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## Integration and the AP test

I was perusing some old AP Calculus exams recently and ran across an interesting problem.  The free-response questions are an interesting bunch.  I won’t analyze or critique them too much except to say that they tend to be kind of the same, without much variety.

The question I was really drawn to presents the graph of a derivative function and asks a series of questions about the maximum/minimum values and points of inflection of the underlying function.  It says that if the graph below is $f(x)$ and $g(x)=\int_2^xf(x)\;dx$, then etc, etc.

The graph of the derivative looks like this:

The test questions based on the graph aren’t all that interesting, but I got really interested in wanting to see the original function.  I suppose you can integrate the piecewise derivative graph and use the identified points to build a piecewise function, but I did this geometrically, since these are all triangles.  Really I was just interested in what the original function looked like – which will appear after the jump for those of you who want to think about this for a minute…

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## Recipe for a Leibniz Quarter Pi

Ingredient: $\frac{1}{1+x^2}$

Divide gently in a long division sauce pan:

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}...$

Integrate briskly over a low flame:

$\int (1-x^2+x^4-x^6+x^8-x^{10}...)dx=$

$=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...$

Evaluate for $x=1$ and let stand at room temperature for 1000 terms for accuracy to three decimal places.

$=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...$

Add lemon zest to taste.

But this doesn’t look like a pi! or taste like a pi!

or a quarter pi!

Ah, but if we return to the original integral

$\int \frac{1}{1+x^2}dx$

and make the trigonometric substitution $x=tan \theta$ so that $dx=sec^2 \theta d \theta$ then:

$\int \frac{1}{1+x^2}dx=\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta$

and

$\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta=\int \frac{1}{sec^2 \theta}sec^2 \theta d \theta$

$=\int d \theta=\theta$

which, if we return to the original substitution $x=tan \theta$, we see that $\tan^{-1} x=\theta$

So, $\int \frac{1}{1+x^2}dx=\tan^{-1}x$, which means that:

$\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...$

and since $\tan^{-1}(1)=\frac{\pi}{4}$, then

$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...$

Enjoy your pi!

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## The Many Solutions of the Population Equation

I never studied the logistic equation as a student.  I first encountered this relationship as an instructor in one of the College Algebra textbooks I was reviewing and/or teaching from and was intrigued by the application of this “growth with constraints” model to a natural resource.  In researching applying the logistic model to natural resource consumption, I immediately ran into M. King Hubbard’s work on Peak Oil.

Then, when I was teaching integral calculus last winter, we began a unit on separable differential equations.  I was poking around looking for good application problems that would utilize separable ODEs and ran into the fundamental population differential relationship $\frac{dP}{dt}=kP$, followed by the relationship I had used for the logistic $\frac{dP}{dt}=kP(1-\frac{P}{N})$ with $N$ defined as the “carrying capacity” or maximum growth for the population.

We went through the procedures for solving each of these ODEs (as well as the continuous mixing problems which follow a similar pattern) and then we moved on.

This year when I was teaching this topic again I was reminded of a paper my thesis advisor had given to me back in 2003 about the application of differential equations to modeling fish populations.  I was intrigued by the profusion of models that could be generated by changing the constraints for a given relationship.

Since I didn’t teach integral calculus for another 10 years after I read that paper, I had essentially forgotten most of the equations, formulas and relationships that generated the graphs that had stuck with me.  This year, while covering the separable ODEs with their applications, I began to look into the application of these relationships to fish populations and population in general.

I found two great resources that go through the set up of these relationships in a very clear manner, and each of them includes wonderful graphs showing the multiple solutions that result when the same differential relationship is solved with different initial conditions.

The opening section of Robert Borelli and Courtney Coleman’s Differential Equations: A Modeling Approach can be read here.

A student project from James Madison University written by Bailey Steinworth, Yuhui Wang and Xing Zhang can be read here.

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## The Farmer and the Fencing (Lagrangian)

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.

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## The Farmer and the Fencing

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 $m$by $n$ 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.

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).

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.

The 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.

This 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:

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.

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## Infinite Series Approximations for Trigonometric functions from 14th century India

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.

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## Deriving the Area of a Circle

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 $A=\frac {1}{2} ab\sin C$.

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!

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