Optimization and Newton s method

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1 Chapter 5 Optimization and Newton s method 5.1 Optimal Flying Speed According to R McNeil Alexander (1996, Optima for Animals, Princeton U Press), the power, P, required to propel a flying plane at constant speed, u > 0 is P = Au 3 + BL2 u, where L > 0 is lift, and A, B > 0 are positive constants. The first term is power needed to push the aircraft through the air, whereas the second term is the power needed to push air downwards, so that the aircraft can remain up high. (Recall that power represents energy per unit time.) (a) Find the speed that minimizes the power P. (b) Reformulate the problem to determine how energy per unit distance depends on the flight speed. (c) Compute the speed that minimizes the energy per unit distance. (a) To minimize the power, find critical points Solve algebraically for u: dp du = 3Au2 BL2 u 2 = 0 u 4 = BL2 3A. Thus ( ) BL 2 1/4 u =. 3A Determine if this is a minimum or maximum using the second derivative test: d 2 P = 6Au + 2BL2 > 0 du2 u 3 since all quantities in the expression are positive. Hence, by the second derivative test, the solution is a local minimum. v November 26,

2 Science 1 Problems (set by L. Keshet) Chapter 5 (b) Power= Energy per unit time. Hence, dividing both sides by speed leads to Energy per unit time / distance per unit time = Energy per unit distance. Thus, the quantity of interest, that we will call Q, is Q = P u = Au2 + BL2 u 2. (c) We now need to minimize Q, so Solving for u leads to the solution: dq du = 2Au 2BL2 u 3 = 0. ( ) BL 2 1/4 u =. A To check that this, too, is a local minimum, use the sign of the second derivative. d 2 Q = 2A + 6BL2 > 0 du2 u 4 Hence, the solution we found here minimizes the energy spent per unit distance. Note: We can also reason about the solution by sketching a graph of the two functions of interest, as shown in Figure Power (Energy per unit time) Energy per unit distance Figure 5.1: Figure for solution to problem 5.1. The optimal flight speed is higher to minimize the energy per unit distance than to minimize the energy per unit time. v November 26,

3 Science 1 Problems Chapter Optimal Foraging I: To optimize the efficiency of energy intake, it is postulated that an animal would chose a foraging time t for which the ratio of energy intake to total time spent in searching and foraging for food would be maximized. Suppose that F(t) is the energy gained (in calories) by foraging in a single patch for a time duration t. Further, suppose that it takes an additional (constant) time t 0 > 0 to travel back and forth to the food patch. The total time available to the animal, daylight hours must satisfy 0 t + t We will define the Efficiency of foraging, R(t), as the ratio Answer the following questions R(t) = F(t) t + t 0. (a) It is commonly assumed that F(0) = 0 and that F(t) is a nonnegative function. What do these assumptions mean, and is either of them ever wrong? (b) Show that critical points of R(t) correspond to values of t for which F (t) = F(t)/(t + t 0 ). (c) Under what condition on the function F(t) is this solution an optimum? (Hint: find the second derivative of R(t) and simplify your expression to deduce what must be true about F (t).) (d) Now consider the function F(t) = t 3. based on part (c), what do you conclude about the optimal foraging time? (Note: if you understand parts (b) and (c) you do not need to repeat any calculations.) (a) F(0) = 0 means that no energy is gained if the animal spends no time at all in the food patch. F(t) is nonnegative means that the energy gain is positive, i.e. that the animal does not lose energy by spending time in the patch. If the food patch is empty (e.g. over-exploited) then there is a possibility that searching for food will consume energy, rather than lead to a gain. In that case, F(t) might be negative. (b) Critical points occur when R (t) = 0. By the quotient rule, R (t) = F (t)(t + t 0 ) F(t)(1) (t + t 0 ) 2 = F (t)(t + t 0 ) F(t) (t + t 0 ) 2 = 0. Thus so F (t)(t + t 0 )) F(t) = 0, F (t) = F(t) (t + t 0 ). (c) We can check that this solution an optimum using the second derivative test, R (t) = [F (t)(t + t 0 ) F(t)] (t + t 0 ) 2 2(t + t 0 )[F (t)(t + t 0 ) F(t)] (t + t 0 ) 4 v November 26,

4 Science 1 Problems (set by L. Keshet) Chapter 5 Expanding the derivative in the numerator leads to R (t) = [F (t)(t + t 0 ) + F (t) F(t) ](t + t 0 ) 2 2(t + t 0 )[F (t)(t + t 0 ) F(t)] (t + t 0 ) 4 R (t) = F (t)(t + t 0 ) 3 2(t + t 0 )[F (t)(t + t 0 ) F(t)] (t + t 0 ) 4 At the critical point, since F (t)(t + t 0 ) F(t) = 0, this expression can be simplified further to R (t) = [F (t)(t + t 0 ) + F (t) F(t) ](t + t 0 ) 2 = [F (t)(t + t 0 )] = F (t) (t + t 0 ) 4 (t + t 0 ) 2 (t + t 0 ). We see from the above that the sign of the second derivative of R is the same as the sign of the second derivative of F at the critical point. (Restated, the concavity of r is the same as the concavity of F at the critical point.) For a local maximum of R, we need the second derivativer (t) < 0, which implies (by the above calculation) that F (t) < 0. This means that the function F(t) must be concave down at the optimal foraging time. (d) If F(t) = t 3 then F is concave up everywhere, so this contradicts the condition for a local optimum derived in (c). Thus, the optimum would occur at an endpoint of the time interval, i.e. at t + t 0 = Optimal Foraging II: Suppose that the energy gained by foraging for a time t is given by the function F(t) = Kt a + t, (in calories), and let t 0 be the time it takes to travel to the food patch and back. Explain the meaning of the positive constants K, a. Find the time that maximizes the efficiency, R, (where R is as defined in Problem 5.2). Your answer should be in terms of constants that appear in the problem. Once you have obtained the answer, indicate how the optimal foraging time would vary if the travel time is longer, or if it takes more time to find food in the patch. The constant K represents the largest amount of energy that can be obtained from the patch. The constant a represents the time it takes to get half the energy from the patch. (To see this, plug t = a into F, and see that F(a) = K/2.) It is significant that the function F(t) is concave down in this problem. (This can be verified easily by sketching the curve, or by taking the second derivative and noting that it is negative.) By Problem 5.2, the time at which the efficiency is maximal corresponds to the solution of the equation F (t) = F(t) (t + t 0 ). v November 26,

5 Science 1 Problems Chapter 5 Since F(t) = Kt/(a + t), we compute the derivative Then the equation to solve is F (a + t) 1(t) (t) = K = K a (a + t) 2 (a + t) 2. K a (a + t) = K t 1 2 (a + t) (t + t 0 ) a (a + t) = t 1 (t + t 0 ) a(t + t 0 ) = t(a + t) at + at 0 = at + t 2 Canceling common terms and simplifying leads to t 2 = a t 0, t = (a t 0 ) 1/2. By Problem 5.2, this is a critical point of the function R(t). Also by Problem 5.2, R(t) will be concave down when F(t) is concave down. This reasoning establishes that we have found a value of t that maximizes the efficiency, R(t), i.e. it guarantees that we have the right type of critical point. Thus the optimal time is an increasing function of the travel time, t 0, and the time to extract food from the patch, a. If either of these constants get larger (e.g. local food is depleted, or harder to find), than the optimal foraging time also gets larger (i.e. the animal should spend more time looking for food for maximal efficiency.) 5.4 Optimal Foraging III: While an animal is active, it uses energy at some rate, ɛ > 0 per unit time. We will here assume that this rate is constant, and is the same when the animal is foraging or traveling to the food patch. Suppose we redefine the efficiency of the animal as R 2 (t) = Energy gain Energy consumed. Total time Let the Total time = t + t 0, as before, where t 0 > 0 is constant travel time to a food patch and t is time foraging in the patch. Show that the optimal foraging time is still the same as in Problem 5.2, despite the new formulation of efficiency. Would this be true even if the animal consumes energy at a higher rate while foraging than while traveling? The energy consumed during the active period of the individual is the rate of consumption multiplied by the time, i.e. ɛ(t + t 0 ). Therefore, the new definition of efficiency is R 2 (t) = F(t) ɛ(t + t 0) (t + t 0 ) = F(t) (t + t 0 ) ɛ(t + t 0) (t + t 0 ) = F(t) (t + t 0 ) ɛ. v November 26,

6 Science 1 Problems (set by L. Keshet) Chapter 5 But ɛ is a constant. This means that the new definition of efficiency is simply a constant subtracted from the old one. Thus R 2(t) = R (t), as before, i.e. the derivative of this new function is the same as the derivative of the previous efficiency function, and therefore the critical points are the same as well. If the animal consumes energy at a higher rate while foraging, the above cancellation will not work, and the critical points will be different. 5.5 Optimal foraging IVa: In some cases, it takes time for an animal to start extracting energy from a food patch, e.g. if burrowing, or digging is required before the food is obtained. Suppose that the energy gain function is given by F(t) = Kt2 a 2 + t 2. We will assume that all other conditions (i.e. constant travel time t 0, definition of efficiency R =Energy gain / Total time spent, etc) are the same as in Problem 5.2, and that the goal is to maximize efficiency, as before. We will investigate this problem in a number of steps. (a) Explain why the function F(t) selected here might represent the scenario described in this problem. Explain the meanings of the positive constants K, a. (b) Show that maximizing the efficiency with respect to foraging time leads to a cubic equation for t. (c) With K = 3 and a = 1 hrs, and t 0 = 1 hrs, use the spreadsheet to find a graphical solution, i.e. draw the graph of F(t). On the same graph, draw a straight line from the point ( t 0, 0) with some positive slope. Adjust the slope until the line meets the curve y = F(t) at a point of tangency. (This line is called a rooted tangent. You will have to find the slope of this tangent line by trial and error using your spreadsheet, since solving a cubic equation is not very easy to do analytically.) Use your graph to read off the value of t, at which the tangent line meets the curve (to two significant digits). Explain why this value of t corresponds to the optimal solution. See Problem 5.6 for a more accurate solution to this optimization problem using Newton s Method for approximating zeros of polynomials. (a) The function F(t) in this problem is sigmoidal, i.e. it has low values for small positive t and then increases sharply for intermediate values of t. This means that initially the energy gain is not proportional to time spent, it is lower. After a while, the energy gain increases with time and later on, as energy is depleted, F(t) saturates, i.e. approaches a constant K. The constants K, a have the same meanings and units as in Problem 5.2. v November 26,

7 Science 1 Problems Chapter 5 (b) We now compute the derivative of F and find: F(t) = Kt2 a 2 + t 2. F (t) = K 2t(a2 + t 2 ) t 2 (2t) (a 2 + t 2 ) 2 = K 2ta2 (a 2 + t 2 ) Since we are still optimizing the efficiency, R(t), the critical point still satisfies the general equation F (t) = F(t) (t + t 0 ). Plugging in the derivative and the new function into this equation leads to Simplifying this leads to the cubic equation K 2ta2 (a 2 + t 2 ) = Kt2 1 2 (a 2 + t 2 ) (t + t 0 ) 2a 2 (a 2 + t 2 ) = t (t + t 0 ) 2a 2 (t + t 0 ) = t(a 2 + t 2 ) t 3 a 2 t 2a 2 t 0 = 0. In general, this type of equation is not convenient to solve analytically. (c) The graph produced by the spreadsheet is shown in Figure 5.2. To produce this graph, the function F(t) given in this problem was plotted on a coordinate system scaled so that the horizontal axis shows the point t = t 0. A straight line through this point has an equation of the form y = m(t + t 0 ) where m is the slope. Various values of m were used until the line approximately touches the graph. The value of m that gave the desired slope was m From the graph it can be seen that the tangent line meets the curve at roughly t = t 1.5 hrs, which is thus the optimal foraging time. This graphical solution can be understood by interpreting the ratio F (t) = F(t) (t + t 0 ). The left hand side (LHS) is tangent line slope. The RHS can also be interpreted as a slope, i.e., the ratio of height to width of a triangle with height F(t) (green line) and base (t + t 0 ) (blue line segment connecting t 0 and t along the horizontal axis). The tangent has been so arranged that the two slopes are the same. v November 26,

8 Science 1 Problems (set by L. Keshet) Chapter Optimal Foraging <= Rooted tangent F(t)=(K t^2)/(a^2 + t^2) <= Point of tangency t_0 t* <= Optimal foraging time Figure 5.2: Figure for solution to problem 5.5. This is the graphical solution to the problem of maximizing the efficiency, R(t) of foraging. t 0 = 1 is the time to travel to the patch. The red curve shows F(t), the energy gained by foraging for a time t in the patch. The optimal foraging time is t, the time at which the tangent line to the curve intersects the point (t 0, 0). 5.6 Optimal foraging IVb and Newton s method Solution In this question, you are asked to use Newton s Method and the spreadsheet to find the optimal foraging time in Problem 5.5 part (c) to 5 digits of accuracy. Assume that K = 3 and a = 1 hrs, and t 0 = 1 hrs, and use the energy gain function F(t) in Problem 5.5. If you have already done Problem 5.5, you need not redo your derivative calculations. Simply find a solution to the cubic equation derived in that problem. In Problem 5.5, we showed that the optimal foraging time satisfies the cubic equation t 3 a 2 t 2a 2 t 0 = 0. v November 26,

9 Science 1 Problems Chapter 5 Plugging in the values of the constants a = 1, t 0 = 1, and calling the resulting polynomial P(t), leads us to the cubic equation P(t) = t 3 t 2 = 0, or restated with the variable x, P(x) = x 3 x 2 = 0. We will need the derivative of this function, namely P (x) = 3x 2 1. For Newton s method, we need an initial estimate x 0 (We avoid using t or calling this initial guess t 0 to prevent confusion with the constant travel time.) We also use the Newton Formula recipe to generate successive values, i.e. x 1 = x 0 P(x) P (x). Pick x 0 = 1 for an arbitrary positive initial guess for the root. Then x 1 = x 0 x3 0 x 0 2 3x = = 2. The successive values are best determined using the spreadsheet. We input x 0 = 1 in row 0 of some column, and implement the recipe to determine the successive values. They are x 2 = , x 3 = , x 4 = x 5 = = Thus the optimal foraging time, to 5 significant figures is t = Iterate x 0 f(x 0 ) f (x 0 ) x 1 = x 0 f(x 0 )/f (x 0 ) Table 5.1: Decimal value for solution to the optimal foraging time using Newton s method, starting from the initial guess x 0 = 1. v November 26,

MATH 114 Calculus Notes on Chapter 2 (Limits) (pages 60-? in Stewart)

Still under construction. MATH 114 Calculus Notes on Chapter 2 (Limits) (pages 60-? in Stewart) As seen in A Preview of Calculus, the concept of it underlies the various branches of calculus. Hence we