Discover the Material (Total) Derivative BRIEF ABSTRACT How does a cyclist perceive temperature changes as she pedals between two cities of different and changing temperatures? This classroom or homework exercise offers a fairly realistic and understandable application of the chain rule for partial derivatives. GENERAL INFORMATION FileName: MatDeriv Full title: Discover the Material (Total) Derivative! Last Update: 6/5/96 Developer: Jerry Fine Department of Mechanical Engineering Rose-Hulman Institute of Technology Terre Haute, IN 47803 Phone: 812-877-8353. Email: fine@nextwork.rose-hulman.edu FAX: 812-877-3198 Contact: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Klebanoff@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM Two cities are separated by a distance of 100 km. The temperatures (in degrees Celcius) in the cities varied as follows during the morning hours. Time 7 a.m. 8 a.m. 9 a.m. 10 a.m. 11 a.m. City A temp. 10.0 12.0 13.5 15.5 18.2 City B temp. 14.0 17.0 19.5 21.7 22.9 A bicyclist leaves City A at 7. a.m. headed towards City B at a speed of 30 km/hr. At 9 a.m she is passing a rest area 60 km from City A. Your task is to find the RATE OF CHANGE OF TEMPERATURE OBSERVED BY THE CYCLIST under the following conditions: (a) as she pedals past the rest stop going 30 km/hr towards City B (b) if she halted at the rest stop at 9. am. (c) if she suddenly reversed direction at 9.a.m. and started pedaling at 30 km/hr back to City A. IMPORTANT: You may assume that the temperature always varies linearly as a function of position between the cities. KEYWORDS Partial differentiation with respect to time, and position. Total derivative, material derivative, derivative following the motion, Numerical approximations to partial derivatives, chain rule for partial derivatives. TEACHER NOTES ISSUES RELATED TO THE PROBLEM This problem is designed to illustrate an important use of partial derivatives. It focuses on a quantity that is often encountered when there is a particle moving through a field: the material derivative, (also called the substantial derivative, or total derivative.) Suppose that we think of a temperature field, T, which depends on time, t, and on a spatial position x. Suppose also that x may be thought of as a function of time. The functionality is T(t, x(t) ). The total derivative is expressed using the chain rule for differentiating functions. Material partial derivative partial derivative derivative of Derivative of = of T with respect + of T with respect x with respect to t T with respect to t to x to t DT/Dt = Partial T/Partial t + Partial T/Partial x dx/dt Note that the derivative dx/dt may be thought of as the velocity, v, of a particle which is moving through the temperature field. The total derivative, then, is the time rate of temperature that this particle would experience. Material partial derivative partial derivative Derivative of = of T with respect + of T with respect velocity T with respect to t to x to t DT/Dt = Partial T/Partial t + Partial T/Partial x v This derivative is the one that the student is required to compute. Prerequisites This exercise requires some knowledge of partial derivatives and of the chain rule, although clever students not familiar with these concepts may "discover" them by thinking the problem through. Time allotment - time management The problem is quite short. Forty minutes should suffice. Expectations The student will recognize that there is a special derivative that can be used to characterize the rate of change experienced by an observer moving through a non uniform, non steady field. Finding this derivative is a consequence of the chain rule, plus some thinking. Future payoffs Understanding this problem will give the student a foot hold in understanding many subjects which involve field variables. Fluid mechanics and heat transfer are two examples. The material derivative is commonly used in expressing the governing equations in these subjects. Extensions Extensions to this problem would be created by considering more complex fields. For example the field could be a function of time and two, or maybe three spatial dimensions. An interesting extension would be to let the field be velocity instead of temperature. Then the total derivative would be the actual acceleration of a particle at a particular point in the field at a certain time. This acceleration would have two parts: (1) local acceleration associated with changes in the field itself at the point, and (2) convective acceleration caused by the particle being carried to a place in the field where the velocity is different. References and Sources Fundamentals of Fluid Mechanics by Bruce Munson, Donald Young and Theodore Okiishi, Second Edition, John Wiley and Sons, New York, 1994. ISBN 0-471-57958-0 POSSIBLE SOLUTION(S) Let x denote the position between the cities. At A, we have x=0 and at B we have x = 100. Data Time 7 a.m. 8 a.m. 9 a.m. 10 a.m. 11 a.m. City A temp. 10.0 12.0 13.5 15.5 18.2 City B temp. 14.0 17.0 19.5 21.7 22.9 In[2]:= tA = Interpolation[ { {7,10.}, {8,12.0}, {9,13.5}, {10, 15.5}, {11,18.2} } ]; tB = Interpolation[ {{7,14.}, {8,17.}, {9,19.5}, {10, 21.7}, {11,22.9}}]; See plots of time vs. temp at the cities. In[4]:= Plot[tA[t],{t,7,11}] Out[4]= -Graphics- In[5]:= In[5]:= Plot[ tB[t], {t,7,11}] Out[5]= -Graphics- Here is a plot of the temperature as a function of time and position. In[6]:= Temp[t_,dis_] := tA[t] + dis (tB[t] - tA[t]) / 100 In[7]:= Plot3D[ Temp[t,x], {x,0,100}, {t,7,11}, AxesLabel->{"A to B","Time","Temp"}, ViewPoint->{-1.926, 0.676, 0.874}] Out[7]= -SurfaceGraphics- Solution using partial derivatives -- from interpolations above Partial Derivative of Temp. wrt time - using derivatives of time from interpolants Evaluated at the rest stop at 9 am. Units: deg per hour. In[8]:= pdTdt1 = (100-60)/100 * tA'[9] + 60/100 * tB'[9] Out[8]= 2.02333 Partial derivative with respect to x Evaluated at 9 am. Since temperature is a linear function of x, the derivative will not depend on x. In[9]:= pdTdx = -1/100 tA[9] + 1/100 tB[9] Out[9]= 0.06 Answers Part (a) -- bicyclist moving at 30 km/hr past the rest area Units: deg/hour. In[10]:= DTDt = pdTdt1 + v pdTdx /. v -> 30 Out[10]= 3.82333 Part (b) -- bicyclist stopped at the rest area. Units: deg/hr. In[11]:= DTDt = pdTdt1 + v pdTdx /. v -> 0 Out[11]= 2.02333 Part (c) -- bicyclist moving at 30 km back towards A. In[12]:= DTDt = pdTdt1 + v pdTdx /. v -> -30 Out[12]= 0.223333 Solution which does not use Interpolation function Use Fit instead -- get a fourth deg. polynomial In[13]:= tta[t_] =Fit[{ {7,10.}, {8,12.0}, {9,13.5}, {10, 15.5}, {11,18.2} }, {0,t,t^2,t^3,t^4},t]; ttb[t_] = Fit[{{7,14.}, {8,17.}, {9,19.5}, {10, 21.7}, {11,22.9}}, {0,t,t^2,t^3,t^4},t]; partial derivative of temp with respect to time In[15]:= pdTdt2 = (100-60)/100 * tta'[9] + 60/100 * ttb'[9] Out[15]= 2.10349 Answers Part (a) -- bicyclist moving at 30 km/hr past the rest area Units: deg/hour. In[16]:= DTDt = pdTdt2 + v pdTdx /. v -> 30 Out[16]= 3.90349 Part (b) -- bicyclist stopped at the rest area. Units: deg/hr. In[17]:= DTDt = pdTdt2 + v pdTdx /. v -> 0 Out[17]= 2.10349 Part (c) -- bicyclist moving at 30 km back towards A. In[18]:= DTDt = pdTdt2 + v pdTdx /. v -> -30 Out[18]= 0.303494 ISSUES IN SOLUTION A major question which will arise is how to compute the numerical derivative given a table of data. The basic idea is to compute an interpolating function based on the data, and then to differentiate this interpolant at the needed time. The actual meanings of the derivatives were discussed in the earlier issues section. Why is the temperature still rising with respect to time when the cyclist turns around? The temperature rises as the day progresses faster than the cyclist can compensate for by returning towards the cooler city.