Calculus for the Life Sciences I
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1 Calculus for the Life Sciences I ions Joseph M. Mahaffy, mahaffy@math.sdsu.edu Department of Mathematics and Statistics Dynamical Systems Group Computational Sciences Research Center San Diego State University San Diego, CA jmahaffy Spring 2013 (1/29)
2 Outline Weak Acids 1 Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid 2 3 Vertex Intersection of Line and Parabola 4 Height of Ball Lambert-Beer Law (2/29)
3 Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Many of the organic acids found in biological applications are weak acids (3/29)
4 Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Many of the organic acids found in biological applications are weak acids Weak acid chemistry is important in preparing buffer solutions for laboratory cultures (3/29)
5 Formic Acid Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Ants (4/29)
6 Formic Acid Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Ants Formic acid (HCOOH) is a relatively strong weak acid that ants use as a defense (4/29)
7 Formic Acid Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Ants Formic acid (HCOOH) is a relatively strong weak acid that ants use as a defense The strength of this acid makes the ants very unpalatable to predators (4/29)
8 Acid Chemistry Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid The Chemistry of Dissociation for formic acid: HCOOH k 1 H + + HCOO. k 1 (5/29)
9 Acid Chemistry Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid The Chemistry of Dissociation for formic acid: HCOOH k 1 H + + HCOO. k 1 Each acid has a distinct equilibrium constant K a that depends on the properties of the acid and the temperature of the solution (5/29)
10 Acid Chemistry Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid The Chemistry of Dissociation for formic acid: HCOOH k 1 H + + HCOO. k 1 Each acid has a distinct equilibrium constant K a that depends on the properties of the acid and the temperature of the solution For formic acid, K a = (5/29)
11 Acid Chemistry Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid The Chemistry of Dissociation for formic acid: HCOOH k 1 H + + HCOO. k 1 Each acid has a distinct equilibrium constant K a that depends on the properties of the acid and the temperature of the solution For formic acid, K a = Let [X] denote the concentration of chemical species X (5/29)
12 Acid Chemistry Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid The Chemistry of Dissociation for formic acid: HCOOH k 1 H + + HCOO. k 1 Each acid has a distinct equilibrium constant K a that depends on the properties of the acid and the temperature of the solution For formic acid, K a = Let [X] denote the concentration of chemical species X Formic acid is in equilibrium, when: K a = [H+ ][HCOO ] [HCOOH] (5/29)
13 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 1 Based on K a and amount of formic acid, we want to find the concentation of [H + ] (6/29)
14 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 1 Based on K a and amount of formic acid, we want to find the concentation of [H + ] If formic acid is added to water, then [H + ] = [HCOO ] (6/29)
15 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 1 Based on K a and amount of formic acid, we want to find the concentation of [H + ] If formic acid is added to water, then [H + ] = [HCOO ] If x is the normality of the solution, then x = [HCOOH] + [HCOO ] (6/29)
16 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 1 Based on K a and amount of formic acid, we want to find the concentation of [H + ] If formic acid is added to water, then [H + ] = [HCOO ] If x is the normality of the solution, then x = [HCOOH] + [HCOO ] It follows that [HCOOH] = x [H + ] (6/29)
17 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 1 Based on K a and amount of formic acid, we want to find the concentation of [H + ] If formic acid is added to water, then [H + ] = [HCOO ] If x is the normality of the solution, then x = [HCOOH] + [HCOO ] It follows that [HCOOH] = x [H + ] Thus, K a = [H+ ][H + ] x [H + ] (6/29)
18 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 2 The previous equation is written [H + ] 2 + K a [H + ] K a x = 0 (7/29)
19 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 2 The previous equation is written [H + ] 2 + K a [H + ] K a x = 0 This is a quadratic equation in [H + ] and is easily solved using the quadratic formula [H + ] = 1 2 ( K a + ) Ka 2 + 4K a x (7/29)
20 Formic Acid Equilibrium Constant, K a Concentration of Acid Concentration of [H + ] 2 The previous equation is written [H + ] 2 + K a [H + ] K a x = 0 This is a quadratic equation in [H + ] and is easily solved using the quadratic formula [H + ] = 1 2 ( K a + ) Ka 2 + 4K a x Only the positive solution is taken to make physical sense (7/29)
21 Example for [H + ] Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Find the concentration of [H + ] for a 0.1N solution of formic acid (8/29)
22 Example for [H + ] Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Find the concentration of [H + ] for a 0.1N solution of formic acid Solution: Formic acid has K a = , and a 0.1N solution of formic acid gives x = 0.1 (8/29)
23 Example for [H + ] Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Find the concentration of [H + ] for a 0.1N solution of formic acid Solution: Formic acid has K a = , and a 0.1N solution of formic acid gives x = 0.1 The equation above gives [H + ] = 1 2 ( ) ( ) 2 + 4( )(0.1) or [H + ] = (8/29)
24 Example for [H + ] Weak Acids Formic Acid Equilibrium Constant, K a Concentration of Acid Find the concentration of [H + ] for a 0.1N solution of formic acid Solution: Formic acid has K a = , and a 0.1N solution of formic acid gives x = 0.1 The equation above gives [H + ] = 1 2 ( ) ( ) 2 + 4( )(0.1) or [H + ] = Since ph is defined to be log 10 [H + ], this solution has a ph of (8/29)
25 Review of Quadratic Equation: The general quadratic equation is ax 2 + bx + c = 0 (9/29)
26 Review of Quadratic Equation: The general quadratic equation is ax 2 + bx + c = 0 Three methods for solving quadratics: 1 Factoring the equation 2 The quadratic formula 3 Completing the Square x = b ± b 2 4ac 2a (9/29)
27 Example of Factoring a Quadratic Equation Consider the quadratic equation: x 2 + x 6 = 0 Find the values of x that satisfy this equation. Skip Example (10/29)
28 Example of Factoring a Quadratic Equation Consider the quadratic equation: x 2 + x 6 = 0 Find the values of x that satisfy this equation. Skip Example Solution: This equation is easily factored (x + 3)(x 2) = 0 (10/29)
29 Example of Factoring a Quadratic Equation Consider the quadratic equation: x 2 + x 6 = 0 Find the values of x that satisfy this equation. Skip Example Solution: This equation is easily factored (x + 3)(x 2) = 0 Thus, x = 3 and x = 2 (10/29)
30 Example of the Quadratic Formula Consider the quadratic equation: x 2 + 2x 2 = 0 Find the values of x that satisfy this equation. Skip Example (11/29)
31 Example of the Quadratic Formula Consider the quadratic equation: x 2 + 2x 2 = 0 Find the values of x that satisfy this equation. Skip Example Solution: This equation needs the quadratic formula x = 2 ± 2 2 4(1)( 2) 2(1) = 1 ± 3 (11/29)
32 Example of the Quadratic Formula Consider the quadratic equation: x 2 + 2x 2 = 0 Find the values of x that satisfy this equation. Skip Example Solution: This equation needs the quadratic formula x = 2 ± 2 2 4(1)( 2) 2(1) = 1 ± 3 or x = and x = (11/29)
33 Example with Complex Roots Consider the quadratic equation: x 2 4x + 5 = 0 Find the values of x that satisfy this equation. Skip Example (12/29)
34 Example with Complex Roots Consider the quadratic equation: x 2 4x + 5 = 0 Find the values of x that satisfy this equation. Skip Example Solution: We solve this by completing the square (12/29)
35 Example with Complex Roots Consider the quadratic equation: x 2 4x + 5 = 0 Find the values of x that satisfy this equation. Skip Example Solution: We solve this by completing the square Rewrite the equation x 2 4x + 4 = 1 (12/29)
36 Example with Complex Roots Consider the quadratic equation: x 2 4x + 5 = 0 Find the values of x that satisfy this equation. Skip Example Solution: We solve this by completing the square Rewrite the equation x 2 4x + 4 = 1 (x 2) 2 = 1 or x 2 = ± 1 = ±i (12/29)
37 Example with Complex Roots Consider the quadratic equation: x 2 4x + 5 = 0 Find the values of x that satisfy this equation. Skip Example Solution: We solve this by completing the square Rewrite the equation x 2 4x + 4 = 1 (x 2) 2 = 1 or x 2 = ± 1 = ±i This has no real solution, only the complex solution x = 2 ± i (12/29)
38 Weak Acids Vertex Intersection of Line and Parabola The general form of the is f(x) = ax 2 + bx + c, where a 0 and b and c are arbitrary. (13/29)
39 Weak Acids Vertex Intersection of Line and Parabola The general form of the is f(x) = ax 2 + bx + c, where a 0 and b and c are arbitrary. The graph of y = f(x) produces a parabola (13/29)
40 Vertex Weak Acids Vertex Intersection of Line and Parabola Write the quadratic function (recall completing the squares) y = a(x h) 2 + k (14/29)
41 Vertex Weak Acids Vertex Intersection of Line and Parabola Write the quadratic function (recall completing the squares) y = a(x h) 2 + k The Vertex of the Parabola is the point (x v, y v ) = (h, k) (14/29)
42 Vertex Weak Acids Vertex Intersection of Line and Parabola Write the quadratic function (recall completing the squares) y = a(x h) 2 + k The Vertex of the Parabola is the point (x v, y v ) = (h, k) The parameter a determines the direction the parabola opens If a > 0, then the parabola opens upward If a < 0, then the parabola opens downward As a increases the parabola narrows (14/29)
43 Finding the Vertex Weak Acids Vertex Intersection of Line and Parabola Given the quadratic function y = ax 2 + bx + c There are three common methods of finding the vertex (15/29)
44 Finding the Vertex Weak Acids Vertex Intersection of Line and Parabola Given the quadratic function y = ax 2 + bx + c There are three common methods of finding the vertex The x-value is x = b 2a (15/29)
45 Finding the Vertex Weak Acids Vertex Intersection of Line and Parabola Given the quadratic function y = ax 2 + bx + c There are three common methods of finding the vertex The x-value is x = b 2a The midpoint between the x-intercepts (if they exist) (15/29)
46 Finding the Vertex Weak Acids Vertex Intersection of Line and Parabola Given the quadratic function y = ax 2 + bx + c There are three common methods of finding the vertex The x-value is x = b 2a The midpoint between the x-intercepts (if they exist) Completing the square (15/29)
47 Vertex Intersection of Line and Parabola Example of Line and Parabola 1 Consider the functions f 1 (x) = 3 2x and f 2 = x 2 x 9 Skip Example (16/29)
48 Vertex Intersection of Line and Parabola Example of Line and Parabola 1 Consider the functions f 1 (x) = 3 2x and f 2 = x 2 x 9 Skip Example Find the x and y intercepts of both functions (16/29)
49 Vertex Intersection of Line and Parabola Example of Line and Parabola 1 Consider the functions f 1 (x) = 3 2x and f 2 = x 2 x 9 Skip Example Find the x and y intercepts of both functions Find the slope of the line (16/29)
50 Vertex Intersection of Line and Parabola Example of Line and Parabola 1 Consider the functions f 1 (x) = 3 2x and f 2 = x 2 x 9 Skip Example Find the x and y intercepts of both functions Find the slope of the line Find the vertex of the parabola (16/29)
51 Vertex Intersection of Line and Parabola Example of Line and Parabola 1 Consider the functions f 1 (x) = 3 2x and f 2 = x 2 x 9 Skip Example Find the x and y intercepts of both functions Find the slope of the line Find the vertex of the parabola Find the points of intersection (16/29)
52 Vertex Intersection of Line and Parabola Example of Line and Parabola 1 Consider the functions f 1 (x) = 3 2x and f 2 = x 2 x 9 Skip Example Find the x and y intercepts of both functions Find the slope of the line Find the vertex of the parabola Find the points of intersection Graph the two functions (16/29)
53 Vertex Intersection of Line and Parabola Example of Line and Parabola 2 Solution: The line f 1 (x) = 3 2x Has y-intercept y = 3 (17/29)
54 Vertex Intersection of Line and Parabola Example of Line and Parabola 2 Solution: The line f 1 (x) = 3 2x Has y-intercept y = 3 Has x-intercept x = 3 2 (17/29)
55 Vertex Intersection of Line and Parabola Example of Line and Parabola 2 Solution: The line f 1 (x) = 3 2x Has y-intercept y = 3 Has x-intercept x = 3 2 Has slope m = 2 (17/29)
56 Vertex Intersection of Line and Parabola Example of Line and Parabola 3 Solution (cont): The parabola f 2 = x 2 x 9 Has y-intercept y = 9, since f 2 (0) = 9 (18/29)
57 Vertex Intersection of Line and Parabola Example of Line and Parabola 3 Solution (cont): The parabola f 2 = x 2 x 9 Has y-intercept y = 9, since f 2 (0) = 9 By quadratic formula the x-intercepts satisfy x = 1 ± 37 2 or x 2.541, (18/29)
58 Vertex Intersection of Line and Parabola Example of Line and Parabola 3 Solution (cont): The parabola f 2 = x 2 x 9 Has y-intercept y = 9, since f 2 (0) = 9 By quadratic formula the x-intercepts satisfy x = 1 ± 37 2 or x 2.541, Vertex satisfies x = 1 2 and y = 37 4 (18/29)
59 Vertex Intersection of Line and Parabola Example of Line and Parabola 4 Solution (cont): The points of intersection of f 1 (x) = 3 2x and f 2 = x 2 x 9 (19/29)
60 Vertex Intersection of Line and Parabola Example of Line and Parabola 4 Solution (cont): The points of intersection of f 1 (x) = 3 2x and f 2 = x 2 x 9 Find the points of intersection by setting the equations equal to each other 3 2x = x 2 x 9 or x 2 + x 12 = 0 (19/29)
61 Vertex Intersection of Line and Parabola Example of Line and Parabola 4 Solution (cont): The points of intersection of f 1 (x) = 3 2x and f 2 = x 2 x 9 Find the points of intersection by setting the equations equal to each other 3 2x = x 2 x 9 or x 2 + x 12 = 0 Factoring (x + 4)(x 3) = 0 or x = 4, 3 (19/29)
62 Vertex Intersection of Line and Parabola Example of Line and Parabola 4 Solution (cont): The points of intersection of f 1 (x) = 3 2x and f 2 = x 2 x 9 Find the points of intersection by setting the equations equal to each other 3 2x = x 2 x 9 or x 2 + x 12 = 0 Factoring (x + 4)(x 3) = 0 or x = 4, 3 Points of intersection are (x 1, y 1 ) = ( 4, 11) or (x 2, y 2 ) = (3, 3) (19/29)
63 Vertex Intersection of Line and Parabola Example of Line and Parabola 5 Solution (cont): Graph of the functions 25 Intersection: Line and Quadratic y = x 2 x 9 ( 4, 11) y y = 3 2x (3, 3) x (20/29)
64 Height of Ball Lambert-Beer Law Height of a Ball 1 A ball is thrown vertically with a velocity of 32 ft/sec from ground level (h = 0). The height of the ball satisfies the equation: h(t) = 32t 16 t 2 Skip Example (21/29)
65 Height of Ball Lambert-Beer Law Height of a Ball 1 A ball is thrown vertically with a velocity of 32 ft/sec from ground level (h = 0). The height of the ball satisfies the equation: h(t) = 32t 16 t 2 Skip Example Sketch a graph of h(t) vs. t (21/29)
66 Height of Ball Lambert-Beer Law Height of a Ball 1 A ball is thrown vertically with a velocity of 32 ft/sec from ground level (h = 0). The height of the ball satisfies the equation: h(t) = 32t 16 t 2 Skip Example Sketch a graph of h(t) vs. t Find the maximum height of the ball (21/29)
67 Height of Ball Lambert-Beer Law Height of a Ball 1 A ball is thrown vertically with a velocity of 32 ft/sec from ground level (h = 0). The height of the ball satisfies the equation: h(t) = 32t 16 t 2 Skip Example Sketch a graph of h(t) vs. t Find the maximum height of the ball Determine when the ball hits the ground (21/29)
68 Height of Ball Lambert-Beer Law Example of Line and Parabola 2 Solution: Factoring h(t) = 32t 16 t 2 = 16 t(t 2) (22/29)
69 Height of Ball Lambert-Beer Law Example of Line and Parabola 2 Solution: Factoring h(t) = 32t 16 t 2 = 16 t(t 2) This gives t-intercepts of t = 0 and 2 (22/29)
70 Height of Ball Lambert-Beer Law Example of Line and Parabola 2 Solution: Factoring h(t) = 32t 16 t 2 = 16 t(t 2) This gives t-intercepts of t = 0 and 2 The midpoint between the intercepts is t = 1 (22/29)
71 Height of Ball Lambert-Beer Law Example of Line and Parabola 2 Solution: Factoring h(t) = 32t 16 t 2 = 16 t(t 2) This gives t-intercepts of t = 0 and 2 The midpoint between the intercepts is t = 1 Thus, the vertex is t v = 1, and h(1) = 16 (22/29)
72 Height of Ball Lambert-Beer Law Example of Line and Parabola 3 Solution (cont): The graph is Height of Ball 15 Height (ft) 10 5 h(t) = 32 t 16 t t (sec) (23/29)
73 Height of Ball Lambert-Beer Law Example of Line and Parabola 3 Solution (cont): The graph is Height of Ball 15 Height (ft) 10 5 h(t) = 32 t 16 t t (sec) The maximum height of the ball is 16 ft (23/29)
74 Height of Ball Lambert-Beer Law Example of Line and Parabola 3 Solution (cont): The graph is Height of Ball 15 Height (ft) 10 5 h(t) = 32 t 16 t t (sec) The maximum height of the ball is 16 ft The ball hits the ground at t = 2 sec (23/29)
75 Height of Ball Lambert-Beer Law Lambert-Beer Law 1 Concentration and Absorbance (24/29)
76 Height of Ball Lambert-Beer Law Lambert-Beer Law 1 Concentration and Absorbance A spectrophotometer uses the Lambert-Beer law to determine the concentration of a sample (c) based on the absorbance of the sample (A) (24/29)
77 Height of Ball Lambert-Beer Law Lambert-Beer Law 1 Concentration and Absorbance A spectrophotometer uses the Lambert-Beer law to determine the concentration of a sample (c) based on the absorbance of the sample (A) The ion dichromate forms an orange/yellow that has a maximum absorbance at 350 nm and is often used in oxidation/reduction reactions (24/29)
78 Height of Ball Lambert-Beer Law Lambert-Beer Law 1 Concentration and Absorbance A spectrophotometer uses the Lambert-Beer law to determine the concentration of a sample (c) based on the absorbance of the sample (A) The ion dichromate forms an orange/yellow that has a maximum absorbance at 350 nm and is often used in oxidation/reduction reactions The Lambert-Beer law for the concentration of a sample from the absorbance satisfies the linear model c = ma where m is the slope of the line (assuming the spectrophotmeter is initially zeroed) (24/29)
79 Height of Ball Lambert-Beer Law Lambert-Beer Law 2 Spectrophotometer data for an redox reaction Data collected on some known samples A c (mm) (25/29)
80 Height of Ball Lambert-Beer Law Lambert-Beer Law 2 Spectrophotometer data for an redox reaction Data collected on some known samples A c (mm) Determine the quadratic function J(m) that measures the sum of the squares of the error of the linear model to the data (25/29)
81 Height of Ball Lambert-Beer Law Lambert-Beer Law 2 Spectrophotometer data for an redox reaction Data collected on some known samples A c (mm) Determine the quadratic function J(m) that measures the sum of the squares of the error of the linear model to the data Sketch a graph of J(m) and find the vertex of this quadratic function (25/29)
82 Height of Ball Lambert-Beer Law Lambert-Beer Law 2 Spectrophotometer data for an redox reaction Data collected on some known samples A c (mm) Determine the quadratic function J(m) that measures the sum of the squares of the error of the linear model to the data Sketch a graph of J(m) and find the vertex of this quadratic function Sketch a graph of the data and the line that best fits the data (25/29)
83 Height of Ball Lambert-Beer Law Lambert-Beer Law 2 Spectrophotometer data for an redox reaction Data collected on some known samples A c (mm) Determine the quadratic function J(m) that measures the sum of the squares of the error of the linear model to the data Sketch a graph of J(m) and find the vertex of this quadratic function Sketch a graph of the data and the line that best fits the data Use this model to determine the concentration of two unknown samples that have absorbances of A = 0.45 and 0.62 (25/29)
84 Height of Ball Lambert-Beer Law Lambert-Beer Law 3 Solution: Given the linear model c = ma, the sum of square errors satisfies J(m) = e e e e 2 4 = ( m) 2 + ( m) 2 + ( m) 2 + ( m) 2 = m m (26/29)
85 Height of Ball Lambert-Beer Law Lambert-Beer Law 3 Solution: Given the linear model c = ma, the sum of square errors satisfies J(m) = e e e e 2 4 = ( m) 2 + ( m) 2 + ( m) 2 + ( m) 2 = m m The vertex has m v = (0.8024), so (m v, J(m v )) = ( , ) (26/29)
86 Height of Ball Lambert-Beer Law Lambert-Beer Law 4 Solution (cont): Graph of J(m) 5 Sum of Square Errors 4 3 J(m) m (27/29)
87 Height of Ball Lambert-Beer Law Lambert-Beer Law 5 Solution (cont): Since the vertex has m v = , the best linear model is c = m (28/29)
88 Height of Ball Lambert-Beer Law Lambert-Beer Law 5 Solution (cont): Since the vertex has m v = , the best linear model is c = m For an absorbance A = 0.45 c(0.45) = 0.441(0.45) = (28/29)
89 Height of Ball Lambert-Beer Law Lambert-Beer Law 5 Solution (cont): Since the vertex has m v = , the best linear model is c = m For an absorbance A = 0.45 c(0.45) = 0.441(0.45) = The best model predicts a concentration of nm (28/29)
90 Height of Ball Lambert-Beer Law Lambert-Beer Law 5 Solution (cont): Since the vertex has m v = , the best linear model is c = m For an absorbance A = 0.45 c(0.45) = 0.441(0.45) = The best model predicts a concentration of nm For an absorbance A = 0.62 c(0.62) = 0.441(0.62) = (28/29)
91 Height of Ball Lambert-Beer Law Lambert-Beer Law 5 Solution (cont): Since the vertex has m v = , the best linear model is c = m For an absorbance A = 0.45 c(0.45) = 0.441(0.45) = The best model predicts a concentration of nm For an absorbance A = 0.62 c(0.62) = 0.441(0.62) = The best model predicts a concentration of nm (28/29)
92 Height of Ball Lambert-Beer Law Lambert-Beer Law 6 Solution (cont): Graph of Best Linear Model and Data Lambert Beer Law Concentration (mm) c(a) = 0.441A Absorbance (A) (29/29)
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