Self Awareness of Memory

At some point, most teachers will have commented on removing distractions to improve the impact of revision. For example, phones away and music off. But do students appreciate why this helps revision? Do they understand how memory works?

Students are the main factor behind their learning. So why not teach students how their memory works and let them improve how they learn?

This presentation does exactly that. It's an investment of time, but I've found that it helps students appreciate why we ask them to do things.

A few examples:

A student does 50 questions one a skill but next lesson has no clue what they did. It could be that conversations alongside their work meant that they weren't storing information in long term memory.

A student needs constant reminders to show working out for their answers. They think it's a waste of time because they can do it without. Through discussions of working memory, they can begin to appreciate that by showing their working, they have more cognitive load available which can be reallocated to encoding in long term memory. By showing working, they remember things better in the long term and it's more efficient overall.

Radian Angles

At A-Level, trigonometry becomes rather advanced in a short space of time. Are students given enough time to form a solid understanding of radians on which to build? Compare it to degrees, where students learn:

• how to measure angles
• how to estimate angles
• angle facts in triangles and quadrilaterals
• angles in parallel lines
• interior and exterior polygons
• trigonometric ratios, Sine Rule, Cosine Rule and areas of triangles
• trigonometric graphs
• (in year 12) trigonometric equations

I'm not suggesting that students dedicate the same amount of time to radians, as some existing schema will already be in long term memory from working in degrees. I'm suggesting that students get some experience working with radians in these areas to make them less abstract, fully embed new information alongside existing schema and develop a solid foundation to build upon, with the aim of reducing misconceptions on more complicated problems later.

Here is a worksheet where angles are fractions of $\pi$ in radians and students either use radian protractors or they use regular protractors and convert to radians.

 

Radians

When students get used to radian angles which are fractions of $\pi$ they sometimes focus so heavily on the fraction that they forget the $\pi$ needs to be there. It's as if the "$\pi$ radians" becomes it's own unit of measure.

Here's a nice little check to drop in a lesson as it serves as an accuracy reminder.

Share the first half and then reveal the second half. I'm always surprised to see how many people draw a 135 degree angle for the first (instead of a ~43 degree angle) and then correct it when they see the second part.

Standard Form Discussion

I've taught Standard Form to be when a number is written as $n\times 10^x$ where $1\leq n\lt 10$ and $x$ is an integer, but on a few occasions I've been asked about numbers where I'm not sure. Consider this:

I've seen alternative definitions to allow for negatives. That is, $n\times 10^x$ where $1\leq |n|\lt 10$ but I haven't seen anything about the others.

I know they can be expressed in Standard Form, but I'm wondering if the coefficient ($n$) and power of 10 including the index have to be explicitly stated for it to be considered Standard Form.

Percentage Change - Link Up

Percentages can be merged with a range of other topics. This worksheet has 22 questions that do just that, and there's fully worked solutions.

As a taster, here are two additional questions.

Solution

Let $p$ be the number of purple sections, $g$ be the number of green sections and $n$ be the total number of sections. Now, $P(orange\:or\:purple)=\frac{2+p}{n}$ and $P(orange\:or\:green)=\frac{2+g}{n}$.

Use the percentage to form the equation

$$\frac{2+p}{n}=1.05\left(\frac{2+g}{n}\right)$$ $$2+p=1.05(2+g)$$ $$20p=2+21g$$ At this point, we can see that the LHS is a multiple of 20, so the RHS must be too. This means that $21g$ must end in 8. Trial $g=8$ and we get $20p=170$ which is not an integer value for $p$. Trialling $g=18$ gives $p=19$. So the minimum total number of sections is $18+19+2=39$

 

Solution

Let the radius be $r$ then the area is $\pi r^2$ and the circumference is $2\pi r$. Set up an equation using the percentage.

$$\pi r^2=1.92(2\pi r)$$ $$r=3.84$$ The diameter is therefore 7.68.

Find the Centre of a Circle using a Rectangle

The Geogebra applet below shows a rectangle and a circle. Using the rectangle and some basic tools, you can find the centre of the circle.

A solution

We know that the angle in a semicircle is 90 degrees. The converse is also true. Use the right angle of the rectangle to identify a diameter. Two diameters intersect at the centre. See below.

Note: This demonstration wasn't accurate. It was more of an exploration into the possible methods for finding the centre of the circle. With more tools, you can construct precise diagrams.

Quadrilaterals Investigation

This is another investigation classes like to explore and can be proven rather easily using vectors.

The investigation

Choose points P, Q, R and S and connect them to form a quadrilateral.

Construct the midpoints of each side.

Connect those midpoints in order with straight lines.

What do you notice?

I encourage students to constuct accurate diagrams and measure side lengths or angles carefully to determine whether their claims seem correct. To keep things simple, I've asked students to find the midpoints by measuring side length and halving it. I've also challenged students with using perpendicular bisectors to find midpoints, but be warned that diagrams get cluttered pretty quickly!

The midpoints of any quadrilateral form a parallelogram, or a straight line.

The proof

Let $2\mathbf{a}=\vec{PQ}$, $2\mathbf{b}=\vec{QR}$ and $2\mathbf{c}=\vec{RS}$

Let $A$, $B$, $C$ and $D$ be midpoints of $PQ$, $QR$, $RS$ and $SP$, respectively.

First, $\vec{AB}=\mathbf{a}+\mathbf{b}$

Secondly, $\vec{DC}=\frac{1}{2}\vec{PS}-\mathbf{c}=\frac{1}{2}(2\mathbf{a}+2\mathbf{b}+2\mathbf{c})-\mathbf{c}=\mathbf{a}+\mathbf{b}$

Now we have that $\vec{AB}=\vec{DC}$. These vectors are parallel and equal in length, therefore the midpoints form a quadrilateral, unless they are collinear in which case they form a straight line.