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Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164553 times)  Share 

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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9345 on: November 11, 2018, 06:22:51 pm »
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Question 4 of the 2017 VCAA exam
https://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/2017specmath2-w.pdf

(a+b)/2=-2+yi when a=x+yi gives the correct answer b in terms of a when solved.
However when solve((x+y*i+b)/2=-2+y*i,b) is put into the TI-Inspire the answer comes out as 'false.' Why is the calc not able to solve something that seems this simple?

If I use CAS, I get b = –x + yi – 4, which gives b = –4 – conjugate(a), if we let a = x + yi.\

Make sure you use csolve, rather than solve.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9346 on: November 11, 2018, 06:30:33 pm »
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Hey guys,

Just wondering why my result came out negative for V? The actual answer is 7.8, but I'm not sure if I've made an error somewhere or am just supposed to drop the negative since its speed (has no direction). I restricted V and t when doing all the working out on CAS, but wouldn't a negative V value reverse the j position vector? Can someone explain this to me?

Cheers
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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9347 on: November 11, 2018, 07:55:01 pm »
+1
Hey guys,

Just wondering why my result came out negative for V? The actual answer is 7.8, but I'm not sure if I've made an error somewhere or am just supposed to drop the negative since its speed (has no direction). I restricted V and t when doing all the working out on CAS, but wouldn't a negative V value reverse the j position vector? Can someone explain this to me?

Cheers

See the posts just above from mzhao and myself – the j component of the ring's position vector is not 3, since the origin is the point of release, not on the ground.

Dais_Deorum

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9348 on: November 11, 2018, 08:04:49 pm »
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See the posts just above from mzhao and myself – the j component of the ring's position vector is not 3, since the origin is the point of release, not on the ground.

Sorry, I forgot to mention that I changed the original function to make the point of reference the ground for j. I was just redoing it; I think in my working out I somehow removed the positive option... who knows though.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9349 on: November 15, 2018, 07:58:56 pm »
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Hi everyone!
I was hoping someone could provide me with a set out answer to this question. I know what the answer is and I can get it with my cas but I am unsure of how to do it by hand...
Thanks.
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mzhao

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9350 on: November 15, 2018, 09:11:05 pm »
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Hi everyone!
I was hoping someone could provide me with a set out answer to this question. I know what the answer is and I can get it with my cas but I am unsure of how to do it by hand...
Thanks.

Using the quadratic formula does the trick.

« Last Edit: November 15, 2018, 09:12:58 pm by mzhao »
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studyingg

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9351 on: December 15, 2018, 11:41:24 am »
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sorry abt the bad quality, but I have a question about sketching rational functions... are we required to label axis intercepts? (just like in methods), because I think that (-1,0) and (0,-1/6) belong on this graph but in the solutions in my textbook, they aren't labelled...
also how are we supposed to know the shape of the function? like I practically drew a similar graph but between the vertical asymptotes the shape of my (cubic? -- not sure what to call it), is not as flat with respect to the left and right sides of (0,-1/6)



S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9352 on: December 15, 2018, 01:00:37 pm »
+3
sorry abt the bad quality, but I have a question about sketching rational functions... are we required to label axis intercepts? (just like in methods),

Yes, it is always good practice to find and label intercepts. It is also important for helping with your second question:

Quote
also how are we supposed to know the shape of the function?

There are a few methods:

1. Using calculus. The sign of the first derivative tells you where the graph is sloping upwards / downwards. The sign of the second derivative tells you where the graph is concave up / concave down. Using calculus is the most accurate way to determine shape, but unfortunately with rational functions this can be quite tedious without CAS. One option is to resolve into partial fractions first, before differentiating.

2. Substituting in "test points" to know whether the graph is above / below the x-axis, between each vertical asymptote and x-intercept. You can then "join the dots" because rational functions are continuous wherever they are defined.

3. Check if the function is odd / even. If you have f(x) = f(-x) or f(x) = -f(-x) you really only need to draw half the graph, and the rest follows by symmetry.

For the example you've given, I would factorise to (x+1)/((x–3)(x+2)), and then substitute in test points to work out the shape, after you've got asymptotes and intercepts. From there, I'd resolve into partial fractions. This makes it obvious that the first derivative is always negative, so there are no stationary points and the graph is always sloping downwards. Finding the point of inflection by hand is very tedious – the x-coordinate is not too bad, but substituting in to find the y-coordinate is annoying. I would not expect to have to do this without technology assistance.

studyingg

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9353 on: December 15, 2018, 01:12:08 pm »
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Yes, it is always good practice to find and label intercepts. It is also important for helping with your second question:

There are a few methods:

1. Using calculus. The sign of the first derivative tells you where the graph is sloping upwards / downwards. The sign of the second derivative tells you where the graph is concave up / concave down. Using calculus is the most accurate way to determine shape, but unfortunately with rational functions this can be quite tedious without CAS. One option is to resolve into partial fractions first, before differentiating.

2. Substituting in "test points" to know whether the graph is above / below the x-axis, between each vertical asymptote and x-intercept. You can then "join the dots" because rational functions are continuous wherever they are defined.

3. Check if the function is odd / even. If you have f(x) = f(-x) or f(x) = -f(-x) you really only need to draw half the graph, and the rest follows by symmetry.

For the example you've given, I would factorise to (x+1)/((x–3)(x+2)), and then substitute in test points to work out the shape, after you've got asymptotes and intercepts. From there, I'd resolve into partial fractions. This makes it obvious that the first derivative is always negative, so there are no stationary points and the graph is always sloping downwards. Finding the point of inflection by hand is very tedious – the x-coordinate is not too bad, but substituting in to find the y-coordinate is annoying. I would not expect to have to do this without technology assistance.

Okay, wow thanks for the explanation!

What I was doing was sketching the individual graphs (x+1) and x^2-x-6 on the same axis, and determiningwhere the graph would be positive or negative (eg the numerator is negative from neginifity to -1 while denominator is positive from neg infinity to -2 ) dividing the values means the graph would be negative from negative infinity to - 2. I had trouble using calculus bc my knowledge is limited to unit1/2 methods calc, but i'm going to try your method from now on as it seems much more reliable.
Thank youu

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9354 on: December 16, 2018, 01:18:41 am »
+1
Okay, wow thanks for the explanation!

What I was doing was sketching the individual graphs (x+1) and x^2-x-6 on the same axis, and determiningwhere the graph would be positive or negative (eg the numerator is negative from neginifity to -1 while denominator is positive from neg infinity to -2 ) dividing the values means the graph would be negative from negative infinity to - 2. I had trouble using calculus bc my knowledge is limited to unit1/2 methods calc, but i'm going to try your method from now on as it seems much more reliable.
Thank youu

1/2 methods should cover quotient rule and differentiation of polynomials, which is all you need to differentiate these functions (unless I'm mistaken with whether quotient rule is in the course; if not, learn it yourself. It's not difficult, just be careful with the subtraction). Let's see how much we can do without calculus though.


Asymptotes are at x = 3, x = -2, y = 0. The third one is a little more complicated to see, as if we let x->inf, we get inf/inf. To see this, divide the top and bottom by the highest power of x, which is x2:

and it is readily seen that this vanishes as x becomes larger; the numerator goes to 0 and the denominator goes to 1.

Next, we want to see the shape of this function. The x-intercept is at x=-1, so we have to check from -inf to -2, -2 to -1, -1 to 3, 3 to inf (consider x intercepts and asymptotes).

If x > 3, then all three brackets are positive, and f(x) > 0. So, f(x) is a large positive number to the right of the asymptote x = 3 and approaches 0 for large positive x.
If -1 < x < 3, then x + 1 > 0, x + 2 > 0, but x - 3 < 0, so f(x) < 0. f(x) is large but negative to the left of the asymptote x = 3.
If -2 < x < -1, then x + 1 < 0, x - 3 < 0, x + 2 > 0 so f(x) > 0. f(x) is large and positive to the right of the asymptote x = -2.
If x < -2, then x + 1 < 0, x - 3 < 0, x + 2 < 0 so f(x) < 0. f(x) is large and negative to the left of the asymptote x = -2 and approaches 0 for large negative x.

Hopefully you can draw a graph with just this knowledge. It doesn't tell you if there are turning points, however.
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studyingg

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9355 on: December 16, 2018, 09:12:52 am »
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1/2 methods should cover quotient rule and differentiation of polynomials, which is all you need to differentiate these functions (unless I'm mistaken with whether quotient rule is in the course; if not, learn it yourself. It's not difficult, just be careful with the subtraction). Let's see how much we can do without calculus though.


Asymptotes are at x = 3, x = -2, y = 0. The third one is a little more complicated to see, as if we let x->inf, we get inf/inf. To see this, divide the top and bottom by the highest power of x, which is x2:

and it is readily seen that this vanishes as x becomes larger; the numerator goes to 0 and the denominator goes to 1.

Next, we want to see the shape of this function. The x-intercept is at x=-1, so we have to check from -inf to -2, -2 to -1, -1 to 3, 3 to inf (consider x intercepts and asymptotes).

If x > 3, then all three brackets are positive, and f(x) > 0. So, f(x) is a large positive number to the right of the asymptote x = 3 and approaches 0 for large positive x.
If -1 < x < 3, then x + 1 > 0, x + 2 > 0, but x - 3 < 0, so f(x) < 0. f(x) is large but negative to the left of the asymptote x = 3.
If -2 < x < -1, then x + 1 < 0, x - 3 < 0, x + 2 > 0 so f(x) > 0. f(x) is large and positive to the right of the asymptote x = -2.
If x < -2, then x + 1 < 0, x - 3 < 0, x + 2 < 0 so f(x) < 0. f(x) is large and negative to the left of the asymptote x = -2 and approaches 0 for large negative x.

Hopefully you can draw a graph with just this knowledge. It doesn't tell you if there are turning points, however.

Thanks man! I I’ll have to learn the quotient rule (is it like differentiating a function which comprises of one polynomial divided by another? ) For turning points.  But i'm making notes out of you and S_R_K's explanations bc they have realllly helped (especially the explanation about the horizontal asymptote and stuff). Thanks for improving my understanding :)

Just one thing --  my question about sketching the graphs is more to do with the 'width' of the graphs. For example, for this case I managed to draw a graph that resembles the one in my text book, however the middle section (between asymptotes -2 and 3) was more compressed and steep in comparison to that in the text book. My question is more focused on knowing how wide or compressed to draw the graph of a rational function.
« Last Edit: December 16, 2018, 09:59:01 am by studyingg »

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9356 on: December 16, 2018, 10:11:14 am »
+1
Thanks man! I I’ll have to learn the quotient rule (is it like differentiating a function which comprises of one polynomial divided by another? ) For turning points.  But i'm making notes out of you and S_R_K's explanations bc they have realllly helped (especially the explanation about the horizontal asymptote and stuff). Thanks for improving my understanding :)

Just one thing --  my question about sketching the graphs is more to do with the 'width' of the graphs. For example, for this case I managed to draw a graph that resembles the one in my text book, however the middle section (between asymptotes -2 and 3) was more compressed and steep in comparison to that in the text book. My question is more focused on knowing how wide or compressed to draw the graph of a rational function.

The quotient rule for differentiation is a rule for finding the derivative of any function of the form f(x)/g(x), where both f and g are differentiable. (So this includes cases where, for example, we want to differentiate sin(x) / log(x)).

With respect to your second question, it depends on whether you are provided with a scale or not.

If you are given a scale you must ensure that your shape is reasonably accurate. That means it must go roughly through the right points, even if you don't calculate / label a table of values. For instance, if the graph goes through (1, 1) (but this isn't a key point like a stationary point or point of inflection, and so you aren't required to label), then your graph should go through (1, 1), or at least be very close to it.

If you aren't given a scale, then you just need to make sure that your graph is consistent with the signs of the first and second derivative. That means that your graph should be increasing when first derivative is positive; decreasing when first derivative is negative; concave up when second derivative is positive; concave down when second derivative is negative (if you don't yet know what concavity means, just look it up). Don't worry about whether your graph is increasing "too quickly" or whether its concavity is "too steep". The generic shape is fine.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9357 on: December 16, 2018, 01:08:17 pm »
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Could someone please explain linear independence and dependence to me? What exactly is the point of vectors being dependent or independent and how can we discern whether vectors or independent or not? Are only parallel vectors dependent in 2D and in 3D only those vectors are dependent that lir on the same 'line segment'?

Also, I just want to ask, how are the strategies for linear dependence and independence different for 2 and 3 vectors?

Thanks. I know I am asking for a lot, even if you answer one question, it will be really appreciated.
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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9358 on: December 16, 2018, 03:03:39 pm »
+3
Could someone please explain linear independence and dependence to me? What exactly is the point of vectors being dependent or independent and how can we discern whether vectors or independent or not? Are only parallel vectors dependent in 2D and in 3D only those vectors are dependent that lir on the same 'line segment'?

Also, I just want to ask, how are the strategies for linear dependence and independence different for 2 and 3 vectors?

Thanks. I know I am asking for a lot, even if you answer one question, it will be really appreciated.

The starting point is the idea of a linear combination of vectors. We say that c is a linear combination of a and b if c = xa + yb for some scalars x and y (this generalises to an arbitrary number of vectors, but in Spesh 3/4 you are only required to show linear dependence for 3 vectors in 3 dimensions). Notice that if c is a linear combination of a and b, then c is redundant.

A set of vectors is linearly dependent if at least one of those vectors is a linear combination of some of the others. Again, the point is that a linearly dependent set of vectors contains some redundant vectors. A linearly independent set contains no redundancy – it is a maximally efficient way of describing any position vector.

Geometrically, in 2D, a set of linearly dependent vectors all line on the same line through the origin. In 3D, a set of linearly dependent vectors all line on the same plane that includes the origin.

To show that a set of 3 vectors is linearly dependent, write one of them as a linear combination of the others, but where the scalars are unknowns. That is, write c = xa + yb, and show that there is a simultaneous solution to x and y. It is pretty much the same strategy for showing that a set of 2 vectors is linearly dependent, except now you only have to show that there is a solution for x to the equation c = xa. Conceptually, it is the same method; the difference is that the algebra tends to be more difficult when you have 3 vectors vs. 2.

If you want to show that a set of vectors is linearly independent then use the same strategy, but show there is no solution.

There are a couple of useful shortcuts. In 2D, any set of 3 or more vectors must be linearly dependent (geometrically, in 2D, any set of three vectors must include at least one that lies on the same line as a combination of the others). In 3D, any set of 4 or more vectors must be linearly dependent (geometrically, in 3D, any set of four vectors must include at least one that lies in the same plane as a combination of the others). A set of 2 vectors in 2D space can be linearly dependent or independent - you have to check. Similar for a set of 3 vectors in 3D space.
« Last Edit: December 16, 2018, 03:30:38 pm by S_R_K »

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9359 on: December 16, 2018, 07:48:32 pm »
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Hey Guys. I just finished my year 12 VCE but I'm deciding to take Specialist Maths 3/4 via Distance Education next year to boost my atar to 98-99 so I have a better chance of getting to Medicine. I'm decent at maths (not excellent by any means); I got a raw 39 in Methods without that much studying for it. Do you guys believe I might be able to survive Spesh 3/4 and get 40+ without doing 1/2 at all? I will take a gap year so I would have a lot of time in my hands. Thanks!!
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