MitPY 13: Trigonometry and Wolfram Alpha

This MitPY comes from frequent commenter, John Friend:

Dear Colleagues,

I gave a CAS-FREE question to my Specialist students whose first part was to solve (exactly) the equation \boldsymbol{\cot 2x = \sec x}. I solved it two different ways and got two different answers that are equivalent. I’ve attached my calculations.

I checked my answers using Mathematica, which lead to my question: Mathematica gives a third different but equivalent answer (scroll down to real solutions). How has Mathematica got this answer?

It may be that Mathematica ‘used’ my Method 2, got my tan answer and then for some arcane reason ‘manipulated’ this answer into the one it finally gives. If so, I can ascribe the answer to a Mathematica quirk. But it may be that Mathematica is using a method unclear to me that leads to its answer. If so, I’m curious.

Any thoughts are appreciated.

Click to access Calculations.pdf

MitPY 11: Asymptotes and Wolfram Alpha

This MitPY comes from frequent commenter, John Friend:

Dear colleagues,

I figured this was as good place as any to ask for help. I’m writing a small test on rational functions. One of my questions asks students to consider the function \displaystyle f(x) = \frac{x^3 + x}{x^2 + ax - 2a} where a \in R and to find the values of a for which the function intersects its oblique asymptote.

The oblique asymptote is y = x - a so they must first solve

\displaystyle \frac{x^3 + x}{x^2 + ax - 2a} = x - a … (1)

for x. The solution is \displaystyle x = \frac{2a^2}{(a+1)^2} and there are no restrictions along the way to getting this solution that I can see. So obviously a \neq -1.

It can also be seen that if a = 0 then equation (1) becomes \displaystyle \frac{x^3 + x}{x^2} = x which has no solution. So obviously a \neq 0.

When I solve equation (1) using Wolfram Alpha the result is also \displaystyle x = \frac{2a^2}{(a+1)^2}. But here’s where I’m puzzled:

Wolfram Alpha gives the obvious restriction a + 1 \neq 0 but also the restriction 5a^3 + 4a^2 + a \neq 0.

a \neq 0 emerges naturally (and uniquely) from this second restriction and I really like that this happens as a natural part of the solution process. BUT ….

I cannot see where this second restriction comes from in the process of solving equation (1)! Can anyone see what I cannot?

Thanks.

MitPY 10: Square Roots

This MitPY comes from a student, Jay:

I have a question relating to polynomial equations. For context I have just finished Y11 during which I completed Further 3&4, Methods 1&2 and Specialist 1&2.

This year during my maths methods class we covered the square root graph, however I was confused as to why it only showed the positive solutions. When I asked about it I was told it was because the radical symbol meant only the positive solution.

However since then I have learnt that the graph of \boldsymbol{y=x^{0.5}} also only shows the positive solution of the square root, while \boldsymbol{y^2=x} shows both. I am quite confused by why they aren’t the same. The only reason that I could think of is that it would mean \boldsymbol{y=x^2} would be the same as \boldsymbol{y^2=x^4}, and while the points (-2,-4) and (2,-4) fit the latter they clearly don’t fit former.

Could you please explain why these aren’t the same?

MitPY 9: Team Games

This MitPY is from commenter HollyBolly, who asked on the previous MitPY for some advice on diplomacy.*

Can you guys after all the serious business give me some advice for this situation: on a middle school Pythagoras and trig test, for a not very strong group of students. Questions are to be different from routine ones provided with the textbook subscription. I try “Verify that the triangle with sides (here: some triple, different from 3 4 5) is right, then find all its angles”. After reviewing, the question comes back: “Verify by drawing that a triangle with sides…”

How do you respond if that review has come from:

A. The HoD;

B. A teacher with more years at the school than me but equal in responsibilities in the maths department;

C. A teacher fresh from uni, in their20s.

Regards.

*) Yeah, yeah. We’ll stay right out of the discussion on this one.

MitPY 8: Verification Code

It’s a long time since we’ve had a MitPY. But, the plague goes on (including the plague of right-wing Creightons).

This one comes from frequent commenter Red Five, and we apologise for the huge delay in posting. It is targeted at those familiar with and, more likely, struggling with Victoria’s VCE rituals:

VCAA uses some pretty strange words in exam questions, and the more exam papers I read, especially for Specialist Mathematics 34, the more I can’t get a firm idea of how they distinguish between the meanings of “show that“, “verify that” and “prove that“.

Verify” seems to mean “by substitution”, “show that” seems to mean “given these very specific parameters” and “prove that” seems to be more general, but is it really this simple?

MitPY 7: Diophantine Teen Fans

This MitPY is a request from frequent commenter, Red Five:

I’d like to ask what others think of teaching (mostly linear) Diophantine equations in early secondary school. They are nowhere in the curriculum but seem to be everywhere in competitions, including the AMC junior papers on occasion. I don’t see any reason to not teach them (even as an extension idea) but others may have some insights into why it won’t work.

MitPY 6: Integration by Substitution

From frequent commenter, SRK:

A question for commenters: how to explain / teach integration by substitution? To organise discussion, consider the simple case

    \[\boldsymbol{ \int \frac{2x}{1+x^2}dx \,.}\]

Here are some options.

1) Let u = 1 + x^2. This gives \frac{du}{dx} = 2x, hence dx = \frac{du}{2x}. So our integral becomes \int \frac{2x}{u}\times \frac{du}{2x} = \int \frac{1}{u}du. Benefits: the abuse of notation here helps students get their integral in the correct form. Worry: I am uncomfortable with this because students generally just look at this and think “ok, so dy/dx is a fraction cancel top and bottom hey ho away we go”. I’m also unclear on whether, or the extent to which, I should penalise students for using this method in their work.

2)  Let u = 1 + x^2. This gives \frac{du}{dx} = 2x. So our integral becomes \int \frac{1}{u}\times \frac{du}{dx}dx = \int \frac{1}{u}du. Benefit. This last equality can be justified using chain rule. Worry: students find it more difficult to get their integral in the correct form.

3) \frac{2x}{1+x^2} has the form f'(g(x))g'(x) where g(x)=1+x^2 and f'(x) = \frac{1}{x}. Hence, the antiderivative is f(g(x)) = \log (1+x^2). This is just the antidifferentiation version of chain rule.  Benefit. I find this method crystal clear, and – at least conceptually – so do the students. Worry. Students often aren’t able to recognise the correct structure of the functions to make this work.

So I’m curious how other commenters approach this, what they’ve found has been effective / successful, and what other pros / cons there are with various methods.

UPDATE (21/04)

Following on from David’s comment below, and at the risk of splitting the discussion in two, we’ve posted a companion WitCH.

MitPY 4: Motivating Vector Products

A question from frequent commenter, Steve R:

Hi, interested to know how other teachers/tutors/academics …give their students a feel for what the scalar and vector products represent in the physical world of \boldsymbol{R^2} and \boldsymbol{R^3} respectively. One attempt explaining the difference between them is given here. The Australian curriculum gives a couple of geometric examples of the use of scalar product in a plane, around quadrilaterals, parallelograms and their diagonals .

Regards, Steve R