking is when I find this, I just make it something simple like 1/u..
- I set up the integrand so that it resembles the original equation, so I take the reciprocal of the additional factors caused by finding du/dx and factor out other constants so we don't lose them. For example in (l), I factor out 3/2, so we still get 3 at the top and 2 cancels the 2 from du/dx

You're mostly right, just a few comments
We use substitution so we can evaluate it more easily, finding the integral of something involving u with respect to u, instead of finding a difficult integral of something involving x with respect to x.
You use a u that's convenient, often either what's in the denominator of a fraction, or what's inside a trig function (probably not up to there in the course).
du/dx * dx = du. We find du/dx and put it in place of another part of the function to be integrated. Also, you can split the variables, so eg du/dx=2x, dx=du/2x and sub it in (less mathematically correct and I'm not sure they'd give it full marks, but it's an alternate way)
Also, the x (or u) cannot be taken out of the integral, only constants can be taken out.
I'm not entirely sure what you're trying to do with the f(u), but you just replace the part with respect to x with u - eg replace 2-x² with u wherever it appears. You want to replace everything involving an x with something involving u in order to evaluate it. I'll do a worked solution for m)


Let me know if you want any further clarification

[Gogo14]

Confused for this question. Dont understand why

[Syndicate]

Confused for this question. Dont understand why

x has a domain of [-Pi, Pi], therefore 2x has a domain of [-2Pi, 2Pi]
Turning points will occur when cos(2x) = -1, 1.

cos(2x)=1
2x = -2pi, 0, 2pi
x = -Pi, 0, Pi

cos(2x) = -1
2x = -Pi, Pi
x = -Pi/2, Pi/2

T.P.

(-Pi, 5), (-Pi/2, 3), (0, 5), (Pi/2, 3), (Pi, 5)

[Gogo14]

x has a domain of [-Pi, Pi], therefore 2x has a domain of [-2Pi, 2Pi]
Turning points will occur when cos(2x) = -1, 1.

cos(2x)=1
2x = -2pi, 0, 2pi
x = -Pi, 0, Pi

cos(2x) = -1
2x = -Pi, Pi
x = -Pi/2, Pi/2

T.P.

(-Pi, 5), (-Pi/2, 3), (0, 5), (Pi/2, 3), (Pi, 5)

But the answer is c

[Syndicate]

But the answer is c

I believe the assessor may have considered -Pi and Pi as end points, rather than turning points, however, it should be considered as both afaik. So C would be right in this case, however, I think E should be correct not C (definitely confirm this with your teacher) because the gradient of a turning point is 0, and as -Pi and Pi are included in the domain, they should be included in the answer (some assessors approve it and some don't)

[Shadowxo]

But the answer is c

Turning points are when the gradient changes sign, ie from + to -, or - to +. The domain is [-pi,pi] and as the calculated turning points include x=pi and x= -pi values, they can't be included as it ends there - the gradient doesn't change sign.
It's just a technicality/terminology issue