Assignment 1, SOLUTIONS

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1 MAT 10: Calculus for the Life Sciences I Pawel Lorek Assignment 1, SOLUTIONS University of Ottawa Problem 1: [4 points] Suppose that every morning a patient receives the same dose of drug. From the dose, the drug concentration in his blood increases by 2. Over the course of 24 hours between doses, 75% of the drug in the blood is removed. (a) Write the linear DTDS for the drug concentration, x t+1 = f(x t ), and find x 4 when x 0 = 88. (b) Draw the updating function and start the cobwebbing process at x = 0.2. (c) Find the equilibrium explicitly. (d) Is the equilibrium stable? Use the stability criterion from class and compare with your cobwebbing. Because formulation of the Problem was ambiguous two solutions are accepted: with DTDS x t+1 = 0.25x t and with DTDS x t+1 = 0.25x t + 2 with x t+1 = 0.25x t (a) Just after receiving the medication the concentration is x n + 2, but after 24 hours, before next dose is applied, 75% of drug is removed, i.e. we have = (x n + 2) 0.75(x n + 2) = 0.25 x n We have x 0 = 88 x 1 = 0.25 x = = 22.5 x 2 = 0.25 x = = x = 0.25 x = = x 4 = 0.25 x = = (b) See Figure 1. (c) Equilibrium x must fulfills x = f(x ) i.e. x = 0.25 x substract 0.25 x from both sides 0.75 x = 0.5 divide both sides by 0.75 x = = = 2 (d) The slope of updating function is equal to 1 4. And 1 4 < 1, so the Slope Criterion implies that x = 2 is a stable equilibrium. Cobwebbing, we should obtain the same equilibrium starting at x

2 0.7 = 0.25 x n y=f(x) = 0.25 x y = x x n Figure 1: Problem 1: Cobwebbing with x 0 = 0.2 with x t+1 = 0.25x t + 2 The analysis is similar, here are just results: (a) x 0 = 88,x 1 = 24,x 2 = 8,x = 4,x 4 =, (b) similar, (c) x = 8, (d) the same. 2

3 Problem 2: [4 points] A group of patients is given a certain dose of a drug once. Two measurements of concentration of the drug in the blood are taken 24 hours apart to determine the rate at which the drug is removed from the blood stream. The measurements are given below. patient initial measurement final measurement (a) Write a DTDS of the form x t+1 = ax t for drug removal and find the value of a. (b) For patient 1, how long will it take until the drug concentration is below 0.1? (c) How long does it take for the initial concentration to decrease by 50%? (d) Now patients are given a dose every 24 hours, i.e., we have the DTDS x t+1 = ax t +b with a as in part (a). How much of the drug has to be given so that the steady state concentration is 6? (a) We can see that a is exactly equal to 1 for each patient, because 1 = 1, = 1.5, = 0.2, = 0.6. Thus x t+1 = 1 x t (b) We have two different (correct) solutions: I: Notice, that the concentration of medication always decrease. Thus it is enough to calculate consecutive steps till first k such that x k < 0.1 It will take 4 days. x 0 = > 0.1 x 1 = 1 = 1 > 0.1 x 2 = 1 1 = 1 = 0... > 0.1 x = 1 1 = 1 9 = > 0.1 x 4 = 1 1 = 1 27 = < 0.1 II: At each step the drug concentration is 1 of what was step before. Thus the solution of DTDS for patient 1 (with x 0 = ) is x t = ( ) 1 t

4 We want to find such t, so that x t 0.1. Apply ln to both sides: ( ) 1 t t t 1 ln(10) (t 1)ln() ln(10) ln() t 1 t 1 + ln(10) ln().0959 It is in days, so we can recalculate it into hours: hours (if we let only whole days, then, as in I, 4 24 = 96 hours is needed) (c) Again, 2 solutions are possible: I: One day is enough, because it will decrease by over 66% which is of course > 50% II: For initial value x 0 the solution is x t = ( 1 t ) x0. We want to find t such that x t 0.5x 0 : ( ) 1 t x 0 0.5x 0 ( ) 1 t 0.5 ( ) 1 t ln ln(0.5) Now we divide both sides by ln( 1 ), notice that it is a negative number, so we have to change into t ln(0.5) ln( 1 ) = So not all day is needed: hours is needed. (d) x t+1 = 1 x t + b Let us find equilibrium, i.e. x such that x = 1 x + b 4

5 x = 1 x + b substract 1 x from both sides 2 x = b multiply both sides by 2 x = 2 b It means, that for any b DTDS x t+1 = 1 x t + b has equilibrium x = 2b. We want this equilibrium to be equal to 6, i.e. 2 b = 6 b = 6 2 = 4 Problem : [4 points] Use a calculator to guess the following limits. (a) e x 1 cos(x) lim, (b) lim. x 0 x x 0 x (a) f(x) = ex 1 x. 5

6 n x n = 1/n f(x n ) n x n = 1/n 2 f(x n ) We see that if series {x n } approaches 0, then f(x n ) approaches. We can conclude that e x 1 lim = 0 x 0 x 6

7 (b) f(x) = cos(x) x n x n = 1/n f(x n ) n x n = 1/n f(x n ) We see that when {x n } approaches 0 from left side, then f(x n ) goes and when the series approaches 0 from right side, then f(x n ) goes to +. Thus we conclude: lim x 0 cos(x) x cos(x) cos(x) does not exist, but we have lim = and lim = +. x 0 x x 0 + x Problem 4: [4 points] The tides at Deep Cove (Nova Scotia) on September 19, 2008 are given below. Time level 2:0 am 7.41m (high tide) 8:0 am 0.41m (low tide) Assume that the tides have a period of 12 hours. Find the parameters in the standard cosine description of these tides, i.e., f(t) = A + B cos(2π(t Φ)/T), where time t is measured in hours and t = 0 indicates midnight. 7

8 First of all let us rewrite the table, so that Time is given in hours from midnight: 2:0 am means that 2.5 hrs passed and 8:0 means that 8.5 hrs past. Time level 2.5 hrs 7.41m (high tide) 8.5 hrs 0.41m (low tide) The value of high tide 7.41 is the maximum, and value of low tide 0.41 is the minimum. Period is given: T = 12 average A: Average is exactly halfway between minimum and maximum: amplitude B: A = 1 2 ( ) = =.91 2 Amplitude is the distance between maximum and average: phase ϕ B = =.5 Phase is the distance between occurrence of first peak and y axis: ϕ = = 2.5 Finally ( 2π f(t) = A + B cos(2π(t Φ)/T) = cos ( π ) = cos 6 (t 2.4), (t 2.4) 12 ) see Figure H T = 2.5 P = 12 A =.5 M =.91 t Figure 2: Problem 4: Tide - sinusoidal approximation 8

9 Problem 5: [4 points] Explain why the following function is continuous. f(x) = cos(e 2x ) x + 7. Evaluate the limit lim x 0 f(x). Explain which rule(s) you use. We will use fact that the following functions are continuous: Type of fun. Example Where cont T1 Polynomial ax + bx 2 + cx + d All x T2 Exponential e ax All x T Cosine cos(x) All x and fact, that the following combinations of continuous functions f(x) and g(x) are continuous: Combination Formula Example C1 Sum f(x) + g(x) 2x e x C2 Composition f(g(x)) e 2x+1 Let f 1 (x) = cos(e 2x ) and f 2 (x) = x + 7 and we have f(x) = f 1 (x) + f 2 (x) f 2 (x) = x + 7 is continuous, because it is a polynomial (T1) f 1 (x) can be rewritten: f 1 (x) = h(g(x)), where g(x) = e 2x and f(x) = cos(x). Note that both of these functions are continuous (T2, T) and f 1 (x) is a composition of them, i.e. also continuous (C2) f(x) is a sum of two continuous functions f 1 (x) and f 2 (x), thus it is also continuous (C1) To calculate lim x 0 f(x) it is enough to plug in the value 0 to f(x), because the function is continuous. lim x 0 cos(e2x ) x + 7 = cos(e 0 ) = cos(1) Bonus problem: [2 points, no part marks] Consider the updating function f(x) = x 2 x + 1 of the DTDS x t+1 = f(x t ). Find the steady state. Carefully draw the updating function and the diagonal. Start cobwebbing from x 0 = 1/2 and again from x 0 = 2. Would you consider the steady state to be stable or unstable? First, let us find the equilibrium, i.e. such x that x = f(x ) x = (x ) 2 x + 1 9

10 0 = (x ) 2 2x + 1 Note that (x ) 2 2x + 1 = (x 1) 2, so we want: (x 1) 2 = 0 0 is the only number which if squared is still 0, thus x 1 = 0 x = 1 The curve y = f(x) = x 2 x + 1 touches the diagonal at point (x,x ) = (1,1). The slope of f(x) at equilibrium seems to be 1, and curve and diagonal do not cross-over. None of the criterions we saw at the lecture apply. 1.5 = x n 2 xn y=f(x) = x 2 x y = x x n If we start cobwebbing from x 0 = 1/2 we approach equilibrium, whereas 10

11 10 = x n 2 xn y=f(x) = x 2 x y = x x n for initial point x 0 = 2 we move further and further away (to + ). We can call this case half-stable. 11