kimbk_891523_53809698_HW10

Numerical Methods HOMEWORK 10 – Due April 17, before 11:59 pm 10.1 Application: Finite Difference Rules Go back to problem 9.4 (from your last HW) which had data for the height y i (m) of a pumpkin as a function of time t i (sec). a) Use the 3-point centered difference rule to calculate the acceleration (d 2 y /d t 2 ) of the pumpkin at t = 1.5 seconds. (Compare this value to what you already know the acceleration due to gravity should be in m/s 2 .) 3 pts Time, t i Height, y i 0 1 2 3 0 10 20 30 b) Use the 4-point backward difference rule to calculate the acceleration (d 2 y /d t 2 ) of the pumpkin at t = 3.0 seconds. Hint : you can find this rule in Table 8-1 (pg 318) of the textbook, or the back of your “Class 30/31” handout. c) IF you obtained and used measurements every 0.1 seconds (instead of 0.3 seconds, like in the table of data), quantify how much you’d expect the error in your acceleration calculation from part (a) to change. 10.2 Fundamentals: Taylor Series This is the equation for the Taylor Series expansion of f ( x + d ), as a function of f ( x ) and all its derivatives at x , that will be given to you on your midterm and exam cheat-sheets: Start with that equation and use it to derive the expression for f ( x i – 5 D x ) ( i.e. f ( x ) evaluated five nodes to the left of x i ) as a function of f ( x i ) and all the derivatives at x i (up to the FIFTH derivative). 2 pts f ( x + δ ) = f ( x ) + δ ʹ f ( x ) + δ 2 2 ʹʹ f ( x ) + δ 3 3! ʹʹʹ f ( x ) + δ 4 4! ʹʹʹʹ f ( x ) + ! 10.3 Application: Error Order and Precision The following is a 5-point difference scheme, over equally-spaced x i , for d 3 f /d x 3 at x = x i : Write out Taylor Series expressions for each of the four f i -3 , f i -2 , f i -1 , f i +1 to the FIFTH derivative, like you did in 8.4, and then combine them using the given difference scheme above to … a) Calculate the discretization error order ( i.e. write the error = O ( D x p ) for some integer p ). b) Calculate the precision of the scheme. ʹʹʹ f ( x i ) = 1 2 Δ x 3 f i − 3 − 6 f i − 2 + 12 f i − 1 − 10 f i + 3 f i + 1 ( ) + Error 7 pts

10.1 Bibel laser hand let drink 18 ala = F"M = A2fXi)+f(X:- = f(1,2) -2f(1.5) + f(1.8) 329.9 - 2(22.1) + 23.4 = -10m/2 ze . 9 b) a = - f (Xi - z) + 4f(xi - 2) - 5f(xi - 1) + 2 f(xi) hi =4-(5x.1) + 12 x 0) 3 - 7.78 mO 2) 29.9-2(27.7 + 23,4 - - 9 · I 12 = Ol - 9% 32 = .09 10.2 f(xi - 5xx) = f(xi) - 56x+(xi) + f() - f(xi) + + (xi)- fN Using the given equation: f(xi)-50xf'(Xi) + "(i)- + (xi) + f(xi)-f(xi) 10.3 1.) f(xi - 3xx) = f(xi) - 30x+(xi) + f"(x ! )- *f(xi)+ f(xi) - xi 2.1 f(xi - 2xx) = f(xi) - 20x+(xi) + f(xi) - ** (xi) + f(xi) - xi 3.) f(xi - ox) = f(xi) - oxf(xi) f(xi)- ** (i) f(xi)- fil 4) f(xi - x) = f(xi) + oxf(xi) f(xi) + * i) *fvi)+i Plug into scheme to get: f"(xi) = (ctx) + (xi) + oxf" " (Xi) - Fox of (Xi)+Error a) Error = 0 (0x b) Lowest order; 3

10.4 1.1 f(xi + 1) = f(xi) + f'(Xi)-x + If "(X:)4x + 5+ (Xi)0x + 0(10x) 2.) f(xi + 2) = f(xi) + 2f(xi)0x + 2(i) + "(Xi)0x + 2(j)f"(xi)0x + 0((0x)) 3.) f(xi + 3) = f(xi) + 3f'(xi)-x + (3)f"(X:)4x + 5(3)+"(Xi)0x + 0(10x) 4.) f(xi + 2) = f(xi) + 4f'(Xi)-x + I()+ "(X:)4x + 5(y)+"(Xi)0x + 0(10x) Want to eliminate its 3rd derivative Mult. Eg1. by 8. Subtract Eq2. 8(fXi + 3) - f(xi + 2) = 7f(xi) + 6f(xi)8x + I f" (Xi)x 5. Mult. Eg. 3 by 2 subtract Eq.H 2 f(xi + x) - f(Xi + n) = - f(xi) + 2f(Xi10x + 2 f"(Xi)0x + 011ox)) 6. Add Eq.S, Eq.6 to get rid 1sder. (fXi + 3) - f(xi + z) + 2 f(xi + x) - f(Xi + n) = 6f(xi) + f"(Xi)(0x) + 0((0x)) divide f"(xi) by x f"(xi) = (8* f(xi - ) - f(xi + z) + 2f(Xi + y) - f(Xi + n) - bf(xi) IX 4 S Ali. Orderik B) i. Order: I ii. BVP ii. BUP iii. = + iii. Y(*)= 0 C) i. Order: 3 + + * = 6

HW10 ( 10.1 – 10.6) due Monday April 17 10.6 Application: (Single-Step, Explicit) Euler Method to Solve an IVP You want to solve the following 1 st -order Initial Value Problem: with the initial condition T ( t = 0) = T 0 = 30. Why?? Because your manufacturing floor is at an excruciating T 0 = 30 o C (about 86 F), when you turn on the air-conditioning. What you want to determine is how long it will take to get the room temperature down to 18 o C. The room has a volume V = (10m) x (20m) x (6m) = 1800 m 3 , so knowing the density of air = 1.2 kg/m 3 there’s m = 2160 kg of air you need to cool. 5 pts dT dt = − T 18 + 43 45 − t 300 m = 2160 kg of air, initially at T 0 = 30 o C m = 120 kg/min of cold air at T AC = 10 o C . mc dT dt = ! mcT AC − ! mcT + ! Q dT dt = ! m m T AC − T ( ) + ! Q mc m = 120 kg/min of room air leaving at T ( t ) . Q = 14400 – 120 t Watts of solar heat Your air-conditioner can only provide 120 kg/min of cold air (at T AC = 10 o C , about 50F). That will mix around with the air in the room, so an equal amount of air will leave the room at whatever the temperature of the room is. Let’s call that T ( t ) . Room temp = T ( t ) That all wouldn’t be too bad, except the sun is still coming in the windows, putting in a lot of heat! It’s near the end of the day, so the amount of solar heat is dropping off with time, according to the formula Q = 14400 – 120 t , where t is in minutes, and Q in Watts (so it will be at 0 after 2 hours). So what does all this mean? You’ll learn in a Thermodynamics class that the temperature of the room T ( t ) will go up and down as the thermal energy in the room goes up and down. The solar heat raises the thermal energy. The difference between the cold air in and room-temp air out lowers the thermal energy. The “energy balance” is given by the 1 st order ODE The “ c ” in the equation represents the “thermal capacity” of the air (how much heat it takes to raise the temperature of 1 kg of air by 1 o C). That’s known to be 1000 J/kg .o C. When I plug all the values for the problem in the equation, and convert units so I’m working with time ( t ) in minutes, I get the equation I started with at the top of the page! So let’s get back to that (since, as you probably figured out by now, you didn’t really have to understand anything in this “Why??” box to solve the problem). , which can be rewritten as (a) Use the (single-step, explicit) Euler method to solve the 1 st -order IVP , where T is the temperature of the room (in o C), and t is time (in minutes). Solve the problem over the range t = [0, 40] minutes, using step size D t = 10 min ( i.e. calculate T i at t 0 = 0, t 1 = 10, t 2 = 20, t 3 = 30, and t 4 = 40 minutes). Show all your work in table form, like we did in class, showing how you start from each ( t i , T i ) to get the next t i+1 , “ slope i ”, and T i+1 . (b) Make a sketch of T i ( t i ) ( i.e. draw and connect the 5 dots), and comment about what time you think (if ever!) the temperature of the room will get down to 18 o C. dT dt = − T 18 + 43 45 − t 300 T (0) = 30