## Tuesday, October 30, 2012

### Pumpkin pi

Inspired by former student Jena’s tradition of carving up some pumpkin pi each year, I decided to do my own variation using the formula

π/4 = 1 – 1/3 + 1/5 – 1/7 + 1/9 …

## Saturday, October 27, 2012

### Will that tree hit my house?

Yesterday my dad and I set about sawing down a large alder tree that was leaning and looking precarious. It was tricky to see whether it was tall enough to hit the house when it fell, but luckily some similar triangles came to the rescue.

I folded a 45º angle in a piece of scrap paper and walked to the side of the house closest to the tree. Sighting up the hypotenuse, I noticed that the topmost branches were still in view – so the tree was taller than the distance between its base and the house.

I used a guide rope attached to another tree to prevent it from falling too close the house, so my windows remained intact. The main problem was that the chainsaw got stuck as we were cutting the notch, so we had to hack at it with an axe for awhile. Warning: the video below is not educational:

## Friday, October 26, 2012

### Similar parabolas

While looking at the graph of a parabola on Geogebra, I noticed that you could "flatten it out" by changing the

*x*-scale on the axes

*or*by zooming in (changing

*x*- and

*y*-scales proportionally) on the vertex. This is intuitive: if you zoom in on any smooth curve, it straightens out - in calculus you call that local linearity.

But this means that any two parabolas can be made to look identical just by zooming (as opposed to scaling only one axis), which is how two similar objects can be made to look identical.

We say two object are similar if their corresponding lengths form proportions (pairs of equal fractions); can we prove that two parabolas are similar in this way?

Say you have two parabolas,

*y = ax*² and

*y = bx*². To show that two shapes are similar, we find a

*scale factor*(

*k*) so that any length on the first shape equals

*k*times the corresponding length on the second shape. For parabolas, let's use the pair of lengths

*x*and

*y*for convenience.

On the first parabola, this pair of lengths is

*x1*and

*ax1*²

*.*On the second parabola, the pair of lengths is

*x2*and

*bx2*²

*.*

We must find a

*k*so that

*x*2 =

*kx1*and

*bx2*² =

*kax1*².

Substituting for

*x2*, we have

*b(kx1)*² =

*kax1*².

So

*bk*²

*x1*² =

*kax1*².

Dividing, we have

*bk*=

*a*, or

*k*=

*a*/

*b*.

So the scale factor between any two parabolas

*y = ax*² and

*y = bx*² is

*a*/

*b*, conveniently.

## Saturday, October 20, 2012

## Monday, October 15, 2012

### Sextants, clocks, and logs

Congrats to my fellow teacher George Christoph, who just released a lesson at TedEd. It gives an overview of nautical navigation and the development of logarithms.