VCE Methods Question Thread!

[Bri MT]

Hey guys, sorry for posting so many questions. For intergers, how many rules do we have to know?

My class wasn't taught what to do when you have to integrate multiples (for example, integrate -4*x*In(x)). Do we need to know how to integrate additions/muliples/divisions? If so, what are the rules?

You do not need to do anything with integration where you have two functions multiplied or divided together, unless it is via anti-differentiation by recognition

[LifeisaConstantStruggle]

is that 2016 exam 1? If it is you have to use the expression in the previous question to do it

[Eric11267]

Hey guys, sorry for posting so many questions. For intergers, how many rules do we have to know?

My class wasn't taught what to do when you have to integrate multiples (for example, integrate -4*x*In(x)). Do we need to know how to integrate additions/muliples/divisions? If so, what are the rules?

Generally you won't have to integrate anything that are the products of two functions because thats in specialist maths. Though some may have to be simplified first and then integrated i.e (x²-1)/x
You may, however, be asked to differentiate something and then use that result to integrate something else (otherwise known as integration by recognition)

[Opengangs]

No, we haven't learned that. Thanks for letting me know! I had a try looking online and applying it to my test but I'm still confused. Can you help me out with the attached file?

So, I've come to the consensus that Methods won't ask you anything of the sort, but if you're curious, here's how to integrate a product.

Suppose: you're asked to integrate -4xlnx.

We set it up as:
f(x) = lnx, f'(x) = 1/x
g(x) = x^2/2, g'(x) = x

The formula for Integration by Parts:


But, Methods wouldn't require you to replicate it

[kiki.]

is that 2016 exam 1? If it is you have to use the expression in the previous question to do it

Thank you so much! It was a head ache trying to figure it out. I can't believe it was right there in my eyes.

Thank you too miniturtle, Erich and Opengangs

[Sine]

Remember if it says "hence" you MUST use your answers from the previous question

[ZNormal]

Hi, I was just wondering how to do this question from the atarnotes book?

Tasmania's model is a good approximation to reality. However, it fails to account for an additional component. The full equation is y(t) = Acos(b*t+c) + 0.001cos(2b/3(t+c/b)). How long does it take for the tidal cycle to repeat itself? State your answer in hours, to one decimal place.

Apparently I needed to find the common period of both functions and find b but I'm not sure how to do that.

Also what is the difficulty of the atarnotes topic tests book relative to the vcaa tests? Imo it's pretty hard but that might be because I'm dumb  , so I would like to know what you guys think.

Any help is appreciated, thank you so much in advance!

[VanillaRice]

Hi, I was just wondering how to do this question from the atarnotes book?

Tasmania's model is a good approximation to reality. However, it fails to account for an additional component. The full equation is y(t) = Acos(b*t+c) + 0.001cos(2b/3(t+c/b)). How long does it take for the tidal cycle to repeat itself? State your answer in hours, to one decimal place.

Apparently I needed to find the common period of both functions and find b but I'm not sure how to do that.

Also what is the difficulty of the atarnotes topic tests book relative to the vcaa tests? Imo it's pretty hard but that might be because I'm dumb  , so I would like to know what you guys think.

Any help is appreciated, thank you so much in advance!

This question has me a bit stumped as well 
I think the common period you are looking for is 6pi/b. The individual periods are 2pi/b and 3pi/b, and the lowest common multiple is 6pi/b, so that is when both trig functions restart a period at the same time. I'm not sure where to go with finding b though, is there any other information given?

I'm not sure about the rest of the questions, but this question looks quite difficult! In my opinion, if this was to appear on an exam, it'd likely be one of those final questions which is usually designed to be difficult.

Sorry I couldn't be of more help though, hopefully someone else has more insight!

[ZNormal]

This question has me a bit stumped as well 
I think the common period you are looking for is 6pi/b. The individual periods are 2pi/b and 3pi/b, and the lowest common multiple is 6pi/b, so that is when both trig functions restart a period at the same time. I'm not sure where to go with finding b though, is there any other information given?

I'm not sure about the rest of the questions, but this question looks quite difficult! In my opinion, if this was to appear on an exam, it'd likely be one of those final questions which is designed to be difficult.

Sorry I couldn't be of more help though, hopefully someone else has more insight!

Hey, thanks for replying

The solutions at the back of the book find the common period like you did, and then says:

Hence the overall period is 6pi/b = 6pi/0.506 = 37.3.

I actually have no idea where the value for b came from, the question is part f of a question, but I can't think of using any of the previous values because they come from a different formula but I could be wrong.

By the way are we supposed to know how to find the common period when they both restart? I've never learnt it before.

[VanillaRice]

Hey, thanks for replying

The solutions at the back of the book find the common period like you did, and then says:

Hence the overall period is 6pi/b = 6pi/0.506 = 37.3.

I actually have no idea where the value for b came from, the question is part f of a question, but I can't think of using any of the previous values because they come from a different formula but I could be wrong.

By the way are we supposed to know how to find the common period when they both restart? I've never learnt it before.

If that's the case, it's likely the value for b is buried somewhere (it might be in the original question stem, or perhaps a diagram).

With respect to solving this type of question, I don't think it's explicitly taught in Methods. But this is the way I thought about it: picture the two trig graphs separately - you'll see that the graphs will only repeat in the exact same way with respect to each other when their periods line up.

Hope this helps

[ZNormal]

If that's the case, it's likely the value for b is buried somewhere (it might be in the original question stem, or perhaps a diagram).

With respect to solving this type of question, I don't think it's explicitly taught in Methods. But this is the way I thought about it: picture the two trig graphs separately - you'll see that the graphs will only repeat in the exact same way with respect to each other when their periods line up.

Hope this helps

Thank you very much! 

[Guideme]

Can anyone help me with this question pls !


Thank in advance

[Eric11267]

Can anyone help me with this question pls !
(Image removed from quote.)

Thank in advance

E(X-1) is the same as E(X)-1 using its linear properties
In the question E(X)= 2.7, so then E(X-1)=1.7
1.7² =2.89=> A

[Shadowxo]

Can anyone help me with this question pls !
(Image removed from quote.)

Thank in advance

I also want to point out that this question is poorly written. It should be clear whether they want
E[(X-1)²] or [E(X-1)]²
If it's the former then you'd subtract one from each of the X's, square it and multiply that by the probability. (Alternatively you could expand it out and solve it that way, but it'd be longer). You'd end up with 5.5 (ie E).  If it's the latter then you'd do what Eric has done.

[VanillaRice]

Can anyone help me with this question pls !
(Image removed from quote.)

Thank in advance

E(X-1) is the same as E(X)-1 using its linear properties
In the question E(X)= 2.7, so then E(X-1)=1.7
1.7² =2.89=> A

I'm not saying that Eric's answer is wrong (100% agree with the solution), but I interpreted the question as E((X-1)²).
If that were the case,

which is also an option (E).

I think an extra set of brackets could've helped make this question less vague, but hopefully OP has the answer?

EDIT: What shadow said