My ability in solving mathematical questions is declining. (1 Viewer)

Mathislife · Jan 15, 2016

Ever since year 12 started I have been struggling to maintain my ability in solving questions. At first I thought it was just stress but it seems like more to me. There are a lot of times now where I will get stuck on questions only to find out they were ridiculously easy. I can't rarely solve advanced questions now without having done one similar to it beforehand. At the end of year 11 this was rarely happening. It's true that I do slightly less math hw now but not enough that it should effect me like this. I do ext1 and ext2 and while it is a bit more understandable for this to happen in ext2, there isn't an explanation as to why the same is happening in ext1 considering I did well in it in year 11. Any help in identifying why this is happening to me would be greatly appreciated as it is really killing me.

leehuan · Jan 15, 2016

1. How often do you practice questions specific to one topic?

If not enough - Stay topic confined, or even just unit confined and keep repeating the same thing (e.g. integration by substitution)
If sufficient - Move to 2.

2. When you do past papers, can you identify WHAT topic is being asked?

If no - You need to start doing some papers open book. You haven't figured out what links to what yet.
If yes - Move to 3.

3. The idea about doing a ton of practice questions for maths, especially for something like MX2 but tbh true for even down to general, is that you get used to question style. The topics change a bit in the HSC despite calling upon preliminary knowledge, so you're probably restricted specifically within the HSC topics.

Try to find any topics you feel weak at.

If you want, post some example questions that you struggle on and what I can do is describe what processes occur in my brain, and how I deduce a method to answer it.

dan964 · Jan 15, 2016

Mathislife said:
Ever since year 12 started I have been struggling to maintain my ability in solving questions. At first I thought it was just stress but it seems like more to me. There are a lot of times now where I will get stuck on questions only to find out they were ridiculously easy. I can't rarely solve advanced questions now without having done one similar to it beforehand. At the end of year 11 this was rarely happening. It's true that I do slightly less math hw now but not enough that it should effect me like this. I do ext1 and ext2 and while it is a bit more understandable for this to happen in ext2, there isn't an explanation as to why the same is happening in ext1 considering I did well in it in year 11. Any help in identifying why this is happening to me would be greatly appreciated as it is really killing me.

practice, practice. attempt the question, then check your answer, and then mark your answer. Repeat process with a similar question.

Trebla · Jan 15, 2016

Like Ext2, the Ext1 course also contains problems which require you to apply what you know in an unfamiliar situation. The course also steps up quite a bit in difficulty in Year 12 compared to Year 11.

The best way to handle it (on top of doing heaps of practice problem solving questions) is actually understanding the concepts properly and the logic behind how they work rather than just rote learning the formulas. That way you can recognise when to call upon these tools when solving an unfamiliar problem.

dan964 · Jan 15, 2016

I think the other thing is you might have to put more effort in, there comes a point where your natural knowledge may not cut it.

porcupinetree · Jan 15, 2016

When first looking at a question that you are unsure of how to attempt, try and notice anything in the question which reminds you of anything else you've seen before. When actually starting to attempt the question, try out all of the ideas that you have, and see where they get you. If you're really stuck, it's alright to look at the answers, but don't let it become a habit.

I think it's also important to have the right attitude, and to make an effort not to get frustrated when you don't know how to answer a question at first. Be determined, but also be patient.

Mathislife · Jan 16, 2016

leehuan said:
1. How often do you practice questions specific to one topic?

If not enough - Stay topic confined, or even just unit confined and keep repeating the same thing (e.g. integration by substitution)
If sufficient - Move to 2.

2. When you do past papers, can you identify WHAT topic is being asked?

If no - You need to start doing some papers open book. You haven't figured out what links to what yet.
If yes - Move to 3.

3. The idea about doing a ton of practice questions for maths, especially for something like MX2 but tbh true for even down to general, is that you get used to question style. The topics change a bit in the HSC despite calling upon preliminary knowledge, so you're probably restricted specifically within the HSC topics.

Try to find any topics you feel weak at.

If you want, post some example questions that you struggle on and what I can do is describe what processes occur in my brain, and how I deduce a method to answer it.

Find the locus of z if: w = z-2/z+2 and arg(w) = pie/3

Drsoccerball · Jan 16, 2016

Mathislife said:
Find the locus of z if: w = z-2/z+2 and arg(w) = pie/3

To vector, or not to vector ?

InteGrand · Jan 16, 2016

Mathislife said:
Find the locus of z if: w = z-2/z+2 and arg(w) = pie/3

$\noindent This says that $\arg \left(\frac{z-2}{z+2}\right) =\frac{\pi}{3}$. So $\arg \left(z-2\right) -\arg \left(z+2\right) =\frac{\pi}{3}$. So the locus is the major arc of a circle starting at the point $z=2$ in the complex plane and going counter-clockwise to the point $z=-2$, with the angle subtended by the chord joining these points in this segment being $\frac{\pi}{3}$. We can use geometry to find the circle's equation if we wish. Also, make sure to put open circles at the points $z =\pm 2$, since $\arg 0$ is undefined and $z\neq -2$, since then $w$ is undefined.$

InteGrand · Jan 16, 2016

$\noindent Basically, the locus of $z$ for $\arg (z-z_0) - \arg (z -z_1) = \alpha$ (where $\alpha$ is fixed with $0\leq \alpha \leq \pi$ and $z_0, z_1$ are fixed complex numbers) is one you should be familiar with. In fact, the shapes are different depending on whether $\alpha = 0$, $\alpha =\pi$, or $\0< \alpha < \pi$.$

$\noindent If $\alpha =0$, then $\arg (z-z_0) = \arg (z-z_1)$, so the locus is line through the points $z_1$ and $z_2$ \textsl{excluding} the line connecting $z_0$ and $z_1$, since $z$ needs to be equally inclined from $z_0$ as from $z_1$.$

$\noindent If $\alpha =\pi$, then the locus is actually just the line segment from $z_0$ to $z_1$, excluding these two points. This is because $\arg (z-z_0) -\arg (z-z_1) =\pi$, so $\arg (z-z_0) = \arg (z-z_1) +\pi$, or $\arg (z-z_0) = -\arg (z-z_1)$. So $z$ must lie on the line segment joining $z_0$ to $z_1$, as the angles it makes with these points and the horizontal must be negatives of each other.$

$\noindent If $0<\alpha < \pi$, then the locus is the circular arc from $z_0$ going counterclockwise to $z_1$ and with angle subtended by this chord at this arc being $\alpha$. Note that this means that if $\alpha < \frac{\pi}{2}$, then the arc is a major arc. If $\alpha =\frac{\pi}{2}$, then the arc is a semicircle (angle in semicircle is $90^\circ$), and if $\alpha >\frac{\pi}{2}$, then the arc is a minor arc. The reason for this shape of locus is seen by drawing vectors and angles, and by using exterior angle of a triangle and angles in the same segment of a circle.$

$\noindent For all these loci, the points $z_0$ and $z_1$ are excluded (draw open circles there) since $\arg 0$ is undefined.$

leehuan · Jan 16, 2016

In the answer InteGrand gave, one key thing to note is that such a locus is required knowledge for the MX2 course.

Admittedly, I never studied it cause I haven't seen it in any HSC paper (it appeared in selective trials) but it's something like arg(z)=pi/4 ---> roted.

dan964 · Jan 16, 2016

leehuan said:
In the answer InteGrand gave, one key thing to note is that such a locus is required knowledge for the MX2 course.

Admittedly, I never studied it cause I haven't seen it in any HSC paper (it appeared in selective trials) but it's something like arg(z)=pi/4 ---> roted.

With ext 2, you definite need to grasp the concepts well. But you need to also practice, i.e. be very rigorous*. The two go hand in hand. Same goes for ext 1, but obviously the concepts are easier. Rote learning doesn't work well for ext 2, in general/most cases; while for ext 1 it can work.

*this is the only reason why my mark is half decent, because towards the later half of the year, I finally decided to put a bit more effort, than I was putting in.

leehuan · Jan 16, 2016

dan964 said:
With ext 2, you definite need to grasp the concepts well. But you need to also practice, i.e. be very rigorous*. The two go hand in hand. Same goes for ext 1, but obviously the concepts are easier. Rote learning doesn't work well for ext 2, in general/most cases; while for ext 1 it can work.

*this is the only reason why my mark is half decent, because towards the later half of the year, I finally decided to put a bit more effort, than I was putting in.

I was specifically referring to the question that OP gave in his comment. It's based off a rote concept.

Obviously to get a high mark you need to practice questions. I already implied that earlier on. But the point of practicing questions is to grasp the concept of it.

InteGrand · Jan 16, 2016

It is actually possible to rote learn some things in HSC maths.

leehuan · Jan 16, 2016

Cough integration by substitution

dan964 · Jan 20, 2016

InteGrand said:
It is actually possible to rote learn some things in HSC maths.

even ext 2? maybe the conic formulae...

leehuan · Jan 20, 2016

dan964 said:
even ext 2? maybe the conic formulae...

Pretty sure the free marks on "write z in modulus-argument form" are roted.

My ability in solving mathematical questions is declining. (1 Viewer)

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