CALCULUS 265 Honors Fall Semester 2000 8:00 to 8:50 MTTF Carver 0002 Instructor: Hentzel Office: 432 Carver Office Hours: 9:00-10:00 MTTF E-mail hentzel@iastate.edu phone 294-8141 Website http://www.public.iastate.edu/~mathclasses/honors.265 From: "Bruce H. Wagner" Subject: Computer Lab schedule this week ------------------------------------------------------------------------------- Math Department: The Math Computer Lab is open this week, but it will be closed during some hours in the morning due to staffing problems. This week's schedule is: Tues 11-5, 7-11 Wed 10-11, 12-5, 7-11 Thurs 9-10, 11-5, 7-11 Fri 9-11, 12-5 Sun 1-11 Hopefully, we can start with the regular schedule next week: Mon-Thurs 9-5, 7-11 Fri 9-5 Sun 1-11 _______________________________________________________________________ The hand-in-assignment is to essentially work the same problem that you did for Thursday with four differences. 1) You have different points. Anyway, one of them is different. 2) You work with the actual partials fx and fy rather than the estimated partials. 3) The use of delta is different. We do not use delta in the approximation of fx and fy. Instead, delta now is the size of the step. Delta is the distance on the plot between successive trials. For finer details, make delta smaller. 4) When you get the answer the first time, use this answer as the starting position instead of {10,10} to see how close you can get to making the partials zero. At this step you may wish to change delta to something smaller to make your steps smaller. (* This is a Mathematic Program to study the partial derivatives *) n[{a_,b_}] := Sqrt[a^2 + b^2]; (* Finds the length of {a,b} *) (* Creates a unit vector in direction of {a,b} *) u[{a_,b_}] := 1/Sqrt[a^2+b^2] {a,b}; a = {1,5}; b = {4,2}; (* the three points are {1,5},{4,2},{5,8} *) c = {5,8}; (* Total distance from a,b,c *) f[{x_,y_}] := n[{x,y}-a]+n[{x,y}-b]+n[{x,y}-c]; (* plot the total distance *) p1 = Plot3D[f[{x,y}],{x,0,10},{y,0,10},PlotPoints->50]; (* Contour plot total distance *) p2 = ContourPlot[f[{x,y}],{x,0,10},{y,0,10},ContourShading->False,Contours->80]; p = f[{x,y}]; (* create the expression to partial differentiate *) fx[{x_,y_}] = D[p,x]; (* Partial wrt x *) fy[{x_,y_}] = D[p,y]; (* Partial wrt y *) grad[{x_,y_}] = {fx[{x,y}],fy[{x,y}]}; (* {fx,fy} *) delta = 0.1; H = Table[{0,0},{n,1,1000}]; (* zero out a table *) H[[1]] = {10,10}; (* start at point {10,10} *) (* Iterate the fill of the table *) Do[ H[[n]] = H[[n-1]]-delta*u[grad[ H[[n-1]]] ],{n,2,1000}]; p3 = ListPlot[H]; (* Plot the table H *) p4 = Show[p2,p3]; (* Show contour plot and path together *) (* Print final point and distance *) Print[ H[[1000]]," ",f[ H[[1000]] ]," ", fx[ H[[1000]] ]," ",fy[H[[1000]] ] ]; Main Idea: If a surface has a tangent plane, then that tangent plane pretty much coincides with the surface for a while. Key Words: dx, dy, delta-x, delta-y, Goal: Approximate values using tangent plane approximations. Tangent plane approximations: 2 2 m s The equation for horsepower is hp = ----------------- 3 550 t Where: m = weight/32; s = distance in feet, and t = time in seconds. If: weight = 3200 +/- 10 lbs. s = 1100 ft +/- 2 ft. t = 10 +/- 0.01 sec. (a) Find the horsepower. (b) how does the horsepower change due to the error in weight? (c) how does the horsepower change due to the error in distance? (d) How does the horsepower change due to the error in time? 2 * (3200/32) * 1100^2 (a) HP = ------------------------- = 440 550 10^3 2 s^2 HP (b) dHP/dm = ------------------------- = ------ = 440/100 = 4.4 550 t^3 m 4 m s 2 HP (c) dHP/ds = ------------------------- = -------- = 880/1100 = 0.8 550 t^3 s -3 m s^2 -3 HP (d) dHP/dt = ----------------------- = -------- = -1320/10 = -132 550 t^4 t The actual errors are approximately (a) 4.4 (10/32) = 1.375 HP (b) 0.8 (2) = 1.6 HP (c) 132 (0.01) = 1.32 HP So if one were to work on a measurement to improve, it would be the distance. That contributes the most error. Write the equation of the tangent plane to z = x^2 + y^2 at the point (101,10,1); ans: z-101 = 20(x-1) + 2(y-1); a1 = Plot3D[x^2 + y^2,{x,-10,10},{y,-10,10}]; a2 = Plot3D[101 + 20(x-10)+2(y-1),{x,9,11},{y,0,3}] a3 = Show[a1,a2]