WitCH 69: Sines of the Times

October 29, 2021

9:47 am

19 Comments on WitCH 69: Sines of the Times

graphing, inverse functions, Specialist Mathematics, trigonometry, VCAA, VCE, WitCH

Yet another old one, from the 2010 Specialist Exam 2 (CAS). We’ll look to semi-update with excerpts from the examination report once people have had a chance to comment. (Readers are of course free to peek at the examination report, anytime they so wish.)

UPDATE (30/10/21)

The following is the (slightly tidied) solution to (c) in the examination report:

$\color{blue}\boldsymbol{ f'(x)&=\frac{4x}{\sqrt{1-\left(2x^2-1\right)^2}}= \frac{4x}{\sqrt{4x^2-4x^4}}= \frac{4x}{\sqrt{4x^2\left(1-x^2\right)}}\,,}$

which gives the stated result. $\color{blue}\boldsymbol{a=1}$

UPDATE (01/11/21)

The question is generally aimless and annoying, but the main issue is (c). It makes absolutely no sense to “find the value of a”, since the number a is in no sense defined. Worse, the report’s method to find the maximum value of a is fundamentally invalid. The domain of the derivative of f’ can only be found by considering when the function f can be differentiated, and definitely not by considering the final form of f’.

19 Replies to “WitCH 69: Sines of the Times”

tom says:

October 29, 2021 at 10:32 am

Not sure how much of the heavy lifting can be done by the calculator allowed. But part c caught my eye. Good thing that the question specifies that one should use calculus. otherwise the candidate might spend ages trying to do it without 😉

Reply
1. Simon says:
  
  October 29, 2021 at 12:26 pm
  
  They expected most to be done or checked by CAS – except for part b and the “show that” aspect of part c, where they clunkily asked for calculus. And the CAS does an ok job on this, making most of it pretty trivial. The skill here (as said in the examiners report) was transferring information from the calculators onto paper.
  
  The percentages in part bi to bii are amusing. Basically, if the student can set the problem up, then the CAS can hammer the integration nail.
  
  The problem I see here is the careless use of domain in defining/restricting/extending functions. But I’m sure Marty has another issue in mind.
  
  Reply
  1. marty says:
    
    October 30, 2021 at 9:20 am
    
    Thanks, Simon. You largely covered it, although the domain thing is worse than you suggest. The worst is the domain thing, and I’ll reply further to John. I also loathed the clunkiness of (d), and the needless use of the (always) clunky term “hybrid function”.
    
    With the CAS thing, I note the report’s comments on (a):
    
    Students are reminded to take care when transferring
    graphical information from technology to a graph with a given scale.
    
    In Specialist Mathematics. This is what students and teachers gotta concern themselves with.
    
    Reply
2. marty says:
  
  October 30, 2021 at 9:13 am
  
  Ah! So “use calculus” really means “don’t use the machine”. For the life of me, I never understood the use of, or at least the origin of, that idiotic expression.
  
  Reply
John (No) Friend (of VCAA) says:

October 29, 2021 at 12:31 pm

With an infinite number of values of $\displaystyle a$ to choose from in part (c), what are the chances a student will give the single value required by VCAA? Of course, the chance of success rises if we assume

“… and find the [ $\displaystyle largest$ ] value of $\displaystyle a.$ ”

is what VCAA $\displaystyle really$ meant.

Hi Tom. I can hear your sarcasm. This is the sort of idiotic instruction required in a $\displaystyle CAS$ -Exam (VCAA’s choosing) to tell a student NOT to use a CAS. This is the lunatic dichotomy VCAA’s technology fetish has created. Stay tuned for “Use algebra to solve the equation” …
As an aside, I would have thought that the question being worth 3 marks together with the global instruction “In questions where more than 1 mark is available, appropriate working must be shown” would be sufficient to tell students that a ‘by hand’ solution was required.

Reply
1. marty says:
  
  October 30, 2021 at 9:47 am
  
  Thanks, John. The “find the value of a” is the main awfulness. And, it is worse than you have suggested. See the update to the post.
  
  Reply
  1. John (No) Friend (of VCAA) says:
    
    October 30, 2021 at 11:24 am
    
    Your update is more of an indictment on the Examination Report than the question (although the question is bad for reasons already discussed). I originally resisted your obvious nudge to look at the Report. But after reading your update – incredulously – I was compelled to confirm that what $\displaystyle you$ said was the $\displaystyle entirety$ of what VCAA said. (By the way, for the benefit of the readers, it is!)
    
    Compounding VCAA’s deception (and that’s what it is), I note that the Report is from the pre-disclaimer era (more recent Reports contain the mealy-mouthed statement:
    
    “This report provides sample answers or an indication of what answers may have included. Unless otherwise stated, these are not necessarily intended to be exemplary or complete responses.”)
    
    For the benefit of students who read these blogs, I’ll add what VCAA $\displaystyle should$ have continued with to Marty’s update:
    
    $\displaystyle = \frac{4x}{2|x| \sqrt{1 - x^2}}$ since $\displaystyle \sqrt{x^2} = |x|$
    
    Substitute $\displaystyle |x| = x$ when $\displaystyle x \geq 0$ :
    
    $\displaystyle = \frac{2}{\sqrt{1 - x^2}}$ , $\displaystyle x \neq 0$
    
    which was to be shown.
    
    The maximal domain of this function is the simultaneous solution to $\displaystyle x > 0$ and $\displaystyle 1 - x^2 > 0$ :
    $\displaystyle 0 < x < 1$ .
    
    Therefore, $\displaystyle assuming$ the largest value of $\displaystyle a$ , $\displaystyle a = 1$ .
    
    It's worth noting that Cambridge had a question asking for the derivative of $\displaystyle f(x) = \cos^{-1}\left( \frac{6}{x}\right)$ (Exercise 6C Q12 (b)) and its answer was wrong (it took $\displaystyle \sqrt{x^2} = x$ ). The error was quite laughable, particularly when part (c) asked for the graph of $\displaystyle y = f(x)$ and the graph is clearly increasing over its entire domain. I don't know if the error's been fixed.
    
    Reply
    1. marty says:
      
      October 30, 2021 at 11:33 am
      
      No, it is an indictment of the question, and not just for the floating a.
      
      Fundamentally, one cannot determine the domain of a derivative by looking at the final form of the differentiated function. The (part) solution in the examination report is fundamentally invalid for determining the value of a, maximal or otherwise.
      
      Reply
marty says:

October 31, 2021 at 10:10 am

Just in case I’m not being pointed enough in my objection to part (c), here is a question for readers to ponder. Consider the function $\boldsymbol{f(x) = \sqrt{x} - \sqrt{x}}$ . What is $\boldsymbol{f'}$ ? What is the domain of $\boldsymbol{f'}$ ?

Reply
1. Sai says:
  
  October 31, 2021 at 2:10 pm
  
  I’d say $f':(69,420) \to \mathbb{R}, f'(x) = 0$ . Since no domain was specified for $f$ , I took the liberty of assuming some domain… Wait is that the issue with the question? That, because we are not told to assume that $f$ is defined on its maximal domain, that means anything following would be rubbish? After all, the derivative is $f'(a) =\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$ and VCAA now expecting $a=1$ are completely idiotic as they have set no prior precedent for the domain of $f$ ?
  
  Edit:
  This question looks worse each time I look at it. When they say “the function with rule…” means that there is one such function with that rule. But clearly, one could define $f:[0,0.0001]\to \mathbb{R} , f(x) = \sin^{-1}(2x^2-1)$ , or choose any domain they like, which fucks up everything down the line.
  
  Reply
  1. marty says:
    
    October 31, 2021 at 3:09 pm
    
    No. Assume the maximal domain for f.
    
    Reply
  2. marty says:
    
    October 31, 2021 at 8:10 pm
    
    Actually, Sai, I answered too quickly. Your answer is simply in incorrect. (Also, be careful with your language.)
    
    Reply
    1. Sai says:
      
      November 1, 2021 at 1:42 pm
      
      I feel conflicted about the answer(s) I arrive at. From one perspective, if I consider the function addition of $\sqrt{x}$ and $-\sqrt{x}$ , both defined on $[0,\infty)$ , then it’s yelling to me that the function proposed above has to have a domain of $[0,\infty)$ , and then it follows that the derivative, which is given by $f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = 0$ where $x > 0$ . But, somewhere I feel obliged to say that the whole representation is a red herring, and that the function $f(x) = 0$ , and consequentially its derivative is the zero function on $\mathbb{R}$ . But I can’t really conclude which is valid.
      
      Reply
      1. marty says:
        
        November 1, 2021 at 1:51 pm
        
        Thanks, Sai. This is exactly the point. So, what is the answer?
        
        Reply
2. Lawrence says:
  
  November 1, 2021 at 4:40 pm
  
  Here’s my opinion. For the definition of $f$ to make sense, the square root must exist, which, for our purposes let that be on $[0,\infty)$ . Then $f$ has maximal domain $[0,\infty)$ also (since to obtain $f(x)$ we must evaluate $\sqrt{x}$ and subtract $\sqrt{x}$ from it). But for every $x$ in that domain, $f(x) = \sqrt{x} - \sqrt{x} = 0$ , so its derivative is $f'(x) = 0$ (immediately on $(0,\infty)$ at least).
  
  Again depending on whether we allow differentiability at boundary points, we could include or exclude $0$ in the domain of $f'$ , so it would be either $(0,\infty)$ or $[0,\infty)$ . But truthfully, $\frac{f(x) - f(0)}{x - 0} = 0 \to 0$ as $x \to 0$ anyways (which is a limit point of the domain), so I’d be happy to say $f'(0) = 0$ .
  
  The derivative is a property of the function (as a mapping $x \to f(x)$ ), not based on its representation, so if it is 0 everywhere, so is its derivative. (Not so relevant in this case, but in the case of something like $g(x) = |x| - |x|$ , $g'(0) = 0$ since $g = 0$ everywhere, even though $x \mapsto |x|$ is not differentiable at 0.)
  
  Reply
  1. marty says:
    
    November 1, 2021 at 7:37 pm
    
    Thanks, lawrence. I forgot to reply to this comment. Yes, my example needlessly brought in the endpoint red herring. I should have chosen my function to be $\boldsymbol{f(x) = \log x - \log x}$ , to avoid the endpoint issue. In any case, although one may or may not include 0 in the domain of f’, the negative reals are definitely not in the domain.
    
    Reply
marty says:

November 1, 2021 at 12:28 pm

Does anybody want to make clear the point of my example in the comment above, or should I simply update the post?

Reply
1. Lawrence says:
  
  November 1, 2021 at 5:02 pm
  
  Ahh, I think I get what your point is? In terms of the maximal domain of $f'$ , it depends on the definition of $f$ . Of course it’s only sensible to ask if $f$ is differentiable at a point in the (maximal) domain of $f$ , which is $[-1,1]$ , Since arcsine is defined on $[-1,1]$ and differentiable on $(-1,1)$ , whenever $2x^2 - 1 \in (-1,1)$ , i.e. $x \in (-1,0) \cup (0,1)$ , we can simply (carefully) use the chain rule to conclude $f$ is differentiable. On $(0,1)$ , the chain rule (i.e. if $g$ is differentiable at $b$ and $h$ is differentiable at $g(b)$ , then $h \circ g$ is differentiable at $b$ with $(h \circ g)'(b) = h'(g(b))g'(b)$ ) gives us that the derivative is what is claimed in part (c). It’s not the form of the derivative, as they seem to suggest.
  
  Reply
  1. marty says:
    
    November 1, 2021 at 5:15 pm
    
    Thanks, lawrence. Yes, that and your other comment above captures it. We determine the domain of f’ purely and simply by deciding where f can be differentiated, and absolutely not by looking at the form of the derivative.
    
    Reply

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