Math 121 - Calculus for Biology I



Math 121 - Calculus for Biology I

Spring Semester, 2001

Chain Rule - Examples | | |

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| |San Diego State University -- This page last updated 16-May-01 | |

[pic]Chain Rule

1. Study of a Beetle Population

2. Chain Rule

3. Analysis of Hassell's Model

4. Worked Examples

5. References

The differentiation in Hassell's model required an additional property of differentiation to complete the computation. Often we encounter composite functions where a function can be written as the composition of two functions. Below we will see that this requires a special rule for differentiation, the chain rule.

Study of a Beetle Population

In the previous section, we examined a special case of Hassell's model for studying the dynamics of insect populations. The more general model, which is applied to insect populations, has the form:

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where a, b, and c are parameters that are chosen to match the data for a population study of some insect. This is a quotient of two functions with the denominator being a composite function, including a linear function, 1 + bPn, which is raised to the c power. The denominator is an example of a function that requires the chain rule.

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In 1946, A. C. Crombie [1] studied several beetle populations to try to better understand their dynamics under the strict control of a constant amount of food regularly supplied. He maintained the amount of food at 10 grams of cracked wheat added weekly, then regularly took census of the beetle populations. These experimental conditions match the assumptions used by the Logistic growth model. One study was on Oryzaephilus surinamensis, the saw-tooth grain beetle. Below is a table of his data (with some minor modifications to fill in times of uncollected data and an initial one week shift).

 

|Week |Adults |Week |Adults |

|0 |4 |16 |405 |

|2 |4 |18 |471 |

|4 |25 |20 |420 |

|6 |63 |22 |430 |

|8 |147 |24 |420 |

|10 |285 |26 |475 |

|12 |345 |28 |435 |

|14 |361 |30 |480 |

By plotting the data of Pn+1 vs. Pn, we can attempt to fit an updating function to the data, then use this updating function to study the population dynamics with an appropriate discrete dynamical model. In this section, we want to study Hassell's model and compare it to the Logistic growth model that we have studied before. Below is a graph showing the best fit to Crombie's data for Oryzaephilus surinamensis for both Hassell's model and the Logistic growth model. (The best fit for Hassell's model was found using the fminsearch routine in MatLab, while the Logistic growth model used Trendline in Excel.)

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With the 3 parameters to fit the curve to the data, Hassell's updating function is capable of fitting the data better than the quadratic function of the Logistic growth model. It also has the added advantage that it never becomes negative, which was a property that we noted was an improvement in Ricker's model. The best model from Hassell's formula is given by:

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while the best Logistic growth model satisfies

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We would like to see how well each of these models simulates the data. With the updating functions, we can readily simulate the discrete dynamical models. Below shows the simulations of the models with the data (assuming the models agree with the data at week 0).

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We can readily see that Hassell's model does appear to simulate the data better, especially having the early rapid rise in the population. All seem to tend toward a similar carrying capacity. For more details on the behavior of these models, we would like to perform our standard analysis of the discrete models, locating equilibria and using the derivative to test the local behavior of these models. Finding the derivative of Hassell's model requires the use of the chain rule. 

Chain Rule

Consider the composite function f(g(x)). Suppose that both f(x) and g(x) are differentiable functions. The chain rule for differentiation of this composite function is given by

f '(g(x))g'(x).

Example 1: Consider the function h(x) = (x2 + 2x - 5)5. Find h'(x).

Solution: This can be considered a composite of the function f(u) = u5 and the function g(x) = x2 + 2x - 5. It is easy to find the derivatives of both f and g. We have f '(u) = 5u4 u' and g'(x) = 2x + 2. From the formula above, we see that

h'(x) = 5(g(x))4 (2x + 2) = 5(x2 + 2x - 5)4 (2x + 2).

Example 2: Consider the function h(x) = exp(2-x2). Find h'(x).

Solution: This can be considered a composite of the function f(u) = eu and the function g(x) = 2 - x2. The derivatives of f and g are f '(u) = eu u' and g'(x) = -2x. From the formula above, we see that

h'(x) = (exp(2 - x2)) (-2x).

See the Worked Examples section for more examples.

Analysis of Hassell's Model

We return to the general Hassell's model to obtain the equilibria and determine stability conditions for the equilibria. Following the usual techniques for studying discrete dynamical models, the equilibria of Hassell's model are found by letting Pe = Pn = Pn+1. Thus, we solve

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This is equivalent to

Pe(1 + bPe)c = aPe .

One of the equilibria is Pe = 0, while the other solves (1 + bPe)c = a. This latter equation is easily solved by taking the cth root of each side, then completing the algebra, so

1 + bPe = a1/c or Pe = (a1/c - 1)/b.

To determine the stability of the equilibria, we differentiate the function H(P). This requires the quotient rule, which in turn requires differentiating the denominator, which is a composite function. Below we use the chain rule to differentiate the term in the denominator.

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Now applying the quotient rule, we find that

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This formula can be used to show that H '(0) = a, but since a > 1 for a positive equilibrium to exist, this condition implies that the zero (or trivial) equilibrium in unstable with solutions monotonically growing away from this equilibrium. At the other equilibrium, we can evaluate H '(Pe) and obtain

[pic]

This calculation shows that the stability of the carrying capacity equilibrium is very dependent on both a and c, but not b. Below we show a series of simulations for Hassell's model with a = 20 and b = 0.01, while c varies. The simulations are for c = 1, 2, 3, and 4. The first three cases are stable, while when c = 4, the model is unstable. We see that for c = 1 the model monotonically approaches its equilibrium at 1900. For c = 2, 3, the solution first oscillates before going to the equilibrium points 347 and 171, respectively. When c = 4, the solution oscillates between 51 and 196.

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References:

[1] A. C. Crombie (1946) On competition between different species of graminivorous insects, Proc. R. Soc. (B), 132, 362-395

Chain Rule - Examples

1. Chain Rule of Differentiation

2. Graphing a Composite Function

3. Hassell's Model

Several examples are presented below to demonstrate the chain rule. Examples show the basic differentiation and applications to extrema in graphing. We also present a couple of applications, including the stability analysis of Hassell's population model and a composite function for studying the effects of energy consumption with increasing population.

Example 1: (Differentiation using the Chain Rule) Differentiate the following functions:

f(x) = (x3 - 4x2 + e-2x)6 and g(x) = (4x + ln(x4 + 4x2 + 1))4

Solution: The function f(x) can be considered a composite of the function f1(u) = u6 and the function f2(x) = x3 - 4x2 + e-2x. It is easy to find the derivatives of both f1 and f2. We have f1'(u) = 6u5 u' and f2'(x) = 3x2 - 8x - 2e-2x. From the chain rule, we see that

f '(x) = 6(f2(x))5f2'(x) = 6(x3 - 4x2 + e-2x)5(3x2 - 8x - 2e-2x).

To analyze the function g(x), we have a couple of composite functions to consider. Let the function g1(u) = u4 and the function g2(x) = 4x + ln(h2(x)). Then the chain rule first gives g1'(u) = 4u3 u' or

g'(x) = 4(g2(x))3g2'(x) = 4(4x + ln(x4 + 4x2 + 1))3 g2'(x).

To find the derivative of g2(x), we need to differentiate the composite h(x) = h1(h2(x)), where h1(v) = ln(v) with h1'(v) = (1/ v)v' and h2(x) = x4 + 4x2 + 1 with h2'(x) = 4x3 + 8x, so

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But g2'(x) = 4 + h'(x). So combining the results above, we see that

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Example 2: (Graphing a Composite Function) An important function in statistics is the normal distribution function, which classically gives the Bell curve. A normal distribution function is given by

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where a is a normalizing factor and σ is the standard deviation. We would like to find the points of inflection for this curve and determine its significance. Also, plot this function for several values of σ.

Solution: To find the points of inflection, we need to take the derivative twice. Its clear that this is an even function, and that its maximum occurs at x = 0 with N(0) = a/σ. We take the first derivative of N(x) and obtain

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Note that N '(x) = 0 at x = 0, as expected.

The second derivative requires the product rule along with the chain rule. The result is given by

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The points of inflection occur when N "(x) = 0, which is easily seen to be when either x = -σ or x = σ. Thus, the points of inflection occur one standard deviation out the normal distribution function. It turns out that 68% of the area under the normal distribution occurs in the interval -σ < x < σ.

Below we show a graph of the normal distribution with σ = 1, 2, 3, and 4.

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Example 3: ( Hassell's Model) Suppose that a study shows that a population, Pn, of butterflies satisfies the dynamic model given by the following equation:

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where n is measured in weeks. Let P0 = 200, then find P1 and P2. Find the intercepts, all extrema of H(P), and any asymptotes for P > 0.

Determine the equilibria and analyze the behavior of the solution near the equilibria.

Solution: The iterations are found in the standard way with

P1= H(200) = 16200/(1.4)4 = 4271.

Similarly, P2= H(4271) = 43. Thus, we see dramatic population swings with this model, suggesting an instability.

Analyzing H(P), we find that the only intercept is (0, 0) and that there is a horizontal asymptote with H = 0. To find where the maximum occurs, we differentiate the function.

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Solving H '(P) = 0, we have 1 - 0.006P = 0 or P = 500/3 = 166.7. With H(500/3) = 4271.5, the maximum occurs at (166.7, 4271.5). A graph of the updating function with the identity function is shown below.

 [pic]

As always, the equilibria are found by solving Pe = H(Pe), which is equivalent to solving

Pe(1 + 0.002Pe)4 = 81Pe.

So either Pe = 0 or (1 + 0.002Pe)4 = 81, which gives 1 + 0.002Pe= 3 or Pe= 1000. The stability of these equilibria can be determined by examining the derivative at the equilibria. At Pe = 0, H '(0) = 81, which implies from our rules that the solutions monotonically grow away from 0. At Pe= 1000, H '(1000) = 81(-5)/243 = -5/3. This implies that the solution near this equilibrium oscillates and goes away from the equilibrium. In fact, this model produces a period 4 solution with the solution asymptotically oscillating from 163 to 4271 to 42 to 2453. A simulation of this model is shown below.

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