I. Calculus can be applied to many aspects of biology, including personal health. Here is a problem of heartbeats; however, it is not medically accurate, so do not act on the advice of the answer.
II. Population change can be modeled mathematically.
III. The discrete version of the Verhulst model provides a cautionary tale about change. Change can be chaotic, in which case the underlying assumptions of calculus may not be appropriate.
Today's Starbird lecture is all about applying Calculus to Biology. He begins with the question of how much we need to exercise. Suppose we have a limited number of heartbeats. When we exercise our heart beats faster, but our resting heartrate is slower. (This immediately reminds me of the "Count the Ways" song about heartbeats.) So he writes:
u = unit of exercise = 3,600 heartbeats extra (This is about one hour of exercise -- our heartrate goes up from its usual 70 to 130.)
Resting heartrate = R(u) = 25/(u + 1) + 50
T(u) = 10,000R(u) + 3,600u (T = total heartbeats in 10,000 minutes, approx. one week)
T(u) = 10,000(25/(u + 1)) + 50 + 3,600u
T'(u) = (250,000)(-1/(u + 1)^2) + 3,600 + 0
250,000/3600 = (u + 1)^2
50/6 = u + 1
7 1/3 = u
Thus we should exercise about 7 1/3 hours per week. (Don't forget the professor's disclaimer -- he's not a doctor, so don't take this seriously.) But now he supposes, if we hate exercise, then how can we maximize our pleasurable weeks? We must ask, does our model reflect reality?
Total beats/beats per week = weeks
Want:
Max pleasurable wks = (wks)(happy heartbeats/wk)/(total heartbeats/wk)
Starbird's next example is about populations. He begins by showing a graph of population over time and assumes that a population doubles every year. This exponential growth can't last forever. So instead, his next graph shows that the growth rate should slow down as we approach the maximum:
dP/dt = KP(Max - P) (where P = population, K = some constant)
The professor now shows us the actual graph of an elephant population. It clearly increases and decreases, and so our differential equation model above doesn't reflect reality. He now writes the following recursive equation for the number of fish in a pond:
P_(n + 1) = P_n + KP_n(1 - P_n)
Starbird refers to this model is called the Verhulst model. He warns us that this model doesn't always reflect reality, as he shows us the following graphs:
y_(n + 1) = 2.6 * y_n * (1 - y_n) (the population stabilizes)
y_(n + 1) = 3.3 * y_n * (1 - y_n) (the population oscillates between two values)
y_(n + 1) = 3.6 * y_n * (1 - y_n) (the population follows a more complex pattern)
y_(n + 1) = 4.0 * y_n * (1 - y_n) (the population is completely chaotic)
The professor shows us all this to point out that now everything can be described using Calculus.
Today on her Daily Epsilon of Math 2020, Rebecca Rapoport writes:
What is the area (as a multiple of pi) of the largest semicircle that can be inscribed within a rhombus of edge length 6 and with one angle 60 degrees?
This is a very tricky question. We know what a rhombus is, as well as a semicircle, and we know how to find their areas. But it's not quite obvious how the semicircle should be drawn inside the rhombus in order to maximize its area (or its radius). Since one of the given angles is 60, we expect this problem to involve 30-60-90 triangles somehow -- but unless we know how the semicircle is oriented we can't even start to solve the problem.
Let's try drawing a rhombus first. We'll label this rhombus ABCD. All of its side lengths are 6, and so we know AB = BC = CD = DA = 6. We assume that Angle A = 60, which makes Angle C also 60, as well as Angle B = D = 120.
At this point, I had to guess where to draw the semicircle to maximize the radius. Here's my guess -- let O be a point somewhere in the interior of the rhombus. Then we choose points E, F, G, H on the rhombus such that A-E-B (that is, E is between A and B), B-F-C, C-G-D, and D-H-A. All four of these points lie on a circle centered at O. We'll let
Once again, I can't be sure that this really is the largest possible semicircle. But when I draw it, it certainly appears to be the largest.
Now we already know that we're looking for 30-60-90 triangles here, so let's start what math blogger Ben Leis calls an "angle chase" (and other posters call an "angle bash") -- we find as many angles in the diagram as we can and hope that 30-60-90 triangles appear.
I've drawn my semicircle such that sides
Now quad. BEOF has three known angles -- Angle B = 120, Angle BEO = BFO = 90. Thus we find that the last angle of our quadrilateral (which turns out to be a kite) is EOF = 60.
Clearly Angles HOE, EOF, and FOG add up to 180, and EOF = 60. Now here's where symmetry comes in -- I know Angle HOE + FOG = 120, but we need to show HOE = FOG. We know that the symmetry diagonal of a rhombus bisects the angle (that is, DO bisects HDG) and that O is the midpoint of
The proof of this is trickier than it would seem at first:
Given: Ray DO bisects Angle HDG, O is the midpoint of
Prove: Triangles HDO and GDO are congruent.
Proof:
From the givens, we have Angle HDO = GDO and OH = OG. And of course, DO = DO. But unfortunately, this gives us only SSA, which doesn't guarantee congruence.
But SSA does give us some information -- if two triangles satisfy SSA, then the angles opposite the middle S must be either congruent or supplementary. (This makes since if we think about the ambiguous case of the Law of Sines -- the two angles must have the same sine.)
So assume that the angles opposite
So Angles DHO and DGO are congruent. So Triangles HDO and GDO are congruent by AAS. QED
Notice that this makes Angles DOH and DOG a congruent linear pair, so each is 90. Then the angles BOH and BOG are right angles as well. And since we can easily show Angles BOE = BOF (as Triangles BOE and BOF are congruent by HL), we subtract to obtain HOE = FOG as desired. We already found their sum above to be 120, so each is 60.
Then we can draw in sides
In particular, Angle HEO = 60, and since Angle AEO = 90, we conclude Angle AEH = 30. Then two of the three angles of Triangle AEH are known -- Angle A = 60 and AEH = 30. So we can conclude that Angle AHE = 90 -- there is our first 30-60-90 triangle. We know that EH = r, the radius of the circle, and so we can find the shortest leg AH = r sqrt(3)/3.
To find another triangle, we note that since Angle AHE + EHO + OHD = 180 and two of the angles are known, AHE = 90 and EHO = 60, we conclude DHO = 30. Then by symmetry again (the symmetry diagonal of a kite or rhombus bisects the angle), HDO = 60. And so Triangle HDO is another 30-60-90 triangle. Once again, the radius r is the side opposite the 60, but this time we want to find the hypotenuse HD = 2r sqrt(3)/3
But of course AH + HD = AD = 6. So now we finally have an equation for r:
r sqrt(3)/3 + 2r sqrt(3) = 6
r sqrt(3) = 6
r = 2sqrt(3)
And now we can find the area of the circle and semicircle:
r = 2sqrt(3)
r^2 = 12
pi r^2 = 12pi
pi r^2/2 = 6pi
Therefore the area of the semicircle is 6pi. Rapport already supplied us the factor of pi, so all we need to provide is the 6 -- and of course, today's date is the sixth.
Since the answer we found matches the date, it means that all of the assumptions we made earlier turn out to be correct. But how can we actually prove that this is the maximum? Since we're still in the middle of our Starbird lectures, we might think we can prove this using Calculus. But I don't see how this is possible -- Calculus allows us to find the maximum value of a function over some parameter, but there is no simple parameter that categorizes "set of all semicircles in rhombus ABCD."
Lesson 10-7 of the U of Chicago text is called "Volumes of Pyramids and Cones." In the modern Third Edition of the text, volumes of pyramids and cones appear in Lesson 10-4.
This is what I wrote last year about today's lesson:
Lesson 10-7 of the U of Chicago text is on the volumes of pyramids and cones. And of course, the question on everyone's mind during this section is, where does the factor of 1/3 come from?
The U of Chicago text provides two ways to determine the factor of 1/3, and these appear in Exploration Questions 22 and 23. Notice that without the 1/3 factor, the volume formulas for pyramid and cone reduce to those of prism and cylinder, respectively -- so what we're actually saying is that the volume of a conic surface is one-third that of the corresponding cylindric surface. So Question 22 directs the students to create a cone and its corresponding cylinder and see how many conefuls of sand fill the cylinder. The hope, of course, is that the students obtain 3 as an answer. This is the technique used in Section 10.6 of the MacDougal Littell Grade 7 text that I mentioned in last week's post as well. (By the way, I just realized that I mentioned four different math texts in last week's post!)
But of course, here in High School Geometry, we expect a more rigorous derivation. In Question 23, students actually create three triangular pyramids of the same base area and height and join them to form the corresponding prism, thereby showing that each pyramid has 1/3 the prism's volume. But this only proves the volume formula for a specific case. We then use Cavalieri's Principle to show that therefore, any pyramid or cone must have volume one-third the base area times height -- just as we used Cavalieri a few weeks ago to show that the volume of any prism, not just a box, must be the base area times height.
I decided not to include either of the activities from Questions 22 or 23. After all, there will be an activity tomorrow and I wish to avoid posting activities on back-to-back days unless there is a specific reason to.
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