VCE Specialist 3/4 Question Thread!

[Shadowxo]

Thanks for that! But for the second question, the answer says that the particle (its position is given by the vector) completes one cycle in one unit of time (not 2 units) - I was wondering why this is the case? How do you know which term to go by? (Sorry, probably should've been clearer with my question) 

So the period is given by 2pi/n (previously made an error, its 2pi/n not n/2Pi)
So the period of the first term is 1, period of second term is 1/2.
From this you can see that the first term completed 1 cycle at t=1, and the second term completed 2 cycles at t=1. The period of the whole equation is essentially the time it takes to get back to where it started, which it is at when t=1
(Alternatively, the period of the first term is 2x the second so you know that it will be back at the original position for every time t is an integer)

[Lavar Big BBB Balls]

5) Recall the definition of the central limit theorem: given a large sample size, the distribution of sample means of a population is normal
Option A implies that ALL distribution are normal, provided the sample size is large enough. While this may be true in your case of binomial distributions, it does not always hold. Take, for example, the income distribution in a region of low socioeconomic class. The data would have most of the population at low income, while a few people will have high incomes. This is not normally distributed.

7) Option D is not the definition of a sample distribution of sample means. The sampling distribution of sample means is the distribution of the means of many samples taken from a population.

It might take awhile to get your head around all these terms which sound the same, but hopefully this helps

EDIT: I find that this video is helpful is explaining the central limit theorem, and also why the sampling distribution of sample means is normal, regardless of the distribution of the population.

Thanks

[Lavar Big BBB Balls]

Ugh, many questions lately 

For these Qs http://imgur.com/a/DMerM

Q1)a) I solved Pr(X<54) with X~N(60, 20^2) which gave me 0.3821, however the answer has 0.3807. Am I doing something wrong or is the answer given wrong?

Q2) For the second lot of information, how do you recognise that they give you the sample mean opposed to the sum of 7 random variables?

I initially thought we had Y = X+X+...X_7 opposed to the sample mean.

Thanks

[Shadowxo]

Ugh, many questions lately 

For these Qs http://imgur.com/a/DMerM

Q1)a) I solved Pr(X<54) with X~N(60, 20^2) which gave me 0.3821, however the answer has 0.3807. Am I doing something wrong or is the answer given wrong?

Q2) For the second lot of information, how do you recognise that they give you the sample mean opposed to the sum of 7 random variables?

I initially thought we had Y = X+X+...X_7 opposed to the sample mean.

Thanks

Your answer for 1.a) should be right assuming you typed it in right etc
2. It gives you the mean daily rainfall which is what I expect you missed.

[Lavar Big BBB Balls]

Tyyyyvmmvmvmvmvmvmv^^

Just for this question http://imgur.com/a/fhu88

Q4b)iii) why is the standard deviation = 2.4 still, even though n=70 now?

[Mattjbr2]

Hey guys,
When you're told to convert a relation to a vector function, is it valid to simply always let x=t and then solve for y then plug t next to i and the value for y (almost always a plus/minus expression) next to j? The answers in the book almost always sub in a trig expression for x or y and then the vector function has trig values for i and j while mine always has t*i +- another expression*j. I test whether my answers are the same as the book's by finding r(1), for example, and I don't end up with the same answer.
I've attached an example. Despite having the same magnitude,they are at different locations on the Cartesian plane. The book's answer has an angle of pi/4 while my answer has an angle of pi/3. ?

[Shadowxo]

Tyyyyvmmvmvmvmvmvmv^^

Just for this question http://imgur.com/a/fhu88

Q4b)iii) why is the standard deviation = 2.4 still, even though n=70 now?

Think about it - you're talking about the mean time for each headache. Whether you have a sample size of 20 or 2000 the standard deviation should be about the same, the graph will just get more accurate with more samples. If you've assumed a SD of 2.4 for both b)i) and ii) then it follows that the SD for iii)  will also be 2.4.

Hey guys,
When you're told to convert a relation to a vector function, is it valid to simply always let x=t and then solve for y then plug t next to i and the value for y (almost always a plus/minus expression) next to j? The answers in the book almost always sub in a trig expression for x or y and then the vector function has trig values for i and j while mine always has t*i +- another expression*j. I test whether my answers are the same as the book's by finding r(1), for example, and I don't end up with the same answer.
I've attached an example. Despite having the same magnitude,they are at different locations on the Cartesian plane. The book's answer has an angle of pi/4 while my answer has an angle of pi/3. ?

In fact, your answers are the equivalent.
You have to remember your t=1 doesn't equal their t=1 (although a certain value for the i component should lead to the same value for the j component in both cases).

Let's take your t=1. Your answer results in 1i ± √3j.
The t value that results in a 1 for the i component of their answer means cos(t) = 1/2, so t = π/3 or -π/3 (plus or minus 2π)
If you plug this in to 2sin(t), you end up with 2* ±√3 / 2 = ±√3 - the same as your answer!

However, their way is preferred as you can plug in any value of t. If for example there was a particle that kept on going around in a circle, t would keep increasing but it would continue in the same path, as their answer implies.

I would recommend always rewriting the equation to make y the subject first. Then, see if you can utilise any trig functions such that you can get a similar answer to them, especially if it's a function that results in a loop like circles, ellipses etc

Hope this helps 

[Mattjbr2]

You have to remember your t=1 doesn't equal their t=1 (although a certain value for the i component should lead to the same value for the j component in both cases).

Let's take your t=1. Your answer results in 1i ± √3j.
The t value that results in a 1 for the i component of their answer means cos(t) = 1/2, so t = π/3 or -π/3 (plus or minus 2π)
If you plug this in to 2sin(t), you end up with 2* ±√3 / 2 = ±√3 - the same as your answer!

However, their way is preferred as you can plug in any value of t. If for example there was a particle that kept on going around in a circle, t would keep increasing but it would continue in the same path, as their answer implies.

I would recommend always rewriting the equation to make y the subject first. Then, see if you can utilise any trig functions such that you can get a similar answer to them, especially if it's a function that results in a loop like circles, ellipses etc

Hope this helps 

But conceptually (or visually, if possible), how is it possible to arrive at multiple vector functions that are equivalent but have different values for t purely based on the method you used to get there? Shouldn't the method not matter when arriving at an answer? How can changing your method alter the speed of the particle while maintaining the path it took?
Note: before today I didn't even know that vector functions were a thing, or that it was possible to graph the motions of particles

[Shadowxo]

But conceptually (or visually, if possible), how is it possible to arrive at multiple vector functions that are equivalent but have different values for t purely based on the method you used to get there? Shouldn't the method not matter when arriving at an answer? How can changing your method alter the speed of the particle while maintaining the path it took?
Note: before today I didn't even know that vector functions were a thing, or that it was possible to graph the motions of particles

The relation they give just tells you what the y value is for a certain x value - it's independent of the speed. You could go around the circle really slowly, or really fast, and it wouldn't matter as for a certain x value the y value/s would still be the same.
t is just a variable, it doesn't really stand for anything (in this question) so you can define it any way you like. You defined it differently to how they did and hence resulted in the two solutions being different for a particular t value. However, in both solutions if you were to graph it with i and j (instead of x and y) you would see they're exactly the same, even though the t's are different

Your answers are predominantly only different in that their function keeps repeating, while your function just makes it go around in a circle once

*Note, I haven't studied relations -> parametric equations / vector functions, only the other way around. Hence I can't necessarily tell you what they want, only why you each got your answer and what they represent

[Mattjbr2]

The relation they give just tells you what the y value is for a certain x value - it's independent of the speed. You could go around the circle really slowly, or really fast, and it wouldn't matter as for a certain x value the y value/s would still be the same.
t is just a variable, it doesn't really stand for anything (in this question) so you can define it any way you like. You defined it differently to how they did and hence resulted in the two solutions being different for a particular t value. However, in both solutions if you were to graph it with i and j (instead of x and y) you would see they're exactly the same, even though the t's are different

Your answers are predominantly only different in that their function keeps repeating, while your function just makes it go around in a circle once

*Note, I haven't studied relations -> parametric equations / vector functions, only the other way around. Hence I can't necessarily tell you what they want, only why you each got your answer and what they represent

Thank you

[tinagranger]

14.
You know the force is at an angle of 75° with the downwards vertical, so that's a bit below horizontal
The length of the string is 2.5m and it is a horizontal distance of 2m from the starting point. This means it's a vertical distance of 1.5m from the top (3,4,5 triangle)
You also have the weight directly downwards of 2g N
From there you should have all the information needed - could you post your working so I can see where you went wrong? (Don't have time to do a full solution atm)
Also for Q15, I can't see it attached. What question are you referring to?

Thanks but still can't figure out where I went wrong! For Q14, could someone please post a worked solution?

Also, here is Q15 which I still don't get

[VanillaRice]

Thanks but still can't figure out where I went wrong! For Q14, could someone please post a worked solution?

Also, here is Q15 which I still don't get

Hi there,
This is how I would approach these questions. Note I have left some gaps for you to have a go for yourself 

Hope this helps

[tinagranger]

Thanks!! Could someone please tell me how to do 4a)i) and ii) in the screenshot?

And also all of the Q4 dynamics question? I just set the tractive force equal to the 2 component weight forces and the 2 resistance forces (since the tensions cancel out) for the first part, but it didn't give me the right answer

I have a sac tomorrow!

[Shadowxo]

For the first screenshot:
a)i) the police boat travels 8 units in the i direction for every 6 in the j direction. To meet the boat, their vertical (and horizontal) components must be the same, so the boat travels 600 units upwards. This means it has travelled 800 units in the i direction. Its final position is therefore (800+400) i +0j = 1200i +0j
The time taken for the motor boat to reach there is distance / velocity = 1200/6 = 200s, using the i component.
ii) Speed in the j direction = 6u = distance / time = 600/200 = 3 m/s. So u=1/2

For the second screenshot:
Did you resolve the weight forces into parallel and perpendicular to the plane? Remember, the engine only has to pull the component parallel to the plane, as the normal force cancels out the part perpendicular to the plane
These plus the resistances should equal the tractive force

Tractive force = resistive force + weight force parallel to the plane
= (50*60 + 30*12) + (12000g sin(a) + 60000g sin(a) )
= 3000 + 360 + 12000g * 1/200 + 60000g * 1/200
= 3360 + 60g + 300g
= 6888 N
Edit: Added a worked solution for tractive force

[tinagranger]

For the first screenshot:
a)i) the police boat travels 8 units in the i direction for every 6 in the j direction. To meet the boat, their vertical (and horizontal) components must be the same, so the boat travels 600 units upwards. This means it has travelled 800 units in the i direction. Its final position is therefore (800+400) i +0j = 1200i +0j
The time taken for the motor boat to reach there is distance / velocity = 1200/6 = 200s, using the i component.
ii) Speed in the j direction = 6u = distance / time = 600/200 = 3 m/s. So u=1/2

For the second screenshot:
Did you resolve the weight forces into parallel and perpendicular to the plane? Remember, the engine only has to pull the component parallel to the plane, as the normal force cancels out the part perpendicular to the plane
These plus the resistances should equal the tractive force

Tractive force = resistive force + weight force parallel to the plane
= (50*60 + 30*12) + (12000g sin(a) + 60000g sin(a) )
= 3000 + 360 + 12000g * 1/200 + 60000g * 1/200
= 3360 + 60g + 300g
= 6888 N
Edit: Added a worked solution for tractive force

Awesome, thanks for the help! If anyone is willing to help with the other parts of the dynamics question I would be forever grateful...