Patterns in nature are visual regularities of form that occur in the natural world. These patterns can be modeled with mathematics and physics. Natural patterns can include symmetries, fractals, spirals, meanders, waves and dunes, foams and bubbles, arrays, cracks, and stripes (some examples are shown below):
Romanesco Broccoli in fractal form
radial symmetry
bilateral symmetry in a zebra's stripes
radial symmetry in a kiwi
logarithmic spiral of a Nautilus
dune meanders
tessellation array of scales
tessellation array of scales
sand dunes at equivalent angles
crack patterns
inelastic crack patterns
meanders
phyllotaxis Fibonacci spirals
phyllotaxis arrangement
Philosophers, mathematicians, and physicists have applied their skills to the natural world across the ages. Early Greek philosophers Plato, Pythagoras, and Empedocles often studied natural form, hoping to explain the ordered occurrence of patterns. In the 19th century, Joseph Plateau developed the theory of minimal surface area as shown in soap bubble films, and was able to mathematically model the concept. Ernst Haeckel painted thousands of Radiolaria (small marine organisms) to show their symmetry in detail. D'Arcy Thompson extensively studied and modeled the growth patterns of flora and fauna, applying mathematics to spiral growth. Alan Turing established methods for predicting morphogenesis in embryos that would eventually become spots and stripes. Benoit Mandelbrot and Aristid Lindenmayer developed the concept of fractals that can be used to approximate plant growth patterns.
While the models are not always spot on, the conceptual process of predicting the patterns of the natural world has broadened our horizons and increased our appreciation of the beauty of nature.
We started out class by talking about how to get a 5 on the AP test this may. Really all you need to get is 30 points on the FRQ. For the free response question, you have 90 minutes to answer 6 questions, which averages to 15 minutes per question. Nine points are allocated for very specific things.
We worked on questions D and C from the midterm exam, starting with question D, which was fairly straightforward in its processing and grading:
To solve for part a, you could receive points for the following:
For part b, you could receive points for the following:
And finally, for part c, you could receive points for the following:
We then went on to work on part C, which was more tricky in understanding what was expected in your answer. The question is posted below:
We started by analyzing what we know about the functions. Function f is a quadratic function, function g is a cubic, and function h is a power function. What determines that function h is a power function is the variables location in the base, not the power. Power functions are fairly easy to solve, like a cubic or a quartic, and you can use the power rules to find the derivative. To answer part a, there were a couple of ways to show that it is not possible to find a value for a so that f meets the second requirement. One way to solve this is shown below:
The points in this problem are allocated for doing some sort of calculation to find a value of a, and some calculation to show that it does not work for another qualification. To solve part b, you need to find a value for c to receive the point.
In part c, the best way to go about this problem is to look at the critical points, because that is where the derivative changes signs. Using some calculus you can find the derivative, shown below.
An important piece of information here is in the explanation. On these free response questions it is really important to justify your answers with calculus. Use of language is very important. Saying something like "the function" or "it" won't be accepted, because it is too ambiguous and could refer to anything. You have to be careful to check over your sentences and justifications to avoid ambiguity, and make sure to use calculus to prove your answer is correct!
Finally, in part d, four points were available. To solve, you need to take the function and evaluate it at four, but also take the derivative at four and set that equal to one. Using the power rule, you can get a system of equations: two equations with two variables. This can be solved with the graphing calculator or using substitution to find n=4. Then you can solve for k with your n value of 4 and get 256. Points are given not only for finding values and setting the equations equal to one, but also in verifying your answers (h(0)=0, h'(0)=0, and h'>0 for 0 < x < 4). This is given below:
If you're worried about these FRQs, don't be! Over the next few months we will be doing a lot of these problems to prepare for the test.
After going over the midterm free response questions, we went over the answers to the multiple choice questions. The multiple choice can't be released, so we won't go over that, but if you have questions about how to solve the problems I'm sure Mr. O'Brien would allow you to come in for help.
To finish off class, we started looking at a more physics oriented side of calculus. In unit four, we're taking time to go back and look at motion, including but not limited to position, velocity, displacement, distance, and acceleration. For this unit, it's important to know the physics students so that the class can make sure who to ask for help and also be aware of those in the class that have never seen this sort of problem before.
There are a few definitions that may be found helpful at this point. Position, x(t) or s(t), is the location of a particle at time t. Velocity, v(t)=s'(t), is how fast the position is changing. Acceleration, a(t)=v'(t)=s''(t), is how fast the velocity is changing.
We did a fairly confusing exercise where Eliot and Connor were particles, and Mr. O'Brien posed several questions that were more confusing then elucidating. What we did determine was at time 0, the position of Eliot was x(0)=7. For Connor at time 5, the position of Connor was x(5)=-4. Position is a particular number representing a particular place in a line, relative to the point, either to the right or left of the origin. We determined that distance divided by time is equal to average speed, or velocity. The average speed in this case is 11/5, which is also the velocity. However if Eliot moved around before moving towards Connor, it doesn't make sense to take the distance and divide it by the time, because we're not talking about a moment in time. We then started to talk about displacement which is the amount by which a thing is moved from its normal position. We tossed around the idea of whether the displacement of Eliot to Connor is any different from the displacement of Connor to Eliot, and determined that the displacement from Eliot to Connor is -11, while from Connor to Eliot it is 11. This is because Connor is moving in a positive direction while Eliot is moving in a negative direction. Finally, we answered the question of whether the velocity is different from the average speed of Eliot moving to Connor. The average velocity of Eliot moving to Connor is thus -11/5, while the average speed is in fact 11/5.
We finished up class with a few minutes to work on 11 questions from the link on iCal. The answers are posted here:
We started class with a two minute quiz on a few trig values, with the promise of another quiz next class.
Our class objective is to "develop an intuitive understanding of limits, including one sided and non-existent limits."
We branched quickly into a warmup where we looked at different equations for limits, and were supposed to try to scrape an answer together. All of these were bad and wrong so we quickly branched into something different.
Taking the first equation of
we went into geogebra to look at the function. The graph of this function looks like:
There is clearly a large gap in the function at x=1. This makes sense, because in the expression the denominator of (x-1) is zero and the numerator of the expression also equals zero. Rewriting the expression, we came up with a simpler form:
geogebra messed it up- x≠1, so for the bottom it's when x>1
When graphed, the piecewise expression looks like:
(On this graph, the blue is the original function and the black is the subsequent parts from which the piecewise function is made.)
As x approaches 1, the expression has no limit because there is a gap in the graph of the expression. However this does not mean that there are no limits for the expression; there are limits as x approaches other values that are not 1.
The only difference shows up in the graph. Using sliders (k,g(k)), we can see that a point does exist at the function g at 1, however the function f does not have a point there. This can be shown by making a second slider of (k,f(k)), where at 1 there is no point. However, this point does not prove that there is or isn't a limit. The limit is still undefined as x approaches 1. The only difference between the two functions is that at x=1 f(1) does not exist and at g(x) there is a point at g(1).
We then looked at the equation h(x) and the piecewise function p(x). At x=1, there a hole in h(x), however in p(x) the value is 2. Even though there is a missing point, the limit value is 3. The function value doesn't exist, the graph has a hole, but the limit value is 3. The functions are shown in the graph below.
Nonexistent Limits: The easiest way to show a limit does not exist is a graph. For an example, we can use the equation:
The denominator is 0 when x is replaced by 2, so we can't use the substitute to find the limit. From either side, the absolute values of the function values get very large, suggesting that the limit does not exist. As x approaches either an asymptote or a gap, the limit does not exist.
The limit also does not exist when the left hand and right hand limits both exist and are not equal. If the two limits are different, the limit does not exist.
We went over the rules for limits briefly towards the end of the class, which can be found both on page 61 of the textbook and also at this website: <http://www.analyzemath.com/calculus/limits/properties.html>
