VCE Methods Question Thread!

[#1procrastinator]

1) If there's a 0.8 probability that person A watches the game, 0.76 chance person B watches the game and 0.72 chance that they both watch the game, how do you determine the probability that they both do not watch the game?
Are the events not independent? I thought I should multiply the probability of each not watching the game but that was wrong. How would we determine it without using a Karnaugh map (this is very unintuitive)

2) Following on from the first question, if a person C has a 0.6 probability of watching the game and that it's independent of whether or not A and B watch the game, what is the probability of all of them watching the game? I tried 0.6*0.72 which was wrong

3) How would you find the probability of two of each of the above people watching the game (so 2 A's, 2 B's, 2 C's)?

4) A probability density function is given by for (that should be an or equal to inequality).
a) Find k in terms of a
b) Find the value of a which gives a standard deviation of 2

I've done (a) but not sure what's the fastest way to do (b). At the moment, the only way I know how is to find E(X) and E(X^2) via integration and then doing E(X^2) - [E(X)]^2 but that seems like a lot of work - is there another way?

[BubbleWrapMan]

1)

Hence

They are not independent because the product of the probabilities is not equal to their intersection.

2) I think it's 0.84*0.6

3) Don't you just square the individual probabilities? What's the answer? I'm not thinking hard enough lol

4) What did you get for E(X)?

[#1procrastinator]

2) That's what I thought, but the answer is 0.6*0.8*0.72 = 0.3456

3) The final answer is 0.36

4) I got 2k...may be wrong though...

[~T]

I'm doing Methods 1/2 but it gets a little tedious so I go looking for other stuff. So I could be missing something fundamental, haha.
Anyways, I found this integral, and thought I'd solve it.

∫√[6x - x^2 - 5] dx

So I figured just to rearrange it and do trig substitution.

Complete the square...
= ∫√[4 - (x-3)^2] dx
Then take out the 4...
= ∫ 2 √[1 - ((x-3)^2)/4] dx

And...
Let (x-3)/2 = cosθ
And using pythag identity:
sinθ = √[1- (cosθ)^2] = √[1 - ((x-3)^2)/4]

So put that into the integral:
= ∫ 2 sinθ dx

Then to find dx in terms of dθ, use the same definition for cosθ back up there and solve for x:
(x-3)/2 = cosθ
x-3 = 2cosθ
x = 2cosθ + 3
Then take the derivative:
dx/dθ = -2sinθ
dx = -2sinθ dθ

So we can substitute that back into the integral:
= ∫ 2 sinθ dx
= ∫ 2 sinθ * -2sinθ dθ
= ∫ -4(sinθ)^2 dθ

Then I got stuck for a while, but worked out I could use the rule for cos(2θ):
cos(2θ) = 1 - 2(sinθ)^2
Rearranging gives...
-2(sinθ)^2 = cos(2θ) - 1
Multiplying by two gives...
-4(sinθ)^2 = 2cos(2θ) - 2

And we can substitute that back into the integral:
= ∫ -4(sinθ)^2 dθ
= ∫ [2cos(2θ) - 2] dθ

And we can finally solve 
= sin(2θ) - 2θ + C

And we just need to get it in terms of x.
So, using the definition of sin(2θ) = 2sin(θ)cos(θ) we get:
= 2sin(θ)cos(θ) - 2θ + C

Then, the first thing we did was show that the integral was ∫ 2 sinθ dx, so:
2 sinθ = √[6x - x^2 - 5]
And we first let:
cosθ = (x-3)/2
And we can rearrange that for θ:
θ = arccos[(x-3)/2]

And now just substitute all three of those things into our result:
= 2sin(θ)cos(θ) - 2θ + C
= √[6x - x^2 - 5] * (x-3)/2 - 2arccos[(x-3)/2] + C

[~T]

YAY! Not quite.

Then I checked the integral in the calculator and got a slightly different answer. In fact, very similar, just instead of - 2arccos(blah) it came out as +2arcsin(blah):

= √[6x - x^2 - 5] * (x-3)/2 + 2arcsin[(x-3)/2] + C


So I checked both by subbing values in, and the results were different. So I figured I got it wrong. But after madly trying to work out what happened, I decided to take the derivative of my answer and the calculator's answer, and see if it got what was under the original integral. Pretty much, both results gave the same - correct - answers when subbing values in... 

So I have three questions:
1. Is my answer and working correct?
2. If so, how can it give different results when values are subbed in?
3. Even more peculiarly, how can they both come out with the same answer when the derivative was taken?

My head hurts.

[Bhootnike]

I'm just gonna say, doing a trig substitution i heard was a rare thing in spesh! ... in methods.... even rarer!

[luffy]

YAY! Not quite.

Then I checked the integral in the calculator and got a slightly different answer. In fact, very similar, just instead of - 2arccos(blah) it came out as +2arcsin(blah):

= √[6x - x^2 - 5] * (x-3)/2 + 2arcsin[(x-3)/2] + C


So I checked both by subbing values in, and the results were different. So I figured I got it wrong. But after madly trying to work out what happened, I decided to take the derivative of my answer and the calculator's answer, and see if it got what was under the original integral. Pretty much, both results gave the same - correct - answers when subbing values in... 

So I have three questions:
1. Is my answer and working correct?
2. If so, how can it give different results when values are subbed in?
3. Even more peculiarly, how can they both come out with the same answer when the derivative was taken?

My head hurts.

What do you notice about the graph of y = arcsinx and y = arccosx? If the range is unrestricted, they are simply vertical translations of each other (similar to how y = sin(x) and y = cos(x) are horizontal translations of each other). Now, remember that when you differentiate functions that are simply vertical translations of each other, you get the same equation.

The reason for your different answers is probably because when u integrate and you get the "+C", this accounts for all vertical translations (I haven't actually tried to do the question myself though). Therefore, your answer and your 'calculator's answer' are both correct.

It similar to how y = x^2 and y = x^2 + 2 yield the same derivative. But, if you subbed in x = 2, they have different y-values. Its because when you integrate y = 2x, the '+C' accounts for the vertical translation.

If you're wondering why y = arccos(x) and y = arcsin(x) have different derivatives (i.e. negatives of each other), its merely because of the range restrictions.

Hope I helped.

[Jenny_2108]

YAY! Not quite.

Then I checked the integral in the calculator and got a slightly different answer. In fact, very similar, just instead of - 2arccos(blah) it came out as +2arcsin(blah):

= √[6x - x^2 - 5] * (x-3)/2 + 2arcsin[(x-3)/2] + C


So I checked both by subbing values in, and the results were different. So I figured I got it wrong. But after madly trying to work out what happened, I decided to take the derivative of my answer and the calculator's answer, and see if it got what was under the original integral. Pretty much, both results gave the same - correct - answers when subbing values in... 

So I have three questions:
1. Is my answer and working correct?
2. If so, how can it give different results when values are subbed in?
3. Even more peculiarly, how can they both come out with the same answer when the derivative was taken?

My head hurts.

Your ans is correct 

Actually luffy already explained the reason is due to "c". I just think 2 other ways

1. If you see the formula of derivative inverse function, you will realise derivative of arcsin(x)= -arccos(x)
In your situtation, the CAS gives ans 2arcsin[(x-3)/2] which is the same as your ans: -2arccos[(x-3)/2] 

2. Other way to look at is:
You let (x-3)/2 = cosθ

So now if I let (x-3)/2 = sinθ
I just copy your post and change a bit: 

So put that into the integral:
= ∫ 2 cosθ dx

Then to find dx in terms of dθ, use the same definition for cosθ back up there and solve for x:
(x-3)/2 = sinθ
x-3 = 2sinθ
x = 2sinθ + 3
Then take the derivative:
dx/dθ = 2cosθ
dx = 2cosθ dθ

So we can substitute that back into the integral:
= ∫ 2 cosθ dx
= ∫ 2 cosθ * 2cosθ dθ
= ∫ 4(cosθ)^2 dθ
= ∫ [2cos(2θ) +2] dθ

And we can finally solve
= sin(2θ) + 2θ + C
= 2sin(θ)cos(θ) + 2θ + C

Then, the first thing we did was show that the integral was ∫ 2 sinθ dx, so:
2 sinθ = √[6x - x^2 - 5]
And we first let:
sinθ = (x-3)/2
And we can rearrange that for θ:
θ = arcsin[(x-3)/2]

And now just substitute all three of those things into our result:
= 2sin(θ)cos(θ) + 2θ + C
= √[6x - x^2 - 5] * (x-3)/2 + 2arcsin[(x-3)/2] + C

Thus, the difference is you let [(x-3)/2]= cosθ or sinθ but both are correct ans 

[~T]

What do you notice about the graph of y = arcsinx and y = arccosx? If the range is unrestricted, they are simply vertical translations of each other (similar to how y = sin(x) and y = cos(x) are horizontal translations of each other). Now, remember that when you differentiate functions that are simply vertical translations of each other, you get the same equation.

The reason for your different answers is probably because when u integrate and you get the "+C", this accounts for all vertical translations (I haven't actually tried to do the question myself though). Therefore, your answer and your 'calculator's answer' are both correct.

It similar to how y = x^2 and y = x^2 + 2 yield the same derivative. But, if you subbed in x = 2, they have different y-values. Its because when you integrate y = 2x, the '+C' accounts for the vertical translation.

If you're wondering why y = arccos(x) and y = arcsin(x) have different derivatives (i.e. negatives of each other), its merely because of the range restrictions.

Hope I helped.

Thank you so much, that makes a lot of sense. I was thinking it may have been to do with the domain on the original function: x∈[1, 5] that maybe both would be the same within the interval, which is why I checked values, but didn't think about the whole vertical translation thing.

So pretty much, if I understood everything you said, the '+C' in each equation would be different, but such that they would satisfy the difference in answer?

[~T]

OH! Even better!
The difference in results would be somewhere about this old figure:
 3.141592653589793238



Is that logical?

[~T]

And thank you Ennjy as well, that makes a lot of sense Nice demonstration of how it works out if you go the other way, too.

[Jenny_2108]

OH! Even better!
The difference in results would be somewhere about this old figure:
 3.141592653589793238



Is that logical?

Yep, its logical. C is just a number so it doesn't matter what C is

[WhoTookMyUsername]

I don't think you'll ever be expected to make a trig substitution like that unless there's something in the question prompting you to D:

[Lasercookie]

I don't think you'll ever be expected to make a trig substitution like that unless there's something in the question prompting you to D:

I think he's just exploring trig substitutions for fun.

[~T]

I don't think you'll ever be expected to make a trig substitution like that unless there's something in the question prompting you to D:

I think he's just exploring trig substitutions for fun.

Bingo.
I'd seen them before, then when I saw this question I figured it applied because of the radical. Looked pretty fun to be honest, and it turned out so