Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.165.09 Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Wednesday, September 30 2.9 Differentials Assignment: Page 146: Problems 8, 20, 28, (horizon), (drag) ################################################################ (horizon) If you are high enough to see the horizon 10 miles away, (a) How much farther can you see if you climb up 10 more feet? (b) How high are you when the horizon is 10 miles away? ............................................................... (drag) 2 2 m s Hp = ------- where m is in slugs. One slug = 32 lbs. 3 550 t Suppose that the weight is 1408 pounds +/- 16 lb. s is 1100 feet +/- 10 ft. t is 10 seconds +/- 0.1 sec. Figure the horse power and the approximate error in Hp due to each error in measurement. If you were going to improve the accuracy, which measurement would be the one to spend the time, money, and effort on? ################################################################ Main idea: Estimating the effects of errors on the answer. Key Words: /\x, /\y, dx, dy, differentials Goal: Learn to approximate an answer using the derivative. ################################################################ Previous Assignment: Page 141 Problems 2, 6, 16, 20, Page 141 Problem 2 Assuming that a soap bubble retains its spherical shape as it expands, how fast is its radius increasing when its radius is 3 inches if air is blown into it at a rate of 3 cubic inches per second? 3 V = 4/3 Pi r 2 dV/dt = 4 Pi r dr/dt 2 3 = 4 Pi 3 dr/dt 3 1 dr/dt = ----------- = -------- = 0.0265258 inches per second 2 12 Pi 4 Pi 3 ################################################################ Get["font.math"]; vo = 4/3 Pi 3^3; r[v_] := (3 v/(4 Pi))^(1/3); A = Table[ ParametricPlot3D[ r[vo+3 t] {Cos[u] Sin[f],Sin[u] Sin[f],Cos[f]}, {u,0,1 Pi},{f,0,Pi}],{t,-10,+10,5}]; a = ParametricPlot3D[ {3 Cos[u],0,3 Sin[u]}, {u,0,2 Pi}, PlotStyle->{Orange,Thickness[0.005]},PlotRange->All]; b = Table[ ParametricPlot3D[ {r[vo+3 t] Cos[u],0, r[vo+3 t] Sin[u]}, {u,0,2 Pi}, PlotStyle->{Orange,Thickness[0.003]},PlotRange->All], {t,-10,10,5}]; B = Table[ Graphics3D[{RGBColor[1,0,0],PointSize[0.015], Point[ {x/4 ,0,0}]}], {x,1,13}]; p = Show[A,b,a,B, PlotLabel->"P141 p2 Size each 5 seconds; red dots = 1/4 in", AspectRatio->Automatic,PlotRange->All]; Export["/math/www/hentzel/class.165.09/sep30.p2x.pdf",p]; q = Show[p,ViewPoint->{0,-10,0},PlotRange->All]; Export["/math/www/hentzel/class.165.09/sep30.p2y.pdf",q]; delta = 50; r[v_] := (3 v/(4 Pi))^(1/3); A = Table[ ParametricPlot[ r[3 t] {Cos[u],Sin[u]}, {u,-Pi/delta,Pi/delta},PlotRange->All],{t,35,40}]; a = ParametricPlot[ 3{Cos[u],Sin[u]},{u,-Pi/delta,Pi/delta}, PlotStyle->{RGBColor[1,0,0]},PlotRange->All]; B = Table[ ListPlot[{{n/100,0}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.03]},PlotRange->All], {n,290,305}]; p = Show[A,a,B, PlotLabel->"P141 p2 Size each second; red dots 0.01", AspectRatio->Automatic, Axes->False,PlotRange->All]; Export["/math/www/hentzel/class.165.09/sep30.p2z.pdf",p]; ################################################################ Page 141 Problem 6 A woman on a dock is pulling in a rope fastened to the bow of a small boat. If the woman's hands are 10 feet higher than the point where the rope is attached to the boat and if she is retrieving the rope at a rate of 2 feet per second, how fast is the boat approaching the dock when 25 feet of rope is still out? |\ | \ -2 ft/sec | \ | \ | \ 10 | \ s | \ | \ | \ | x \ |__________\ 2 x = Sqrt[ s - 100] 2 -1/2 dx/dt = 1/2 (s -100) 2 s ds/dt s ds/dt dx/dt = ---------------- Sqrt[s^2-100] -50 -50 dx/dt = ---------------------- = ----------------- 2 5 Sqrt[25-4] Sqrt[25 - 100] -10 dx/dt = ---------- = -2.18218 Sqrt[21] ################################################################ Get["font.math"]; f[s_] := Sqrt[s^2 - 100]; A = Table[ ParametricPlot[ w{0,10}+(1-w) {f[50-2 t],0},{w,0,1}, PlotStyle->{Thickness[0.002]}], {t,0,20}]; a = ParametricPlot[ w{0,10} + (1-w) {f[25],0},{w,0,1}, PlotStyle->{RGBColor[1,0,0]}]; b = Table[ ListPlot[ {{2 t,0}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.005]}],{t,0,25}]; p1 = Show[A,a,b,PlotLabel->"P141 p6 red dots = 2 feet, rope each second", AspectRatio->Automatic,PlotRange->All]; Export["/math/www/hentzel/class.165.09/sep30.p6.pdf",p1]; ################################################################ Page 141 Problem 16 An aircraft spotter observes a plane flying at a constant altitude of 4000 feet toward a point directly above her head. She notes that when the angle of elevation is 1/2 radian it is increasing at a rate of 1/10 radian per second. What is the speed of the airplane. /| / | / | / | 4000 / | / | / u | /_______| x x = 4000 Ctn[u] 2 dx/dt = 4000 (-Csc [u] ) du/dt 2 dx/dt = 4000 (-Csc [1/2] ) 1/10 2 dx/dt = -400 Csc [1/2] = -1740.27 ft/sec = -1740.27 60/88 = -1186.55 miles/hr ################################################################ Get["font.math"]; xo = 4000 Cot[1/2]; A = Table[ ParametricPlot[ w {0,0} + (1-w) {xo+1740.27 t,4000},{w,0,1}], {t,-2,8}]; a = ParametricPlot[ w {0,0} + (1-w){4000 Cot[1/2],4000},{w,0,1}, PlotStyle->{RGBColor[1,0,0],Thickness[0.004]}]; b = Table[ ListPlot[ 4000 {{Cos[t/10],Sin[t/10]}}, PlotStyle->{RGBColor[1,0,0]}],{t,1,10}]; c = ParametricPlot[ {xo-1740.27,y},{y,0,4000}, PlotStyle->{RGBColor[0,0,1]}]; d = ParametricPlot[ {xo ,y},{y,0,4000}, PlotStyle->{RGBColor[0,0,1]}]; e = ParametricPlot[ {xo+1740.27,y},{y,0,4000}, PlotStyle->{RGBColor[0,0,1]}]; B = Table[ ParametricPlot[ {500 n,y},{y,0,100}, PlotStyle->{RGBColor[0,1,0],Thickness[0.005]}],{n,1,32}]; p1 = Show[a,A,b,c,d,e,B, PlotLabel-> "P141 p16 position each second; red=1/10 radian,green=500 ft", AspectRatio->Automatic,PlotRange->All]; Export["/math/www/hentzel/class.165.09/sep30.p16.pdf",p1]; ################################################################ Page 141 Problem 20 Water is pumped at a uniform rate of 2 liters (1 liter = 1000 cubic centimeters) per minute into a tank shaped like a frustrum of a right circular cone. The tank has altitude 80 centimeters and lower and upper radii of 20 and 40 centimeters, respectively. How fast is the water level rising when the depth of the water is 30 centimeters? ################################################################ Note: The volume, V of a frustrum of a right circular cone of altitude h and lower and upper radii a and b is 2 2 V = 1/3 Pi h (a + ab + b ) r[h_] := 20 + 20(h/80); 2 2 v[h_] := 1/3 Pi h (20 + 20 r[h] + r[h] ) 2 v[h_] := 1/3 Pi h( 400 + 20(20+h/4) + (20+h/4) ) 2 v[h_] := 1/3 Pi h( 400 + 400 + 5 h + 400 + 10 h + h /16) 2 v[h_] := 1/3 Pi h (1200 + 15 h + h /16) 2 3 v[h_] := Pi( 400 h + 5 h + h /48 ) 2 dv/dt = Pi( 400 + 10 h + 3 h /48 ) dh/dt ################################################################ dv/dt dh/dt = -------------------------- 2 Pi(400 + 10 h + 3 h /48) 2000 dh/dt = ---------------------------- 2 Pi (400 + 10 30 + 3 30 /48 ) 2000 dh/dt = --------------------------------- Pi (400 + 300 + 900/16 ) 2000 dh/dt = --------------------------------- Pi (400 + 300 + 225/4 ) 8000 dh/dt = --------------------------------- Pi (1600 + 1200 + 225 ) 8000 dh/dt = ---------------- = 0.841811 cm/min Pi 3025 ################################################################ Get["font.math"]; p1 = ParametricPlot[ w {20,0}+(1-w){40,80},{w,0,1}]; p2 = ParametricPlot[ w {-20,0}+(1-w){-40,80},{w,0,1}]; r[h_] := 20+20(h/80); v[h_] := 1/3 Pi (20^2 + 20 r[h] + r[h]^2) h; f[v_] := 2 (-40 + (64000+6v/Pi)^(1/3)); A = Table[ Plot[ f[2000 t],{x,-r[f[2000 t]],r[f[2000 t]]}],{t,1,42}]; a = Plot[30,{x,-r[30],r[30]},PlotStyle->{RGBColor[1,0,0]}]; b = Table[ ListPlot[ {{0, 5 t}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.010]}],{t,1,16}]; p3 = Show[p1,p2,A,a,b, PlotLabel->"P141 p20 depth each minute, red are 5 cm ", PlotRange->All, AspectRatio->Automatic,PlotRange->All, AxesOrigin->{0,0}]; Export["/math/www/hentzel/class.165.09/sep31.p20.pdf",p3]; ################################################################ New Material: The tangent line approximation is relatively close to the function, but not totally accurate of course. * | o------ | * | /|\ | * .___ | /\ | * . ' |/|\ |/__\y | *. ' | |dy | | o_'_________|\|/__\|/_ | * dx | -------------------+------*---*---------|-----------|------ | x x+dx | There are four objects under discussion. dx, dy, /\x, /\y. dx and /\x are the same thing. But dy is NOT the same as /\y. By definition: /\y is the actual change in the y-value. dy is only an approximation. /\y = f(x+dx) - f(x) / dy = f (x) dx / It is almost true that the actual change is equal to f (x) dx. So / /\y "===" dy = f (x) dx almost. ################################################################ Examples of the process of using dy to approximate /\y. 3 2 Example 1. Evaluate f(x) = x + 2 x at x = 1.1. We know that f(1) = 3. Then f(1+dx) should approximately equal / f(1) + f (1) 0.1. Here, dx = 0.1 / 2 / f (x) = 3x + 4x and f (1) = 7. Thus f(1.1) = 3 + 7(0.1) = 3.7 approximately. The correct value is 3.751 ################################################################ 1/3 Example 3. Approximate the value of 65 . --------------------------------------------------- 1/3 Let f(x) = x . / -2/3 f (x) = 1/3 x . / f(65) = f(64) + f (64) 1 = 4 + (1/3)(1/16) 1 = 4 + 1/48 = 4.02083 Since f[65] = 4.02073 the approximation is pretty close. ################################################################ These numerical approximations were really important when I took calculus. That was in the days before hand calculators. Now to approximate the cube root of 65 seems quaint because any reasonable person would use the calculator. But there are other uses of this method which are very useful even with the availability of calculators. This is in the estimation of errors. How do the inevitable errors in the initial measurements affect the final answer? ################################################################ For example: You measure the edge of a cube and know its length is 4 +/- 0.1. What is the error in the volume of the cube? 3 V[x] = x . V[4] = 64 / 2 / V [x] = 3x V [4] = 48 error in area is 48 dx. If the error in measurement were 0.1, the error in volume would be at about 4.8. ################################################################ With the calculator the correct volume should be between V[3.9] and V[4.1]. That is between 59.319 and 68.921 the actual error would be between -4.681 and +4.921 which is about (+/-) 4.8. The calculus method tells you a simpler way to think of how the measurement error affects the volume error. It is easier to think of "48 times dx." ################################################################ Horse Power Example. Work = Force * distance Force * distance Power = ------------------ time We want to measure the horsepower of a car. We plan to measure how fast the car can go a distance s from a standing start. This time depends on the horse power and the weight of the car. ################################################################ We now will argue that the best the car can do is to have constant acceleration. If we run the engine too slow, there are not enough pops per minute in the cylinders to get much power. If we run it too fast, it will fall apart, and even before that, the pistons race down the cylinders ahead of the exploding gasses so that we do not get the full benefit of the explosion. The best thing to do is to run the engine at its peak power output and then use gears to transfer the power to the tires. Since the engine is working at its peak speed, the power is a constant. Since the engine has constant output, the force delivered to the tires is a constant. The acceleration is related to the force by F = mA, therefore the acceleration is a constant. ################################################################ v = a t 2 s = 1/2 a t We can measure the distance s, and the time t. The acceleration 2 s 2 m s a = ------------ force = m a = ----------- 2 2 t t The work done is force*distance force*distance The power is -------------- time force*distance The horse power is --------------- Since 550 ft lbs/sec = 1 hp. 550 time ################################################################ 2 2 m s hp = ------- 3 550 t The variables in the hp equation are given by: m = weight/32; s = distance in feet, and t = time in seconds. We have a car and we made the following measurements. 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 2 * (3200/32) * 1100 (a) HP = ------------------------- = 440 3 550 10 2 2 s HP (b) dHP/dm = ------------------------- = ------ = 440/100 = 4.4 3 m 550 t 4 m s 2 HP (c) dHP/ds = ------------------------- = -------- = 880/1100 = 0.8 3 s 550 t 2 -6 m s -3 HP (d) dHP/dt = ----------------------- = -------- = -1320/10 = -132 4 t 550 t ################################################################ The actual errors are approximately (a) 4.4 (10/32) = 1.375 HP (weight) (b) 0.8 (2) = 1.6 HP (distance) (c) 132 (0.01) = 1.32 HP (time) So if one were to work on a measurement to improve, it would be the distance. That contributes the most error. ################################################################ In Class Problems: (a) Approximate Sin[10 degrees] (b) Approximate the volume of cardboard in a box 10x10x10. Suppose that the cardboard is 1/4 inch thick. ans (a). / y = Sin[x] y = Cos[x] The tangent line at (0,0) is the line y=x. Thus Sin[x] is approximately equal to x when x is small. In particular Sin[10 degrees] is approximately equal to 10 Pi/180 = 0.174533. (Compare this to the actual value of Sin[10 Pi/180] = 0.173648). 3 ans(b). V = s 2 dV/dt = 3s | dV/dt| = 300. dV = 300 1/2 = 150 cubic inches. |s=10 Why is 1/2 used instead of 1/4? ################################################################ Class Problem: An NDB (Non Directional Beacon) radiates a signal from the transmitter at the station. An ADF (Automatic Direction Finder) in the airplane points its needle at the station. In still air if the plane follows the needle it will fly to the station. Here is a recommended technique to tell how far the station is. "Fly at right angles to the arrow for a time t and notice how many degrees your heading changed." (a) In 10 minutes the plane's course changed 8 degrees. How far to the station in minutes? (b) A private pilot wants an (easy) formula to use to get the time. What would you tell him? ################################################################ ans: Time for bearing change x 60 Time to station = ------------------------------- Degrees of bearing change