VCE Specialist 3/4 Question Thread!

[lzxnl]

"The position vector of a particle at time t, is given by:
r(t) = i + j,

Find the angle between the vector r(t) at time t = 1 and the vector -i - j."

So after plotting the two vectors out, you can see that there are two possible angles to choose from (duh), but how do we know which one? It doesn't specify obtuse/reflex angle.

This was an exam 1 question and the answer was , so using dot product was a bit tough ( and don't go nicely together .. or do they?). Either way, r(1) and - i - j were nice numbers whose angles were also nice, so I thought it was a bit quicker to just plot the vectors onto an graph and add the angles visually.

Any reason why you'd take the obtuse other than the reflex?

I guess I'm just lucky in that I recognise so many potential problem cosines of weird rational multiples of pi that can be expressed in exact value form.

For cos(11pi/12), I would personally have firstly looked at the angle and work out what quadrant it was in. Generally with dot products if it's positive, you take the first quadrant; negative, second quadrant. That's a rule. And the sign of the dot product is easy to see.

I'd use the double angle formula as a guess to see if it simplifies things.
So here r(1)=sqrt3 i + j
The simpler method would be to note that r(1) makes an angle of pi/6 with the horizontal and that -i-j makes an angle of 5pi/4 with the positive horizontal so subtracting gives 13pi/12 as the reflex or 11pi/12 as the obtuse.
If I didn't see that, I'd do something like
dot product = -sqrt 3 - 1
Product of magnitudes = 2*sqrt2
So cosine angle = (-sqrt(3)-1)/(2sqrt2)

If I got that in the exam to solve though...
First reaction, double angle formula.
Let cos t = -(sqrt(3)+1)/2sqrt2
cos 2t = 2cos^2 t - 1 = (3+1+2sqrt 3)/4 - 1 = sqrt 3 / 2
So cos 2t = cos pi/6 or 11pi/6
t=pi/12 or 11pi/12. Negative dot product=> take 11pi/12

[barydos]

I guess I'm just lucky in that I recognise so many potential problem cosines of weird rational multiples of pi that can be expressed in exact value form.

For cos(11pi/12), I would personally have firstly looked at the angle and work out what quadrant it was in. Generally with dot products if it's positive, you take the first quadrant; negative, second quadrant. That's a rule. And the sign of the dot product is easy to see.

I'd use the double angle formula as a guess to see if it simplifies things.
So here r(1)=sqrt3 i + j
The simpler method would be to note that r(1) makes an angle of pi/6 with the horizontal and that -i-j makes an angle of 5pi/4 with the positive horizontal so subtracting gives 13pi/12 as the reflex or 11pi/12 as the obtuse.
If I didn't see that, I'd do something like
dot product = -sqrt 3 - 1
Product of magnitudes = 2*sqrt2
So cosine angle = (-sqrt(3)-1)/(2sqrt2)

If I got that in the exam to solve though...
First reaction, double angle formula.
Let cos t = -(sqrt(3)+1)/2sqrt2
cos 2t = 2cos^2 t - 1 = (3+1+2sqrt 3)/4 - 1 = sqrt 3 / 2
So cos 2t = cos pi/6 or 11pi/6
t=pi/12 or 11pi/12. Negative dot product=> take 11pi/12

Oh that's a cool way to look at it.

Also, I asked my teacher and she said that generally the preferred angle is the acute/obtuse rather than a reflex angle (in this case the obtuse ).

[BubbleWrapMan]

The angle between vectors is conventionally defined as the angle in the interval .

[availn]

Haha thanks for the clarification, I knew initially that he didn't write it right but I didn't want to tell him off. Again, thank you nliu and brightsky

Just to add on to this post, when trying to find a solutions of a polynomial (in this case ), is it fine to do the working out below:






From the exam answer sheet, they did it differently in that they expanded and used that to divide in order to find the answer.

Thanks!

EDIT: I just made this up while I was doing trial exam 1, but yeah thanks heaps nliu , appreciate it alot

Could someone explain the "By equating each of the respective coefficients to find c" to me? I have no idea what happened there.

Firstly, ask yourself, which direction are the masses going to accelerate in? Since they're both hanging vertically (and there is no external torque on the peg), the heavier mass will accelerate downwards and the lighter mass will accelerate upwards. So if we look at the 3kg mass we have the weight force acting downwards, and the tension force acting upwards. The sum of these  (taking into account direction) will give the net force, i.e. . Since the object is accelerating downwards, the weight force is larger, so for the 3 kg mass we will take down as positive. So we have


Now for the 1 kg mass, we have the weight force acting downwards and the tension force acting upwards, and since it is accelerating upwards we will take up as positive.




The problem that a few people have is that they put the the wrong way around, as they don't define directions as positive, and just take both up as positive. This actually results in a different system. By making sure you have the larger "mg" or "T" first, then you keep your acceleration positive, and account for directions. (You could also flip both, but that would be redundant).

Easiest way to solve these is normally solve for or and then substitute into the other equation.
Solving:

I find it easier to completely get rid of simultaneous equations. If we treat the two particles and the string as one body, then you can ignore the tension when finding acceleration. Just find the net force, which is the larger weight take away the smaller weight, and divide by total mass:
a = F/m = (3g - g) / 4 = g/2

Then tension can be found by subbing in acceleration for only one particle.

[Stevensmay]

Availan,
So if we have a equation we can equate each component to find our unknown values, as this is the only way they can be produced. a must equal 2 as we cannot get a x^2 any other way.

This is the same technique for complex as well.
If I have a complex equation , we can see that x must be 6 and y must be 5. So by equating each component we can see what they are.

The P(z) equation containing c can be equated with the original P(z).

Simplest way to find c is to equate the real only components, as they cannot be produced any other way.

[b^3]

I find it easier to completely get rid of simultaneous equations. If we treat the two particles and the string as one body, then you can ignore the tension when finding acceleration. Just find the net force, which is the larger weight take away the smaller weight, and divide by total mass:
a = F/m = (3g - g) / 4 = g/2

Then tension can be found by subbing in acceleration for only one particle.

Yeah, that works too. You're effectively using the result of the simultaneous equations in the end. You've just started at a later point in the working. You should still know how to do it the simultaneous equation way, since you can be asked for the equations of motion for each block. e.g. Q2 2010 Exam 2 http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010specmath2-w.pdf
If you're not asked for them, doing the other way should be fine, just sometimes there are marks given for showing the equations of motion of each block, so just keep that in mind (really depends on how many marks the question is worth I guess. Although I guess you would get the mark for that for the equation of motion for the whole system, just depends how picky they want to be really.).

I'd also argue that doing it the 'long way' here sets you up for after VCE Yes I know it's a simple problem, and that it works for this simple problem, and that you'd want to use the easiest method you have, but depending on what direction you head in later on, you may not be able to do it this way for more complicated problems. All I'm saying is it's beneficial to know the longer way.

[kinslayer]

Availan,
So if we have a equation we can equate each component to find our unknown values, as this is the only way they can be produced. a must equal 2 as we cannot get a x^2 any other way.

This is the same technique for complex as well.
If I have a complex equation , we can see that x must be 6 and y must be 5. So by equating each component we can see what they are.

The P(z) equation containing c can be equated with the original P(z).

Simplest way to find c is to equate the real only components, as they cannot be produced any other way.

To put it another way, think of each side of the equation as a vector rather than just a number, so that for example you have (a, b, c) = (2, 6, 3) or (x, y) = (6, 5).

[lzxnl]

Imagine you had ax^4+bx^3+cx^2+dx+e = Ax^4+Bx^3+Cx^2+Dx+E
If they're the same function, they must be equal at x=0. So e=E. Easy enough.
Now if they're the same function, their derivatives are also equal at x=0. 4ax^3+3bx^2+2cx+d=4Ax^3+3Bx^2+2Cx+D
Setting x=0, we have d=D
I think you can see what the result is if we keep repeating this process.

[e^1]

Quick question! (refer to attachment)

The answer is D, and they used the fact that to help get to the answer. How did they get this equation?

Thanks

edit: thanks stevens (lol that was easy).

[Stevensmay]

is the sum of the vertical forces. So we have our normal (N) acting up, 60g (weight) acting down and 120N resolved into it's vertical component. We know that since its not lifting these forces must sum to 0.

The above is just a rearrangement of

From here we find our horizontal component. (I'm assuming left is negative)
Sum of forces again.
Acceleration here is 0 as again it is in equilibrium, just moving this time.
Substitute in our value for N that was found above and rearrange to find the coefficient of friction.

[~T]

Anyone have any ideas on some kind of way to think about the asymptote equations for hyperbolae in the form and that you sometimes get as multiple choice questions? When it just gives you a hyperbola equation and asks for the two asymptote equations in the aforementioned form...

I expanded the usual form:



and got:



Which maybe sort of helps a bit. Perhaps if I practise using this... there're a couple of things that you can quickly gather from it to rule out some solutions I guess. The plus/minus bh bit is nice, because it usual rules out most; the difference between the two constants must be 2bh. Stuff like that. Thoughts? Not worth it, just expand every time?

[Jaswinder]

VCAA 2006 exam 2, question, 5aii, would writing {z: Re(z)=Im(z)}, where z=x+yi be the same as the answer or would I get it wrong?

Also on the same paper, how would you solve question 3bii (equations which would enable to R and L to be found)? Thanks

[Stevensmay]

VCAA 2006 exam 2, question, 5aii, would writing {z: Re(z)=Im(z)}, where z=x+yi be the same as the answer or would I get it wrong?

Also on the same paper, how would you solve question 3bii (equations which would enable to R and L to be found)? Thanks

The question asks for it to be in terms of . Seeing as your answer does not contain this then I imagine it would be wrong? Wolfram Alpha also states that the equivalent of the VCAA answer is but I am not sure if I entered it correctly, so this may be wrong.

[sin0001]

VCAA 2006 exam 2, question, 5aii, would writing {z: Re(z)=Im(z)}, where z=x+yi be the same as the answer or would I get it wrong?

Also on the same paper, how would you solve question 3bii (equations which would enable to R and L to be found)? Thanks

You've got the correct equation, but you're not answering in terms of the conjugate of z, as asked for by the question :/

And as for 3bii, just think of it as a plane/surface which is inclined at 10 degrees to the horizontal and resolve it parallel to the surface (85000-R - Wsin(10) =0) and perpendicular to it (L-Wcos(10) = 0). If you resolve it in vertical and horizontal components, instead of parallel/perpendicular to the plane, you're making it harder for yourself!

Beaten!

[ahat]

VCAA 2006 exam 2, question, 5aii, would writing {z: Re(z)=Im(z)}, where z=x+yi be the same as the answer or would I get it wrong?

Also on the same paper, how would you solve question 3bii (equations which would enable to R and L to be found)? Thanks

Like the others said, you'd would lose the mark, because you didn't meet the criteria. Look at the attached solution, it may help.