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cos(108)

I came across an interesting tidbit of math today in solving a problem from the ever-fruitful, continuously engaging book Mathematics Review Exercises by David P. Smith Jr. and Leslie T. Fagan.

Apparently \sqrt{2-2\cos 108^\circ}=\phi.

In my last post on the Farmer and the Fencing optimization problem, I mentioned that separating the pen into two triangular pieces with a diagonal cross-section makes the problem a lot more complex and requires the use of Lagrangian methods to solve.  After conceiving the problem in 1999-2000 and setting up the Lagrangian system that would solve the problem in 2001-2002 I finally solved it this week by using Wolfram Alpha.  If anyone out there sees a good algebraic solution for the Lagrangian system of three equations with three unknowns, please let me know.

Given 100 feet of fencing, a rectangular pen with length =x, width =y and a diagonal cross-section =\sqrt{x^2+y^2}, the perimeter constraint equation would be:

2x+2y+\sqrt{x^2+y^2}=100

The area equation we want to optimize is:

A=x*y

So, the Lagrangian process says that to maximize a function f(x,y), with a given constraint g(x,y)=C, we need to set up a system based on the equality:

\bigtriangledown f(x,y)=\lambda \bigtriangledown g(x,y)

In this problem, f(x,y)=A=xy and g(x,y)=2x+2y+\sqrt{x^2+y^2}.

So the gradient of f(x,y), or \bigtriangledown f(x,y) is:

\bigtriangledown f(x,y)=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})

And the gradient of g(x,y), or \bigtriangledown g(x,y) is:

\bigtriangledown g(x,y)=(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y})

So, going back to the functions we’re working with based on the perimeter and area, the Lagrangian set up would be:

\bigtriangledown f(x,y)=\lambda \bigtriangledown g(x,y)

(y,x)=\lambda (2+\frac{x}{\sqrt{x^2+y^2}}, 2+\frac{y}{\sqrt{x^2+y^2}})

This generates the system of equations:

y=2\lambda+\frac{\lambda x}{\sqrt{x^2+y^2}}

x=2\lambda+\frac{\lambda y}{\sqrt{x^2+y^2}}

2x+2y+\sqrt{x^2+y^2}=100

I must admit once again to resorting to a technological solution for this system.  Using Wolfram Alpha, we get the solution

x=y=\frac{100}{4+\sqrt{2}}\approx 18.47

This makes the cross-piece:

\sqrt{x^2+y^2} =\frac{100}{\sqrt{9+4\sqrt{2}}} \approx 26.12

Which of course means that the optimal area is once again generated by a square!  If we assumed this at the outset of the problem, it becomes much simpler (but less “fun”).  The permimeter in that case is:

4x+x\sqrt{2}=100

and

x=\frac{100}{4+\sqrt{2}}

I’ve been carrying this around with me for nearly 15 years and dug out the work I had originally done a few weeks ago to look at it “with fresh eyes.”  I found a few minor errors and was happy to use Wolfram Alpha to solve the system and finally generate a solution for this problem.

Four years ago I wrote a post about math education that mentioned the problem of “The Farmer and the Fencing.”  The link for the post is here and a sample assignment from an Elementary Algebra (Algebra I) course is here.  I mentioned in that post that I’ve used this problem at a variety of levels of mathematics starting with the 7th grade.

For the 7th grade class, we looked at some possible values for the length and width of the pen that was to be constructed from the fencing and made a table of values showing the length, width and area of the pen for each set of values.

The project from the Elementary Algebra class that is linked above addresses the question from a somewhat more sophisticated level, asking the students to introduce variables for the length and width and then substitute in for one of the variables to arrive at an expression for the area as a function of only one of the dimensions.  The students can then find the maximum area using a graphing calculator.

In a Pre-Calculus class I taught during the 1999-2000 school year, we covered a unit that considered the parabola as a conic section and included applications on projectile motion and maximum and minimum values.  In the Pre-Calculus class, we found the max or min by finding the vertex of the parabola defined at the point x=\frac{-b}{2a} and y=f(\frac{-b}{2a}).  We didn’t need Calculus to determine this, the \frac{-b}{2a} can be defined from the vertex form of the equation of a parabola y=n(x-h)^2+k.  With a little classical algebra, you can show that h=\frac{-b}{2a}.

If you expand y=n(x-h)^2+k

y=n(x^2-2hx+h^2)+k

y=nx^2-2nhx+nh^2+k

Then

\frac{-b}{2a}=\frac{2nh}{2n}=h

I taught Differential Calculus as a grad student at the University of Maine in 2003 and we covered The Farmer and the Fencing from the perspective of Calculus, in which we differentiate the area equation and set the derivative equal to zero to find the maximum area.  The problems can be spruced up a little by dividing the pen into smaller pens using cross-sectional pieces.

In the fundamental problem the maximum area of a simple pen with four sides and no cross-divisions is arrived at by splitting the perimeter equally among the four sides to make a square.  If the pen is cross divided in any way, the maximum area comes from a scenario in which half of the perimeter is allocated to the lengthwise pieces and the other half allocated to the any widths.

If there are m lengths and n widths then the perimeter is represented by ml+nw=P.  The area is A=l*w.  If we isolate the length in the perimeter equation and substitute it into the area equation, we can then differentiate the area equation for an answer for the general problem of maximum area of a pen with mby n crosspieces:

ml+nw=P

ml=P-nw

l=\frac{P-nw}{m}

l=\frac{P}{m}-\frac{nw}{m}

Then substituting into the area equation:

A=l*w

A=(\frac{P}{m}-\frac{nw}{m})*w=\frac{P}{m}*w-\frac{n}{m}*w^2

So, the area equation is then

A=\frac{P}{m}*w-\frac{n}{m}*w^2

Which makes the derivative

A'=\frac{P}{m}-\frac{2n}{m}*w.

Setting this equal to zero and solving for w:

0=\frac{P}{m}-\frac{2n}{m}*w

\frac{P}{m}=\frac{2n}{m}*w

\frac{m}{2n}*\frac{P}{m}=w

\frac{P}{2n}=w

or

\frac{P}{2}=nw

which means that half of the perimeter should be allocated to the n widths and the other half to the m lengths.

For example, if the farmer has 300 feet of fencing, the maximum area is a square that is 75 feet on each side.

FencingSquare

If the pen is split into two smaller pens with a crosswise fence, then the maximum area will allocate 150 feet to the widths (75 feet each) and 150 feet to the three lengths (50 feet each).

FencingCrossPen

When I taught this problem in 1999-2000 I asked myself what might happen if the pen were to be divided on the diagonal creating two triangular pens.  Little did I know how much this would complicate such a simple little problem.

FencingDiagonalThe Briggs and Cochran Calculus textbook includes a similar problem in its section on optimization (problem 10c on pg. 214).  In a key modification that makes the problem solvable for Calculus I students, they eliminate one of the lengths of the figure by placing the pen against a barn.

FencingDiagonalBarnThis eliminates one of the lengths in the problem.  Let’s look at the solution to this problem before we tackle the more general version which requires that both lengths come from the available fencing.  In the problem in the textbook, 200 meters of fencing are to be used, so the constraint on the amount of fencing creates the following equation:

l+2w+\sqrt{l^2+w^2}=200

If we want to maximize the area of the pen then we need to substitute for one of the variables and find the derivative of the area equation:

l*w=A

In order to substitute we need to isolate one of the variables from the constraint equation.  This will involve getting the square root by itself so that we can square both sides of the equation:

l+2w+\sqrt{l^2+w^2}=200

\sqrt{l^2+w^2}=200-2w-l

(\sqrt{l^2+w^2})^2=(200-2w-l)^2

l^2+w^2=(200-2w-l)(200-2w-l)

l^2+w^2=40,000-400w-200l-400w+4w^2+2lw-200l+2lw+l^2

l^2+w^2=40,000-800w-400l+4w^2+4lw+l^2

This is where this problem differs from the general problem, in that the two l^2 terms cancel each other out and allow us to isolate the l

w^2=40,000-800w-400l+4w^2+4lw

We’ll move both terms that contain the l to the same side, so that we can factor out l and then divide through by 400-4w

400l-4lw=40,000-800w+3w^2

l(400-4w)=40,000-800w+3w^2

l=\frac{40,000-800w+3w^2}{(400-4w)}

So now our area equation becomes:

\frac{40,000-800w+3w^2}{(400-4w)}*w=A

OR

\frac{40,000w-800w^2+3w^3}{(400-4w)}=A

Finding the derivative for this is pretty messy and strains the display capabilities of this blog!  We’ll do our best to show the work here:

\frac{(400-4w)(40,000-1600w+9w^2)}{}

\frac{-(40,000w-800w^2+3w^3)(-4)}{16(100-w)^2}=A'

\frac{(16,000,000-640,000w+3600w^2-160,000w+6400w^2-36w^3)}{}

\frac{+160,000w-3200w^2+12w^3)}{16(100-w)^2}=A'

So, combining like terms:

-640,000w-160,000w+160,000w=-640,000w

3600w^2+6400w^2-3200w^2=+6800w^2

-36w^3+12w^3=-24w^3

\frac{(16,000,000-640,000w+6800w^2-24w^3)}{16(100-w)^2}

And then cancel a common factor of 8 in the numerator and denominator:

\frac{(2,000,000-80,000w+850w^2-3w^3)}{2(100-w)^2}

The simplified expression for the derivative is:

\frac{-3w^3+850w^2-80,000w+2,000,000}{2(100-w)^2}=A'

So from this we need to set the numerator equal to zero.  In the textbook, there is no indication that technology will be necessary to complete the problem, but I don’t see any way around using a graphical solution to finding the root of the derivative.  I suppose there is always the cubic formula, but that is truly an anachronism.  Here’s what the graph of the derivative looks like with the root indicated:

GraphicalSolution

From the graph, we can conclude that:

w\approx 38.814

l\approx 55.03

and the crosspiece \sqrt{l^2+w^2}\approx 67.34

We’ll look at the diagonal problem without the barn (which leads to a Lagrangian solution) in another post.

The infinite series approximations that have been used for many years to calculate the values of trigonometric functions have traditionally been attributed to Brook Taylor and Colin Maclaurin, European mathematicians of the early 18th century who were building on the work of Newton, Leibniz, James Gregory and Isaac Barrow among others.

However, I recently discovered that they were not the first to use these techniques.  As author George Gheverghese Joseph points out in the previous link, the work of Newton and Leibniz was tremendous, however the Indian development of infinite series approximations for trigonometric functions was equally amazing and important.  In addition, it came nearly 300 years before the European development of these techniques.

Madhava of Sangamagrama is generally recognized of the founder of the Kerala school of mathematics and astronomy in what is today the state of Kerala in southwest India.  The work of the mathematicians of the Kerala school was based on a desire for accurate trigonometric values for use in navigation.

Madhava lived in the late 1300s and early 1400s and most of his original work has been lost.  However, he is mentioned frequently in the surviving work of later mathematicians from the Kerala school.  Madhava is credited with power series calculations for the sine, cosine, tangent and arctangent, and like Leibniz, he used the arctangent power series to approximate the value of \pi to 13 decimal places.

Victor Katz’ A History of Mathematics (Brief Edition) has a wonderful and detailed derivation of the Kerala school trigonometric series, with diagrams showing how they used the relationships between the angles, radii, chords and arcs in a circle to arrive at these amazing calculations.

Katz also published this derivation in a paper for the MAA (Mathematics Magazine, vol. 68, n. 3, June 1995, pp. 163-174)

The derivation for the infinite series begins on page 169 (pg. 7 in the pdf).

I’ve just begun to unpack this derivation and will post a step-by-step explanation of Katz’ work in the “near” future.

During the 1998-99 school year I was teaching math at Ellsworth High School in Maine.  I got to talking with one of the other math teachers there about deriving the area of a circle.  We started approaching the idea from a similar perspective as Archimedes had 2300 years ago in his work with circles and \pi.  We inscribed a polygon in a circle and then took the limit of the area of the polygon as the number of sides (n) approached infinity.  For the area of the polygon, we divided it into n triangles and used the area formula A=\frac {1}{2} ab\sin C.

CircleArea

In this diagram, both a and b are radii of the circle and are labeled as R.  The central angle \theta which will be used to find the area of the triangle is actually equal to \frac {360^\circ}{n} or (and this will be VERY important later) \frac {2\pi}{n}.

So the area of the polygon is n times the area of the triangle, or:

n*\frac {1}{2} R*R*\sin \frac{2 \pi}{n}

For this to be the area of the circle we want to take the limit of the previous expression as n \to \infty.

A=\lim_{n \to \infty} n*\frac {1}{2} R*R*\sin \frac {2 \pi}{n}

Since the \frac {1}{2}*R*R don’t involve n, we can pull them out from the limit.

A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}

Now comes the calculus – we were just talking about the derivative of the sine function in Calculus I last week, which is what made me think about this.  Remember that:

\lim_{x \to 0} \frac{\sin x}{x}=1

So, we need to establish this limit somewhere in our derivation and then replace it with 1.  The most likely place for this would be in here:

\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}

but we need to have:

\lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac{2 \pi}{n}}

OK – so, let’s multiply our origninal expression by \frac{2 \pi}{2 \pi}.

\lim_{n \to \infty} \frac{2 \pi}{2 \pi}*n*\sin \frac {2 \pi}{n}

or

\lim_{n \to \infty} 2 \pi*\frac{n}{2 \pi}*\sin \frac {2 \pi}{n}

which is equivalent to

\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}

Let’s bring back the whole expression:

A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}

A=\frac {1}{2} R^2*\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}

A=\frac {1}{2} R^2*2 \pi \lim_{n \to \infty} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}

as n \to \infty,

\frac{2 \pi}{n} \to 0

A=\frac {1}{2} R^2*2 \pi \lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}

A=\frac {1}{2} R^2*2 \pi *1

or

A=\pi R^2

When we were working through this the first time, we kept the 360^\circ in the formula and ended up with A=180^\circ R^2, which didn’t make any sense until we realized that 180^\circ is \pi radians!

Archimedes’ conclusion was that the area of the circle was equal to one-half the radius times the circumference:

\frac{1}{2} R*2\pi R = \pi R^2

or, since the mathematicians in the Greek tradition like Archimedes thought in terms of geometry – the area of a circle is the same as the area of triangle with height equal to the radius of the circle and base equal to the circumference of the circle!

Combinatorics

We were covering some elementary combinatorics in the Pre-Calculus course this past week and were discussing some problems from VK Balakrishnan’s Discrete Mathematics textbook.  I took a Discrete Math course at University of Maine in 2000 from Balakrishnan and really enjoyed the class.

I needed to make a quiz question for the Pre-Calculus class and tried to remember some of the more interesting ways Balakrishnan had constructed his problems and came up with this one:

Seven people (A, B, C, D, E, F, G) want to ride into Portland but the car will only hold four.  Person “A” owns the car and will always be driving.  People “F” and “G” don’t get along and won’t ride together.  If it makes no difference where in the car people sit, how many ways are there for four people to ride into Portland together?

Continue Reading »

Confounding Calculus

Another fun Calculus problem we ran into the other day involved using the quotient rule versus using the product rule.  This problem also came from the Briggs and Cochran book in the section on Implicit Differentiation (pg. 162 #15).  So, the problem this time started out as

Use implicit differentiation to find \frac {dy}{dx} .

x^3=\frac {x+y}{x-y}

So, if we start out by differentiating the expressions on each side, we get

3x^2=\frac {(x-y)(1+y')-(x+y)(1-y')}{(x-y)^2}

If we expand and simplify a little -

3x^2=\frac {(x+xy'-y-yy')-(x-xy'+y-yy')}{(x-y)^2}

3x^2=\frac {x+xy'-y-yy'-x+xy'-y+yy'}{(x-y)^2}

3x^2=\frac {2xy'-2y}{(x-y)^2}

So then

3x^2(x-y)^2=2xy'-2y

and

\frac{3x^2(x-y)^2+2y}{2x}=y'

This is the answer that appears in the text – but, if we assume that x\neq y , then we can clear the denominator in the original problem and use implicit differentiation on the result.

x^3=\frac {x+y}{x-y}

x^3(x-y)=x+y

x^4-x^3y=x+y

Then, differentiating

4x^3-(3x^2y+x^3y')=1+y'

or

4x^3-3x^2y-x^3y'=1+y'

Then, to isolate y' :

4x^3-3x^2y-1=x^3y'+y'

4x^3-3x^2y-1=y'(x^3+1)

and

\frac{4x^3-3x^2y-1}{x^3+1}=y'

I spent some time playing around with these, and believe me, they are not the same.  I graphed them and here is what the first answer \frac{3x^2(x-y)^2+2y}{2x}=y' looks like:

And here is the second version, \frac{4x^3-3x^2y-1}{x^3+1}=y'

I tried plugging values for x and y into these different expressions for the derivative of the same function to see if I got the same answer.  This led me to realize the solution to the conundrum!

I started by plugging in some simple values for x and y like x=1, y=1 but quickly realized that this wasn’t valid because x\neq y .  So then I tried x=1, y=0 and for these values, the two expressions were the same!

\frac{3(1)(1-0)^2+2(0)}{2(1)}=y'

\frac{3}{2}=y'

and

\frac{4(1)^3-3(1)^2(0)-1}{(1)^3+1}=y'

\frac{4-1}{1+1}=y'

\frac{3}{2}=y'

These are the same – because the point (1,0) satisfies the original function: x^3=\frac {x+y}{x-y} .

So I got out the Excel spreadsheet and found a series of points that satisfied the original function and tried a few of these points in the two expressions for the derivative and for each one the value of the two expressions is the same!

So next I was determined to show that assuming the original function was true would show the equality of the two different expressions for the derivative.

So, assuming x^3=\frac {x+y}{x-y} we need to turn \frac{4x^3-3x^2y-1}{x^3+1}=y' into \frac{3x^2(x-y)^2+2y}{2x}=y' .

First I substituted \frac{x+y}{x-y} in for x^3 , so:

\frac{4x^3-3x^2y-1}{x^3+1}=y'

\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+1}=y'

\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+\frac{x-y}{x-y}}=y'

or

\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'

Then I cleared the fraction in the numerator and denominator:

(\frac{x-y}{x-y})\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'

\frac{4(x+y)-3x^2y(x-y)-1(x-y)}{2x}=y'

OK, so this is

\frac{3x^2y^2-3x^3y+4x+4y-x+y}{2x}=y'

or

\frac{3x^2y^2-3x^3y+3x+5y}{2x}=y'

I saw that where I was headed with this had a 2y and a 3x^2 in it, so I moved a few things around and ended up with:

\frac{3x^2y^2-3x^3y+3(x+y)+2y}{2x}=y'

Going back to the original function x^3=\frac {x+y}{x-y} I saw that I could substitute x^3(x-y) in for the (x+y) :

\frac{3x^2y^2-3x^3y+3(x^3(x-y))+2y}{2x}=y'

\frac{3x^2y^2-3x^3y+3(x^4-x^3y)+2y}{2x}=y'

\frac{3x^2y^2-3x^3y+3x^4-3x^3y+2y}{2x}=y'

Factor out 3x^2 from the everything except the 2y and this is starting to look good:

\frac{3x^2(y^2-xy+x^2-xy)+2y}{2x}=y'

And there it is:

\frac{3x^2(x-y)^2+2y}{2x}=y'

- sigh of relief -

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