WitCH 70: Troubled Relations

November 1, 2021

12:27 pm

5 Comments on WitCH 70: Troubled Relations

calculus, exams, functions, parametrised curves, Specialist Mathematics, trigonometry, VCAA, VCE, WitCH

This is not an old one. It is from the 2019 Specialist Exam 2 and comes courtesy of student PURJ, who previously contributed to the discussion here. PURJ noted one, glaring, issue with the question below and its grading, but we think there are other issues as well. As with our previous WitCH, we’ll semi-update with excerpts from the examination report once people have had a chance to ponder and to comment.

UPDATE (01/11/21)

To move this along a little, we’ll note that the examination report gives the solution to (b) as

(b) $\color{blue}\boldsymbol{x\in [2,\infty), y\in [0,\infty)}$

UPDATE (01/11/21)

To add to the discussion of (b), the following is the examination report’s comment:

While most students stated the correct range, a significant number gave a domain which did not account for the restriction on t.

5 Replies to “WitCH 70: Troubled Relations”

tom says:

November 1, 2021 at 7:45 pm

Am I right in saying that the domain of $y=\sqrt{x^2-2x}$ is actually $[2,\infty)\cup(-\infty, 0]$ ?
This contrasts with the parametric equation where only the section $[2,\infty)$ applies. So the question either assumes the two relations are identical, or that part b refers to the parametric specification.

By the way, I was taught to use the term “revolve” for the situation in part e – to avoid confusion with “rotate” which could mean something different.

How to fix the (potentially good) question? I would start with some questions about $y=\sqrt{x^2-2x}$ and introduce the parametric function later, asking to show that it corresponds to part of the first relation.

Reply
1. marty says:
  
  November 1, 2021 at 7:58 pm
  
  Thanks, tom. That’s it. No connection has been made between the “rule” in (a) and the “relation” in (b), so there is no reason to consider the domain of the relation to have been restricted. I think the nonsense then percolates through the question, but (b) is the main issue: the answer in the examination report is wrong.
  
  Reply
SRK says:

November 2, 2021 at 1:19 pm

For part e, using $y=\sqrt{x^2-2x} \rightarrow x^2 = \left(\sqrt{y^2+1}+1\right)^2$ gives the volume of the solid as $\pi\int_0^{2\sqrt{2}}\left(\sqrt{y^2+1}+1\right)^2 \mathop{dy}$ . So here’s my “definite integral in terms of t that gives the volume of the solid”: $\pi\int_0^{2\sqrt{2}}\left(\sqrt{t^2+1}+1\right)^2 \mathop{dt}$ .

What’s wrong with this? I guess maybe $0 \leq t < \frac{\pi}{2}$ , so let's just fiddle with the integral a bit, and $2\pi\int_0^{\sqrt{2}}\left(\sqrt{4t^2+1}+1\right)^2 \mathop{dt}$ does the trick.

Reply
1. marty says:
  
  November 2, 2021 at 1:37 pm
  
  Thanks, SRK. It’s exactly the kind of thing I was thinking of.
  
  Reply
2. John Friend says:
  
  November 2, 2021 at 4:23 pm
  
  Very droll, SRK. Nice work.
  
  (Which t …? This t? Or that t?)
  
  Reply

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