Outline. Calculus for the Life Sciences. Average Height and Weight of Girls 1. Chain Rule. Lecture Notes Chain Rule

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1 Outline Calculus for the Life Sciences Lecture Notes Joseph M. Mahaffy, Department of Mathematics an Statistics Dynamical Systems Group Computational Sciences Research Center San Diego State University San Diego, CA jmahaffy Spring Average of Girls 2 for Special Functions 3 Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (1/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (2/31) Average of Girls Average of Girls Average of Girls 1 Average of Girls Functional relationships where one measurable quantity epens on another, while the secon quantity is a function of a thir quantity This functional relationship is a composite function The ifferentiation of a composite function requires the chain rule Over a range of ages the rate of growth of girls in height is constant Height an age are approximate well by a linear function Height an weight of animals satisfies an allometric moel Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (3/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (4/31)

2 Average of Girls Average of Girls 2 Average of American Girls age height weight age height weight (years) (cm) (kg) (years) (cm) (kg) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (5/31) Average of Girls Average of Girls 3 Least Squares Best Fit: Moel of Height as a function of age h(a) = 6.45a+73.9 Moel shows that the average girl grows about 6.45 cm/yr Height (cm) Average Height of Girls h(a) = 6.45a Age (yr) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (6/31) Average of Girls Average of Girls 4 Allometric Moel: An allometric moel for the height an weight of a girl satisfies Weight (kg) W(h) =.72h 2.17 Average Weight of Girls w(h) =.72 h Height (cm) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (7/31) Average of Girls Average of Girls 5 Composite Moel: The linear moel shows that the average girl grows about 6.45 cm/yr How o we fin the rate of change in weight for a girl at any particular age (between 1 an 13)? The Allometric Moel gives the weight as a function of height Create a composite function of the allometric moel an the linear moel to give a function of the weight as a function of age The chain rule gives the rate of change of weight with respect to age Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (8/31)

3 for Special Functions for Special Functions Example : Consier the composite function f(g(x)) Suppose that both f(u) an u = g(x) are ifferentiable functions The chain rule for ifferentiation of this composite function is given by f x = f u ux Alternately, the chain rule is written x (f(g(x))) = f (g(x))g (x) Example - : Consier the the function h(x) = (x 2 +2x 5) 5 Fin h (x) Solution: Consier the composite of the functions f(u) = u 5 an g(x) = x 2 +2x 5 The erivatives of both f an g are f (u) = 5u 4 an g (x) = 2x+2 From the chain rule h (x) = 5(g(x)) 4 (2x+2) h (x) = 5(x 2 +2x 5) 4 (2x+2) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (9/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (1/31) Example 2 for Special Functions for Special Functions for Special Functions Example 2 - : Consier the the function h(x) = e 2 x2 Fin h (x) Solution: Consier the composite of the functions f(u) = e u an g(x) = 2 x 2 The erivatives of both f an g are f (u) = e u an g (x) = 2x General Derivative of Exponential Function x ef(x) = e f(x) f (x) General Derivative of Logarithm Function x ln(f(x)) = f (x) f(x) General Derivative of Sine x sin(f(x)) = f (x)cos(f(x)) From the chain rule h (x) = e 2 x2 ( 2x) General Derivative of Cosine x cos(f(x)) = f (x)sin(f(x)) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (11/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (12/31)

4 for Special Functions Example 3: Derivative of Cosine Function for Special Functions Example 4: of Differentiation Example 3: Consier the function f(x) = e 3x cos(x 2 +4) Fin the erivative of f(x) Skip Example Solution: This erivative uses the prouct an chain rule f (x) = e 3x ( 2xsin(x 2 +4))+cos(x 2 +4)(e 3x ( 3)) f (x) = e 3x (2xsin(x 2 +4)+3cos(x 2 +4)) Example 4: Consier the function Fin the erivative of f(x) Skip Example f(x) = 3x 2 sin(ln(x+2)) Solution: This erivative uses the prouct an chain rule f (x) = ( 3x 2)( ) x sin(ln(x+2)) +6xsin(ln(x+2)) f (x) = 3x2 cos(ln(x+2)) x+2 +6xsin(ln(x+2)) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (13/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (14/31) for Special Functions Example 5: of Differentiation 1 Example 5: Consier the function f(x) = 4e cos(2x+1) Fin the erivative of f(x) Skip Example Solution: This erivative uses the chain rule f (x) = 4e cos(2x+1) (2sin(2x+1)) : The example for the weight an height of a chil given above foun The weight W as a function of height h is W(h) =.72h 2.17 The height as a function of age is h(a) = 6.45a+73.9 The composite function weight as a function of age W(a) =.72(6.45a +73.9) 2.17 Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (15/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (16/31)

5 2 Composite Function: Weight as a function of age W(a) =.72(6.45a +73.9) : Weight as a function of age W(a) =.72(6.45a +73.9) 2.17 Weight (kg) Average Weight of Girls W(h) =.72(6.45a ) Age (yrs) From the chain rule, the erivative of the weight function is W a = W h h a W h = 2.17(.72)h1.17 an h a = 6.45 Combining these an substituting the expression for h W (a) =.18(6.45a +73.9) 1.17 Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (17/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (18/31) 4 5 Change of Weight of Girls Graph of W (a) =.18(6.45a +73.9) 1.17 Change of Weight of Girls Graph of 6 Change of Weight of Girls W (a) =.18(6.45a +73.9) 1.17 Weight Change (kg/yr) W (h) =.18(6.45a ) 1.17 This graph is almost linear, since it is to the 1.17 power The actual average weight changes are given for the ata above We see that the moel uner preicts the weight gain for oler girls Age (yrs) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (19/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (2/31)

6 1 2 : This is an important function in statistics Gives the classic Bell curve The normal istribution function is N(x) = a σ e (x µ)2 /(2σ 2 ) a is the normalizing factor σ is the stanar eviation µ is the mean of the istribution : Consier N(x) = a /(2σ 2 ) σ e (x µ)2 Fin the maximum an points of inflection Plot this function for several values of σ Discuss the importance of the results Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (21/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (22/31) 3 Solution: Consier The erivative is N x N x N(x) = a σ e (x µ)2 /(2σ 2 ) = a ( /(2σ 2 ) σ e (x µ)2 2(x µ) ) 2σ 2 = a(x µ) σ 3 e (x µ)2 /(2σ 2 ) The erivative is zero when x = µ, so there is a maximum at (µ, a σ ) 4 Solution: The erivative is The secon erivative is 2 N x 2 = a σ 3 2 N x 2 = a σ 3 N x = a(x µ) σ 3 e (x µ)2 /(2σ 2 ) ( (x µ)e (x µ)2 /(2σ 2 ) (1 (x µ)2 σ 2 ) e (x µ)2 /(2σ 2 ) The points of inflection occur at x = µ±σ with N(µ±σ) = a σ e 1 2 ( 2(x µ) ) ) 2σ 2 +e (x µ)2 /(2σ 2) 1 Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (23/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (24/31)

7 5 Solution: Graph of the with µ = an σ = 1,2,3,4 N(x) σ = 2 σ = 3 σ = 4 σ = x Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (25/31) 6 Properties of N(x) = a σ e (x µ)2 /(2σ 2 ) As note above, the mean of the normal istribution is µ The normal istribution is a bell-shape curve centere about its mean The points of inflection occur one stanar eviation, σ, from the mean, µ It can be shown that 68% of the area uner the normal istribution occurs in the interval, [ σ, σ] The area uner a istribution function is important in measuring probabilities an confience intervals for statistics Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (26/31) 1 2 is often use in the stuy of insect populations Consier Hassell s Upating function: H(P) = 81P (1+.2P) 4 Fin the intercepts an any asymptotes for P Fin the erivative of H(P) an etermine all extrema Graph H(P) for P Solution: Hassell s moel satisfies: H(P) = 81P (1+.2P) 4 For H(P), the only intercept is (,) The power of P in the enominator (4) excees the power in the numerator (1), so there is a horizontal asymptote with H = Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (27/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (28/31)

8 3 4 Solution: (cont) Next we fin the erivative of Hassell s moel: 81P H(P) = (1+.2P) 4 H P = 1 P 4(1+.2P) (1+.2P)4 (1+.2P) 8 H P = (1+.2P.8P) 81(1+.2P)3 (1+.2P) 8 H (1.6P) = 81 P (1+.2P) 5 Solution (cont): The erivative is H P = 81 (1.6P) (1+.2P) 5 Critical points satisfy H (P) =, so 1.6P = or P = 5 3 = With H(5/3) = , the maximum occurs at (166.7, ) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (29/31) Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (3/31) 5 Solution (cont): The graph of H(P) = 81P (1+.2P) 4 Upating Function of 4 3 H(P) P Joseph M. Mahaffy, jmahaffy@mail.ssu.eu Lecture Notes (31/31)