VCE Specialist 3/4 Question Thread!

[lzxnl]

Sorry I got a bit lost, but now understand that he graph can cross the asymptote.
second question
I know that the definition of an asymptote is x reaching +/- infinity, but for this graph, when it does the oblique asymptote changes as the graph is reflected in y axis. So the oblique asy as x approaches -infinity does not equal the oblique asy as x approaches +infinity.

An asymptote only has to be something you approach as x or y approaches one of the two (signed) infinities. They don't have to approach the same curve at + or - infinity.

For these 2 questions http://m.imgur.com/a/GnFVp

Q7b) I'm abit lost, because looking at it graphically does not make any sense. How does finding the angles the lines make with the x-axis, then taking the difference allow us to determine the angle made between the graphs?

Q5c) My working is what the worked solutions had. Only determined asymptotes and TP. How does this let us be confident in sketching the graph with only this information? I know it'll always be positive and ascend, but what makes us so sure that the graph doesn't do anything dodgy between the -ve asymptote and y-axis and +ve asymptote and the y-axis?

Also, something unrelated. If we're asked to differentiate arctan(2/5x) in the exam,  do we have to also write the restriction of x or is the answer fine?

The function is continuous and differentiable inside its domain and blows up at x = +-2 to positive infinity. Now, you know that the function's derivative is positive for positive x and negative for negative x, with a stationary point at x = 0. This suggests that your function decreases smoothly from the asymptote at x = -2 to the turning point and vice versa for positive x. You don't have any other turning points, so your function can't oscillate or change direction weirdly. It can only decrease for x < 0 and increase for x > 0.

arctan(2/5x)? Do you mean arctan((2/5) x) or arctan(2 / (5x))? If the latter, your derivative will be undefined at x = 0, yes (because the function isn't continuous). If the former, there's no domain restriction.

[Jimmonash1991]

G'day all,

I am a current math major, who is revisiting specialst maths for revision purposes, so please, prioritise those in need first.

I am looking at linear dependence, and knowing that with three vectors a, b and c, having ka + lb = c with k and l scalar quantities, defines this, I wonder how I might go about the following:

Three vectors a = i− j + 2k, b = i + 2j + mk and c = 3i + nj + k are linearly dependent, find m in terms of n in simplest fraction form.

Now, there must be a simple way to go about it. I took the following approach:

ka + lb = c

==> ki - kj + 2kk + li + 2lj + lmk = 3i + nj + k

I then found the following system of equations by equating for i, j, and k unit vectors on both sides of the equality

k + l = 3 ==> equation 1

-k + 2l = n ==> equation 2

2k + lm = 1 ==> equation 3

I decided to solve k in terms of l from equation 1 to create a new equation

k = 3 - l ==> equation 4

Then I substituted into equation 2 to find an expression for l in terms of n

-(3 - l) + 2l = n

3l - 3 = n

l = 1/3(n + 3) ==> equation 5

Thus finding k in terms of n

k = 3 - 1/3(n+3) ==> equation 6

I gathered since the vectors were linearly dependent, then they should satisfy equation 3, so I substitued 5 and 6 into 3

2[3 - 1/3(n+3)] + 1/3(n+3)m = 1

6 - 2/3(n+3) + 1/3(n+3)m = 1

18 - 2(n+3) + (n+3)m = 3

(n+3)m = 2(n+3) - 15 = 2n - 9

m = (2n + 9)/(n+3)

As one can see, this is not the answer at all. I also solved a system of linear equations using Gaussian Elimination, by setting up a 3X3 augmented matrix, and knowing the bottom row must be [0,0,0] and after row reduction, I found the element in row 3 column 3 to be -5 - 1/3(n+3)(m-2), and requiring this to be equal to zero for this to occur, I solved for m in terms of n, and got the same answer as above.

I have double checked my working a number of times, and still arrived at this using the method of ka + lb = c as defined by the textbook. I am using cambridge specialst maths 3&4 2016 if that is of any help.

Thanks again!

[Shadowxo]

Hi,
I tried solving the equations from scratch, similar method except I used simultaneous equations to get rid of variables instead of substitution.
I ended up with the same answer you had: \(\frac{2n-9}{n+3}\) and I don't know how cambridge got their answer, but sometimes they are wrong.
It's likely that they made a mistake or did it a slightly different way so don't worry about that particular question.

[Jimmonash1991]

I don't know how cambridge got their answer, but sometimes they are wrong.
It's likely that they made a mistake

I feel like a naughty scientist for quoting the above phrases, but I have seen already several instances of cambridge being wrong in the 2016 edition, which pains me, because imagine getting an answer, and following all the right steps, but not being sure of your ability. It is a confidence killer is all, and I hope this does not happen from here on in!

Really appreciate the help. I will be posting here in the future me thinks.

Would it be okay if I answered questions as well? My study score in Spesh was only 32, but I do have some college math under my belt 

Cheers!

James

[MightyBeh]

I feel like a naughty scientist for quoting the above phrases, but I have seen already several instances of cambridge being wrong in the 2016 edition, which pains me, because imagine getting an answer, and following all the right steps, but not being sure of your ability. It is a confidence killer is all, and I hope this does not happen from here on in!

Really appreciate the help. I will be posting here in the future me thinks.

Would it be okay if I answered questions as well? My study score in Spesh was only 32, but I do have some college math under my belt 

Cheers!

James

It's absolutely okay to answer questions! Everyone's welcome to contribute, we appreciate it a lot. 

[Jimmonash1991]

Thanks mate. Love the appreciation for my "mad" skills! 

I had another question myself unfortunately. It is in the vectors section of the cambridge 3&4 text for 2016.

As a disclaimer, I want to say there is likely an easy way to solve my problem, but help with this algebraic approach would be most appreciated.

 In triangle OAB, OA = 3i + 4k and OB = i + 2j−2k.

a Use the scalar product to show that ∠AOB is an obtuse angle.

This just required the dot product, and the angle was indeed obtuse, at 109.47 degrees to two decimal places

For convenience, if the angle is X (this becomes important below), then cosX = -1/5

b Find OP, where P is:

(i) the midpoint of AB

Again self explanatory. Find AP = 1/2AB, and then the coordinates of OP = OA + AP
 
(ii) the point on AB such that OP is perpendicular to AB

This one required some thought, but I solved the vector resolute of BO onto AB, and then found the vector rejection, which was perpendicular to AB per the dot product.

(iii) the point where the bisector of ∠AOB intersects AB.

This one is proving a little elusive. I figured make the vector OP a unit vector, so magnitude one, and then find the angle AOB, which was obtuse as per part (a). Then if OP bisects AOB, it means that the angle will be halved between OA and OP, and OP and OB.

I started a noble quest to form three equations:

The dot product between OA*OP = [OA]*[OP]cosx

The dot product between OB*OP = [OB]*[OP]cosx

The magnitudes of [OA] = 5, and [OB] = 3

The magnitude [OP] = 1, as it is a unit vector, thus if OP = (x,y,z), sqrt(x^2+y^2+z^2) = 1

All through, x is half the original angle of AOB X,  which in terms of cosine is cosx = sqrt(10)/5. I solved this using half angle formula

cosX = 1/2cos^2(x) - 1

I went ahead and solved for z in terms of x and y from the first two equations, and substituted into equation three to find the ultimate solution for z. I get complex solutions because the discriminant of a quadratic with very large coefficients, renders the discriminat negative.

I have a hunch that the angle x may differ for the above cases because X was obtuse. Any insight is appreciated. I am going to continue plugging away.

[Quantum44]

Thanks mate. Love the appreciation for my "mad" skills! 

I had another question myself unfortunately. It is in the vectors section of the cambridge 3&4 text for 2016.

As a disclaimer, I want to say there is likely an easy way to solve my problem, but help with this algebraic approach would be most appreciated.

 In triangle OAB, OA = 3i + 4k and OB = i + 2j−2k.

a Use the scalar product to show that ∠AOB is an obtuse angle.

This just required the dot product, and the angle was indeed obtuse, at 109.47 degrees to two decimal places

For convenience, if the angle is X (this becomes important below), then cosX = -1/5

b Find OP, where P is:

(i) the midpoint of AB

Again self explanatory. Find AP = 1/2AB, and then the coordinates of OP = OA + AP
 
(ii) the point on AB such that OP is perpendicular to AB

This one required some thought, but I solved the vector resolute of BO onto AB, and then found the vector rejection, which was perpendicular to AB per the dot product.

(iii) the point where the bisector of ∠AOB intersects AB.

This one is proving a little elusive. I figured make the vector OP a unit vector, so magnitude one, and then find the angle AOB, which was obtuse as per part (a). Then if OP bisects AOB, it means that the angle will be halved between OA and OP, and OP and OB.

I started a noble quest to form three equations:

The dot product between OA*OP = [OA]*[OP]cosx

The dot product between OB*OP = [OB]*[OP]cosx

The magnitudes of [OA] = 5, and [OB] = 3

The magnitude [OP] = 1, as it is a unit vector, thus if OP = (x,y,z), sqrt(x^2+y^2+z^2) = 1

All through, x is half the original angle of AOB X,  which in terms of cosine is cosx = sqrt(10)/5. I solved this using half angle formula

cosX = 1/2cos^2(x) - 1

I went ahead and solved for z in terms of x and y from the first two equations, and substituted into equation three to find the ultimate solution for z. I get complex solutions because the discriminant of a quadratic with very large coefficients, renders the discriminat negative.

I have a hunch that the angle x may differ for the above cases because X was obtuse. Any insight is appreciated. I am going to continue plugging away.

I did find the final part of this question rather challenging but I believe that this answer is a reasonable way to obtain to solve the problem.

[Jimmonash1991]

Thats a quantum leap from where I was!

Curious though. The vector c is arrived at how on the triangle? I gather you were using it to represent the vector OP, but I would love some deeper insight into how you knew to do that.

The rest of it is just linear dependence/ independence/ collinearity stuff.

Great work! Bet that pencil sounded gorgeous on paper.

More mathematics needs to be done with pencil!

Cheers

James

[Quantum44]

Thats a quantum leap from where I was!

Curious though. The vector c is arrived at how on the triangle? I gather you were using it to represent the vector OP, but I would love some deeper insight into how you knew to do that.

The rest of it is just linear dependence/ independence/ collinearity stuff.

Great work! Bet that pencil sounded gorgeous on paper.

More mathematics needs to be done with pencil!

Cheers

James

Haha, I just used vector c to represent the direction of the bisector so I could obtain the ratio. Whenever you work with bisectors you first need to establish the direction of the vector, the magnitude at this point is irrelevant. This can simply be done by adding the two unit vectors. Then it just becomes a simple collinearity problem once you have the ratio of OP, although the variable m does become redundant once a has been established and the problem can be solved.

Mathematics is indeed rather pulchritudinous when performed using a lustrous 2B pencil.

[lzxnl]

Thats a quantum leap from where I was!

Curious though. The vector c is arrived at how on the triangle? I gather you were using it to represent the vector OP, but I would love some deeper insight into how you knew to do that.

The rest of it is just linear dependence/ independence/ collinearity stuff.

Great work! Bet that pencil sounded gorgeous on paper.

More mathematics needs to be done with pencil!

Cheers

James

The physics geek inside me is compelled to say this.

"Quantum" in "quantum mechanics" refers to actually the smallest divisible unit. For instance, the quantum of light of a given frequency is a photon of energy Planck's constant * light frequency. The word refers to a VERY small amount in general and using it in the expression 'quantum leap' reveals a complete lack of understanding of its origins

[Guideme]

Help pls

[de]

The physics geek inside me is compelled to say this.

"Quantum" in "quantum mechanics" refers to actually the smallest divisible unit. For instance, the quantum of light of a given frequency is a photon of energy Planck's constant * light frequency. The word refers to a VERY small amount in general and using it in the expression 'quantum leap' reveals a complete lack of understanding of its origins

My understanding always was that "quantum leaps" referred to quantum jumps- as in the apparently discontinuous transition between energy states in atoms. So the phrase is less about it being a "big" jump and more about it being intuitive and seemingly disconnected from previous thought??
Of course this is all assuming that quantum jumps/leaps aren't just a figment of my imagination or my physics misconceptions...

[RuiAce]

(Image removed from quote.)
Help pls

[lzxnl]

My understanding always was that "quantum leaps" referred to quantum jumps- as in the apparently discontinuous transition between energy states in atoms. So the phrase is less about it being a "big" jump and more about it being intuitive and seemingly disconnected from previous thought??
Of course this is all assuming that quantum jumps/leaps aren't just a figment of my imagination or my physics misconceptions...

On the contrary, energy states in atoms are related to each other because of the Schrodinger equation, so there's nothing disconnected from previous thought about them.

[deStudent]

For these 2 calc questions http://m.imgur.com/a/cwa1L

Q8f) why are they treating e^x as if it is just x? What's the logic behind what the answer did? I was abit lost doing this question.

Q12a) when you flip 6/x -> x/6, why do both of the inequalities flip? I thought you'd only flip the inequality sign of -1 because you'll be dividing by a negative number? Whereas for 1, you won't be so you don't flip the sign? Do you flip both both regardless if you're flipping 'the angle' (6/x)?