SPECIALIZED FILM MAKING BRIEF ABSTRACT A camera tracks an object undergoing sinusoidal motion. We develop equations for the tracking camera and ascertain maximum angular acceleration and maximum focal length acceleration. GENERAL INFORMATION FileName: FILMING Full title: Specialized Film Making Last Revision Date: 27 May 1996. Developer: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.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 We have been hired by a film company to enable them to get close-up shots of moving, studio model objects. We are contracted to assist them in determining the focusing distance (i.e., the distance from the film plane of the camera to the photographed object) required to get focused shots of objects as they move about their preset tracks. Our camera can continuously adjust the f stop, i.e. the distance from the film in the camera to the object, necessary to maintain perfect focus on the object. But we must provide a computer in the camera with information as to where to point the camera, how to move the camera, what the distance from the camera to the object is, and how this distance is changing. This information will enable the clock and the control motors in the camera to give real time response to track and focus on the moving object. Now on this particular day the studio is planning to have an object traverse a sinusoidal path, in front of the camera. The object will move in such a way that at times t = 0, 1, and 2, seconds the object will cross the horizontal x-axis at positions the three points (0, 0), (32, 0), and (64, 0), respectively, , i.e. the horizontal speed is 32 ft/sec. The object will start at time t = 0 and stop at time t = 2 seconds. Furthermore the x coordinate of the object will move with a constant velocity from left to right. All distances are measured in cm. Finally, the furthest distance the object moves away from the horizontal x-axis is 10 cm. The camera is positioned at location C = (25,-50). Consider and answer the following questions: (1) Describe the x and y coordinate of the object, each as a function of time, t in seconds. (2) Describe (b) the distance from the object to the camera as a function of time, t, in seconds. Plot this function in the time interval [0, 2]. (3) Describe (a) the angle that the camera must make with the vertical line x = 25 as it aims at the moving object as a function of time, t, in seconds. Begin with the positive acute angle between the line of sight from camera to object and the vertical line x = 25. Plot this angle in the time interval [0, 2]. (4) Find the time at which the the object is nearest the camera and find the distance between the object and camera at this time. (5) Find the time at which the the object is farthest from the camera and find the distance between the object and camera at this time. (4) and (5) are necessary to ascertain just what lens we are to use, for different lenses have different focal length capabilities. Although we can presume that our equipment is quite good, we might be interested in additional questions. For the answers may be of concern to us with regard to the performance specifications of the camera and the motors in the camera. (6) What is the fastest rate that our camera has to turn? (7) What is the fastest rate that our focal length has to change? (8) Find the maximum angular accelerations for the rotation of the camera. (9) Find the maximum accelerations for the focal length of the camera. KEYWORDS Sine function, parametric equations, distance function, inverse trigonometric function, derivatives (up to third derivative), position, velocity, acceleration, angular velocity and acceleration, plotting, optimization TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites We have successfully used this problem in class for a number of years. There is a preliminary non-calculus component (1) - (3) and a follow-up calculus component (4) - (9). Non-Calculus: The problem asks that the student be able to formulate a mathematical model (32 t, 10 sin(4 pi t)) for the location of an object as a function of time t from a physical description and several data points on the path. The problem expects the student to use an inverse trigonometric function to describe an angle as a function of time. Calculus: The student is then asked to optimize two velocities and two accelerations and the strategy is to examine plots, perhaps zoom in on a region of interest for optimization purposes, and use some root finding algorithm to ascertain when certain derivatives are zero for optimization approaches. Alternative optimization (i.e. setting derivative to zero) in calculus: Use the zoom in feature alone to ascertain optimal values. Time allotment - time management This problem can be formulated in 30 minutes in class and the students can begin to examine some of the issues and plot it often in less time. Expectations We expect the student to be able to writw a parametric description of the motion of the partical being tracked. Future payoffs The real payoff could be the ease with which the CAS addresses the problems which demand manipulation of huge expressions, e.g., maximizing the accelerations. Extensions This could be done as a graphing calculator exercise easily; it could let students familiarize themselves with the zoom function and graphing parametric equations. Ask to place a filter between the camera and the object and keep it at horizontal distance say y = -20. Discuss its motion . Does it ever stop? What is its position, velocity, and acceleration? References and Sources A POSSIBLE SOLUTION: (1) The x coordinate of the function is 32t as the horizontal speed is a constant 32 ft/sec. x[t_] = 32 t; The y coordinate of the function is 10 sin(pi/8 32 t) = 10 sin(4 pi t). y[t_] = 10 Sin[4 Pi t]; (2) Let b denote the distance from the object to the camera as a function of time, t, in seconds. b[t_] = Sqrt[(25 - x[t])^2 + (50 + y[t])^2]; (3) The angle, a(t), the camera makes with the vertical x = 25 can bedescribed as a(t) = arctan[(25 - 32t)/(10 sin(4 pi t) - (-50)]. We use this convention so that the argument of arctan will be in the range -pi/2 to +pi/2 - exactly the domain of arctan. a[t_] = ArcTan[(25 - x[t])/(50 + y[t])]; (4) To determine where the object is nearest to the camera plot b(t) over the entire interval [0, 2], zooming in on what appears to be a reasonable interval ([0.8, 0.9]) at which the the distance b(t) is minimum in the interval [0, 2]. Plot[b[t], {t, 0, 2}]; Plot[b[t], {t, 0.8, 0.9}] -Graphics- We seek to determine where the function is at a local minimum near 0.9, i.e. where b'(t) = 0 near 0.9. sol = FindRoot[b'[t] == 0, {t, 0.87}] {t -> 0.873504} And we ascertain this distance by evaluating this function at the time t where b'(t) = 0. b[t]/.sol[[1]]//N 40.1106 Thus the time at which the object is closest to the camera occurs 0.873504 seconds into the shoot. The distance is 40.1106 cm. (5) To determine where the object is farthest from the camera, zoom in on what appears to be a reasonable interval ([1.6, 1.8]) at which the the distance b(t) is maximum in the interval [0, 2]. Plot[b[t], {t, 1.6, 1.8}]; We seek to determine where the function is at a local maximum near 1.65, i.e. where b'(t) = 0 near 0.9. sol = FindRoot[b'[t] == 0, {t, 1.65}] {t -> 1.63425} And we ascertain this distance by evaluating this function at the time t where b'(t) = 0. b[t]/.sol[[1]]//N 65.8557 Thus the time at which the object is farthest from the camera occurs 1.63425 seconds into the shoot. The distance is 65.8557 cm. (6) To determine the fastest our camera has to turn we need to find the maximum (absolute) value of the angular velocity, a'(t). These will occur when the derivative of a'(t), i.e. a''(t), is zero and again we plot a'(t) to ascertain a good starting value for our root finding. Plot[a'[t], {t, 0, 2}] -Graphics- From the plot we see that a reasonable estimate of t for maximum a'(t) is 1.75. sol = FindRoot[a''[t] == 0, {t, 1.75}] {t -> 1.7745} a'[t]/.sol[[1]]//N -1.65072 Thus the fastest rate the camera has to turn is 1.65072 radians/second at time t = 1.7745 sec. (7) To determine the fastest rate that our focal length has to change we need to find the maximum value of the change in the focal length or distance (b(t)) from the film in the camera to the lens, i.e., maximize b'(t). These will occur when the derivative of b'(t), i.e. b''(t), is zero and again we plot b'(t) to ascertain a good starting value for our root finding. Plot[b'[t], {t, 0, 2}] -Graphics- From the plot we see that a reasonable estimate of t for maximum a'(t) is 0.25 for a negative velocity and 1 for a positive velocity. sol = FindRoot[b''[t] == 0, {t, 0.25}] {t -> 0.249896} b'[t]/.sol[[1]]//N -129.276 sol = FindRoot[b''[t] == 0, {t, 1}] {t -> 1.0002} b'[t]/.sol[[1]]//N 128.887 Thus the fastest the focal length will have to change is 129.276 cm/second at time t = .24896 sec. (8) To determine the fastest our camera has to accelerate its rotation we need to find the maximum value of the angular acceleration, a''(t). These will occur when the derivative of a''(t), i.e. a'''(t), is zero and again we plot a''(t) to ascertain good starting value for our root finding. Plot[a''[t], {t, 0, 2}] -Graphics- From the plot we see that a reasonable estimate of t for maximum a''(t) is 1.75. a3[t_] = a'''[t]; sol = FindRoot[a3[t] == 0, {t, 1.85}] {t -> 1.87618} a''[t]/.sol[[1]]//N 19.9279 Thus the fastest rate the camera has to accelerate its rotation is 19.279 radians/second at t = 1.87618 sec. (9) To determine the maximum acceleration for the focal length of the camera we need to find the maximum value of the acceleration, b''(t). These will occur when the derivative of b''(t), i.e. b'''(t), is zero and again we plot b''(t) to ascertain a good starting value for our root finding. Plot[b''[t], {t, 0, 2}] -Graphics- From the plot we see that a reasonable estimates of t for maximum b''(t) are near t = 0.8 and near t = 0.6. b3[t_] = b'''[t]; sol1 = FindRoot[b3[t] == 0, {t, 0.8}] {t -> 0.873871} sol2 = FindRoot[b3[t] == 0, {t, 0.6}] {t -> 0.625818} b''[t]/.sol1[[1]]//N 1600.26 b''[t]/.sol2[[1]]//N -1556.88 Thus the maximum acceleration for the focal length of the camera occurs is 1600.26 cm/sec^2 at t = 0.873871 sec. ISSUES IN SOLUTION The attendant mathematics is straightforward, i.e., formulating a function and optimizing certain derived functions. One possible difficulty is in determining a suitable starting value for using the root finding algorithm to converge to the actual value.