turskyben_887599_53798211_HW10

\0.I 10.1 Application: Finite Difference Rules 304 % 3 pts Go back to problem 9.4 (from your last HW) which had data for the = -h . . . . £ 20 height y; (m) of a pumpkin as a function of time t; (sec). sy 0 322 Q . . T 1 . a) Use the 3-point centered difference rule to calculate the 0 03 R acceleration (d?y/dt?) of the pumpkin at t = 1.5 seconds. (Compare 0 ) . this value to what you already know the acceleration due to gravity 0 T2 3 09 = Time, t; 12 299 should be in m/s?.) b) Use the 4-point backward difference rule to calculate the acceleration (d?y/dt?) 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. 21 188 c) IF you obtained and used measurements every 0.1 seconds (instead of 0.3 seconds, like in the R table of data), quantify how much you’d expect the error in your acceleration calculation from 27 7t 3 0 part (a) to change. q} ¢ 2 .S AX 0.3 ) ¢ - , 24, -2F.+ Fe - 234 le-lf)j’" = -0 4% (). oo = 25 R A dtz AXZ ‘ &(/L( l,_/afd C)' GPWLe Ye=30 YpenT b i £ MP 2 -6F - y 2% _\@81"1[\3»('()‘5-(?\\)—}2[6) —T¢-5% L~ c - ) e T (03] l{:"(%) = ’?’,??S/ ) £) BAx D) 3f i+ gos fom 0300 Fe B of (WYY, Wod go Souin by e fucks 10.2 Fundamentals: Taylor Series 2pts This is the equation for the Taylor Series expansion of f(x+9), 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: &’ &’ o f(x+0)= f(x)+5f'(x)+Tf"(x)'*'?f”'(x)"'zf””(x)""" Start with that equation and use it to derive the expression for f{x;— 5Ax) (i.e. f(x) evaluated five nodes to the left of x;) as a function of f(x;) and all the derivatives at x; (up to the FIFTH derivative). 7 2 — .\ I 4 A \1\3 f f—A,‘\‘.(

/ L o4a) (aaxJ y 22 £x) = ) g £ () Flxi-8ax) = FO<) - (AXFZX)+ 4 (= 54%3 £ (x) 10.3 Application: Error Order and Precision 7pts The following is a 5-point difference scheme, over equally-spaced x;, for d3f/dx3 at x = ;: n 1 )= (/5675 +12f,, —10f,+3f,,) + Error Write out Taylor Series expressions for each of the four fi.3, fi, fi1, fi1to 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(Ax) for some integer p). b) Calculate the precision of the scheme. (e, = fla) s of )+ b APk B ) 4 FaxE ) 4 T 6P L) 9T 7 ) g s O - Qxfle) v % “1l0) - & a3 Y b L) -t P b ) 6y - Fli)-28x 80 £ 205F D) B AXE )1 347 ) e affle) -G PUe) - 30000 ) + 4PV FU3F et FaF ) - By 4 e po2e -3= (DDA -3() =S Laribe2e t2d = L(o)r 5(12)2 2647 () o (M- z(12) 3( -6)-3(1) - 5% =D+ fi 1)+ ‘5( £)4 '?7(‘) S bem TG - dob - wl " —E—La = 55(3)- (%((fl rs’('é 'qatl) s (fi - Ghigt 12 - loF: 4 i) = M(ZA/’F”(& zAx‘Féx.)) " £1x) of ( ) 47. . "?”‘ AXP)szz)fefl": AR (<) tem: el | ACF (%) fem’ o - g ze Sia:z = 70 < AxtE" () rum: 7.£\><3

105) 10.5 Fundamentals: Characterizing ODEs (Chapter 10) 6 pts For each of the following three ODEs ... i. characterize its order (e.g. 1%:-order, 2"-order, etc.), ii. characterize it as an IVP (Initial Value Problem) or BVP (Boundary Value Problem), iii. write the ODE in standard form. 3 (4) qe’=r%+ 2—2‘2], g(2)=0. ¢'(0)=1 (B) y(d;)=sm(x)+ ) (-4) r (©) 1+@(;f )=—4%, x(3)=3, x'(3)=2, x"(3)=1 o) 2nd oxder k) st ofder ¢) 3 oRr or \._/: _A/Y Cinlw) () Bx 4% o Ty | FYER Y2 - X (' O e ) 10.6)

10.6 Application: (Single-Step, Explicit) Euler Method to Solve an IVP 5pts You want to solve the following 1%-order Initial Value Problem: initial condition T(t = 0) = T, = 30. Yo, %o 1 o L =© ‘?—'o*f"g‘Bw -22.8q fi_fi ve | e T Ew _w.ge a3 29 . L:'Z (< T« 30 17.474 43 30 L=3 T t 9 300 dT T 43 t — =——+———— with the dt 18 45 300 Why?? Because your manufacturing floor is at an excruciating T, = 30°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°C. The room has a volume V = (10m) x (20m) x (6m) = 1800 m?, so knowing the density of air = 1.2 kg/m? there’s m = 2160 kg of air you need to cool. Your air-conditioner can only provide 120 kg/min of cold air (at Tac = 10°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). Q= 14400 - 120t Watts of solar heat m =120 kg/min of cold air at T, = 10°C \\ Room temp m = 2160 kg of air, initially at T, = 30°C =T(t) room air leaving at 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 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%t order ODE mc% =mcT . —mecT + Q , which can be rewritten as % 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°C). That’s known to be 1000 J/kg°C. When | plug all the values for the problem in the equation, and convert units so I’'m working with time (t) in minutes, | get the equation | 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). i formula Q = 14400 — 120t, where t is in minutes, and Q in Watts (so it will be at 0 after 2 hours). —T)+2 m=120kg/minof mc (a) Use the (single-step, explicit) Euler method to solve the 15-order IVP Z—_ — where T is the temperature of the room (in °C), and t is time (in minutes). Solve the problem over the range t = [0, 40] minutes, using step size At = 10 min (i.e. calculate T; at t, =0, t; = 10, t, = 20, t3 = 30, and t, = 40 minutes). Show all your work in table form, like we did in class, showing how you start from each (t; T;) to get the next t;,;, “slope;”, and Tj,;. . /.\ —— (b) Make a sketch of T;(t;) (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°C. ar _ T 43 1 18 45 300 7(0)=30 dt &) Tiv= Te tFlTdt t‘;.” > t(: 'f4t A¢ (o,%]) Tt - Ti + Sleget dx - B-6Al(jo) = 2254 - 22.%9 —0.3%‘{00) - 12.3%96 17474 - O 2010) = 16.322