MAT 1332: Calculus for Life Sciences. A course based on the book Modeling the dynamics of life by F.R. Adler

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1 MT 33: Calculus for Life Sciences course based on the book Modeling the dynamics of life by F.R. dler Supplementary material University of Ottawa Frithjof Lutscher, with Robert Smith? January 3,

2 MT 33: dditional Course Notes The inverse tangent function The tangent function is defined as tan(x = sin(x cos(x, and its derivative can be compute by the quotient rule as d d tan(x = d sin(x cos(x sin(x cos(x cos = cos (x + sin (x (x cos = (x cos (x. In particular, the function is defined for all x that are not odd multiples of π, and the function is monotone increasing, see Figure. as The inverse of the tangent is denoted as arctan or tan and it is defined in the usual way arctan(tan(x = x, tan(arctan(x = x. See Figure. What is its derivative? We differentiate the first equality above by the chain rule and find d [arctan(tan(x] = d dy arctan(y d tan(x =, with y = tan(x, since d x =. Hence, we can divide d dy arctan(y = d tan(x = pplication to integration cos (x cos (x + sin (x = + sin (x cos (x We can now integrate the derivative of the arctan function to get = arctan(x + C. + x = + tan (x = + y.

3 MT 33: dditional Course Notes 5 y=tan(x 5 y!5! x=!!/ x=!/!5!5!!3!! 3 5 x 3 y=arctan(x y=!/ y!! y=!!/!3!!!5 5 x Figure : Graphs of the tangent function and its inverse, the arctangent function

4 MT 33: dditional Course Notes 3 Integrals and volumes First introductory example The E. coli bacterium has a rod-like shape with a cylinder in the middle and two half balls at the ends. What is its volume? The volume of a cylinder with radius r and height h is πr h. The volume of a ball with radius r is πr 3 /3. Now we only have to add the two volumes. For E. coli, the data are approximately r =.8 6 m and h = 6 m. This gives a volume of V = π( m = m. Second introductory example termite mound is approximately cone shaped. What is its volume? If we imagine that we cut a cone into thin horizontal slices (perpendicular to its rotational axis, then each slice is approximately a cylinder. For each of these cylinders, we know how to compute the volume; and then we add the volumes. This procedure is very similar to the Riemann sums for the area under a curve, except that we are now using three-dimensional objects. But it gives us the right idea. General Idea To calculate the volume of an object with rotational symmetry, we need to know the diameter or radius at each point of its rotational axis. This gives the area of the cut surface at each point. Then we find the volume by integrating the area. See Figure. More precisely, assume that the rotational axis is the x-axis and the radius at each point is given by f(x. Then the area of the cut surface at point x is (x = πf (x and the volume of the object between points a and b is given by Example : The cylinder Volume = V = b a (x = π b a (f(x. Rotating a constant function f(x = r around the x-axis gives a cylinder. The volume of the cylinder between x = and x = h is Of course, we knew that before. Example : The ball V = h πr = πr [x] h = πr h. Rotating the function f(x = r x with r x r around the x- axis gives a ball of radius r. The volume is given by r r [ V = π(f(x = π(r x = π r x ] r 3 x3 = r 3 πr3. r r 3

5 MT 33: dditional Course Notes Figure : Illustration of volumes of rotation. The straight line function f(x is rotated around the x-axis. t x = a the radius of the slice is f(a, so the area of the slice is π(f(a. The volume of the cone is given by integrating the area between the base and the top.

6 MT 33: dditional Course Notes 5 Of course, we know that, too. But now we can even compute the volume of a ball with the top and/or bottom removed. For example, what is the volume of the earth (radius km with the north and south pole removed (say 5 km on either end? V = h h π(r x = π [ r x 3 x3 ] h h = π [r h 3 h3 ]. With r = km and h = km we get V = π ( ( ( km 3 = π( km 3 km 3 Example 3: The termite mound Suppose the cone-shaped termite mound has a base diameter of metre and a height of metres. What is its volume? t first, we have to find the radius at any given height. t the bottom, the radius is.5m, at m it is zero. In between, we need a linear function. The function f(x =.5( x/ gives the profile. Then we integrate to get the volume. V = π (f(x = π ( x/ = π ( x+x / = π [ ] x x /+x 3 / = 8π 8.5. The units are, of course, cubic metres. Example : The tree Suppose that a tree trunk is 5 metres high, and that its radius at height x metre is r(x = e x metres. What is its volume? V = π 5 (e x = π The units are, again, cubic metres. Example 5: Trig functions 5 e x = π [ e x] 5 = π( e 5 π. What is the volume obtained by rotating the function f(x = cos ( x with π x π around the x-axis? The volume is given by the formula V = π = π π π π π [f(x] ( cos x. We can t integrate cos θ directly, so we have to use our long-forgotten- but-recently-lookedup-in-the-textbook trigonometric identities. Thus cos θ = cos θ sin θ = cos θ ( cos θ (since cos θ + sin θ = = cos θ. 5

7 MT 33: dditional Course Notes 6 Thus cos θ = cos θ+. Substituting this into the integral, we have π V = π π cos ( x + = π π π (cos x + = π [ ] π sin x + x π = π [(sin π + π (sin( π π] = π (π = π units 3. 6

8 MT 33: dditional Course Notes 7 Partial fractions First introductory example We don t know how to integrate the fraction x x+, but we can write the fraction in a simpler way and use known rules to find the integral as follows: [ x (x + x + = = ] = x ln x + + C. x + x + Second introductory example We don t know integrate the fraction 3x x(x. But if we simplify the fraction as 3x x(x = x + x (check this! then we can integrate as follows: ( 3x x(x = x + = ln x + ln x + C. x General Idea Rational functions are fractions of polynomials, i.e., if P (x and Q(x are polynomials, then P (x/q(x is called a rational function. We already know how to integrate some of them, namely the following building blocks (you need to know these! = ln x + a + C, x + a x + = arctan(x + C = tan (x + C, x x + = ln(x + + C. (You don t have to memorize the last one; you could use substitution to solve it. If we can split a rational function into sums of these building blocks, then we can integrate easily. The goal of this section is to find a technique to integrate (find antiderivatives of all rational functions. We only consider cases where deg(q, i.e., the highest power of x in the denominator is no more than. The idea is to decompose a rational function into a sum of simpler rational functions, namely the three examples above, which we know how to integrate. Recipe for partial fractions To find the integral of a rational function P (x/q(x, follow these steps.. If deg(p deg(q then use long division to split the rational function into several parts. Now assume that deg(p < deg(q. 7

9 MT 33: dditional Course Notes 8. If Q(x = ax + bx + c = a(x x (x x has two distinct real roots, the one can find numbers, B such that P (x Q(x = [ + B ]. a x x x x Then use the natural logarithm to integrate the two terms. 3. If Q(x = ax + bx + c = a(x x has only one real root, the one can find numbers, B such that P (x Q(x = [ ] B + a x x (x x. Then one can integrate using substitution, the logarithm, and direct integration.. If Q(x = ax + bx + c has no real roots, then complete the square to get [ ( Q(x = a x b + c ( ] b a a = a[(x + B]. a Then use the natural logarithm and the arctan to integrate the two terms (potentially substitute first. We illustrate each of these cases with examples. Example P (x = x +, Q(x = x. Then deg(p = > = deg(q, hence we need to do long division. We find x + = (x (x + +. Therefore x [ + x = x + + ] = x + x + ln x + C. x Example P (x = x 3 + 3x + x +, Q(x = x +. gain, since deg(p = 3 > = deg(q, we need to do long division. We find Therefore x 3 + 3x + x + x = + x 3 + 3x + x + = (x + (x [ x ] x = x + 3x + arctan(x + C. + 8

10 MT 33: dditional Course Notes 9 Example 3 P (x = x +, Q(x = x + x. This time, deg(p = < = deg(q, so no long division is necessary. Instead, we factor Q as Q(x = (x (x +, so that x + x + x = x + (x (x +. On the other hand, for two numbers,, B, we find x + B x + ( + Bx + B =. (x (x + Comparing with the expression above, we find that + B = and B =. Hence, = B =. Then we integrate [ ] [ x + x = + x x + ] = ln x + ln x + + C. x + Example P (x = x + 5, Q(x = x x +. gain, deg(p = < = deg(q, so no long division is necessary. But Q(x = (x, has only a single root, i.e., x + 5 x x + = x + 5 (x. On the other hand, for two numbers,, B, we find x + B x + B = (x (x. Comparing with the expression above, we find that = and + B = 5. Hence, =, B = 7. Then we integrate [ ] [ ] x + 5 x = x + x + 7 (x = ln x 7 x + C. Example 5 P (x = 3x +, Q(x = x x + 5. No long division necessary. However, Q has no real roots. We complete the square Q(x = x x + 5 = x x = (x +. Now we write 3x + x x + 5 = 3x + (x + = 3x + ( x +. 9

11 MT 33: dditional Course Notes This is a case for substitution. We choose u = x so that x = u + and = du. Then we get 3x + ( x + = 6u u + du + 5 u + du. The first of these integrals requires another substitution, w = u +, the second is again an acrtan. With this we find 6u u + du + 5 u + du = 3 w dw + 5 u + du = 3 ln w + 5 arctan(u + C. fter back-substituting, we find that the integral with respect to x is given by 3 ln x x ( x arctan + C. Example 6 P (x = x, Q(x = x 3x +. Long division first, or the simpler way x x 3x + = x 3x + + 3x x = + 3x 3x + x 3x +. Now, the denominator is Q(x = (x (x, hence we set the partial fractions as x + B x ( + Bx ( + B = x. 3x + Hence, we need + B = 3 and + B =, which is given by =, B =. Now we can integrate x ( x 3x + = + x + x Practice Problems Find the indefinite integral.. (x+3(x. x x+8 3. x +x+. x x +7x+ 5. x 6 6. x +6 = x + ln x + ln x + C.

12 MT 33: dditional Course Notes 7. x x 6 8. x x 9 x +5x+6 9. x 3 + x +. x +9 x 9

13 MT 33: dditional Course Notes Solutions to Practice Problems. (x+3(x Partial fractions hence we integrate as (x+3(x = (x + 3(x = 5 x+3 + B x = (+Bx+3B (x+3(x ( x + 3 x give = B = /5, and = 5 ln x x + 3. x x+8 Complete the square: x x + 8 = x x + + = (x +, then integrate x x + 8 = (x + = ( x = + 3. x +x+ Complete the square: x + x + = (x + + 9, then integrate x + x + = (x = 9 ( x+ 3 du u + = arctan = ( x arctan 3. x x +7x+ The denominator is x + 7x + = (x + 5(x +. Partial fractions (+Bx++5B (x+5(x+ 5. x 6 give =, B = and then we integrate as ( x + 5 x + x x + 7x + = ( x + C. x+5 + B x+ = = ln x + 5 ln x + + C. The denominator is x 6 = (x (x+. Partial fractions give = B = /8 so that we can integrate as x 6 = ( 8 x = x + 8 ln x x +. x + B x+ = (+Bx+( B (x (x+ +C. 6. x +6 This can be integrated directly x + 6 = 6 ( x = ( x + arctan + C. 7. x x 6 The denominator is x x 6 = (x 3(x+. Partial fractions x+ + B x 3 = (+Bx 3+B (x+(x 3 give = B = /5 and then we integrate as ( x x 6 = 5 x + = x 3 5 ln x 3 x + + C.

14 MT 33: dditional Course Notes 3 8. x x 9 x +5x+6 First of all long division, or write the numerator as x + 5x + 6 9x 5 to see that x ( x 9 x + 5x + 6 = 9x + 5 x + 5x + 6. Then do partial fractions with x +5x+6 = (x+(x+3 and x+ + B x+3 = (+Bx+3+B (x+(x+3 so that = 7 and B =. Then integrate as x ( x 9 x + 5x + 6 = 7 x + = x 7 ln x + ln x C. x x 3 + x + Long division or rewrite the numerator as x 3 + = x(x + x +. Then (with the substitution u = x + in the second term x 3 ( + x + = x x x + + x = x du + u + = x ( x ln x + + arctan + C. ( x + =. x +9 x 9 First of all the numerator: x +9 = x 9+8 and the denominator x 9 = (x 3(x+3. This gives x ( + 9 x 9 = + x 3 + B x+3 = (+Bx+3( B (x 3(x+3 Then partial fractions integrate as x ( + 9 x 9 = + 3 x 3 3 x (x 3(x + 3. Hence = B = 3. Then we = x + 3 ln x 3 x C. 3