Some Interesting Mathematical Modeling Problems

Some interesting and unusual mathematical modeling problems and their solutions Problem 1. The graph of a certain function y = f (x) passes through the origin. If (x, y) is a point (6= (0, 0)) on the graph in the first quadrant, then the graph divides the rectangle with vertices (0, 0), (x, 0), (x, y), and (0, y) into two parts A and B so that the area of A is n times the area of B (n > 1, integer). Find f (x). Y y=f(x) A y B x X Figure 1. Illustration to Problem 1. From Figure 1 we have that area of A ∪ B is given by (area of A ∪ B) = xy = xf (x) = (area of A) + (area of B) = n Zx 0 f (t) dt + Zx f (t) dt = (n + 1) 0 Zx f (t) dt. 0 After differentiating both sides of the identity xf (x) = (n + 1) Zx f (t) dt, 0 we see that f (x) satisfies the differential equation f (x) + xf 0 (x) = (n + 1)f (x), =⇒ xf 0 (x) − nf (x) = 0 The last equation is the first order linear differential equation that can be solved explicitly. Its solution is f (x) = cxn . 1 2 Remark 1. If one interchanges the roles of A and B in Figure 1, i.e., set the area of A ∪ B equal to xf (x) = Zx 1 f (t) dt + n 0 Zx f (t) dt, 0 then one gets the answer f (x) = cx1/n . Problem 2. The graph of a certain function y = f (x) passes through (3, 2). Let L(x, y) be the segment of the tangent line to the graph at (x, y) in the first quadrant. Suppose that each point (x, y) on the graph is the midpoint of L(x, y). Find f (x). Y y=f(x) y0 (x 0 , y0 ) L(x 0 , y 0) x0 X Figure 2. Illustration to Problem 2. From Figure 2 we see that the tangent line has the x-intercept equal to 2x0 and the y-intercept equal to 2y0 (since (x0 , y0 ) is the midpoint of the line segment L(x0 , y0 )). Furthermore, the slope of the tangent line is 2y0 equal to − = f 0 (x0 ). This true at any point (x, y) on the graph of y = f (x). This also means that f (x) 2x0 satisfies the equation y y 0 = − , or equivalently xy 0 + y = 0. (1) x The solution of (1) is y = c/x. Now, if (3, 2) is on the graph, then c = 6; thus f (x) = 6/x. 3 Problem 3. A large snowball is shaped into the form of a sphere. Starting at some time, which we can designate as t = 0, the snowball begins to melt. We assume for the sake of discussion that the snowball melts in such a manner that its shape remains spherical. Discuss the quantities that change with time as the snowball melts. If possible, construct a mathematical model that describes the state of the snowball at any time. The radius, volume, and surface are of the snowball change in time. As the snowball melts the volume decreases. In other words, the process of melting can be interpreted as the rate of change of the volume of the snowball in time.. Now, it seems reasonable to assume (WHY?) that the rate at which the snowball melts is proportional to the surface area. That is, dV /dt = −kS, where k > 0 is a proportionality constant. Now, V = 4πr 3 /3 and S = 4πr 2 , so dV dr dr = 4πr 2 = S = −kS. dt dt dt Thus, dr/dt = −k is the differential equation describing the above process of snowball melting. Problem 4. Ralph Palmer Agnew’s snowplow problem (from the text Differential Equations by Ralph Palmer Agnew, McGraw-Hill Book Co.) One day it started snowing at heavy and steady rate. A snowplow started out at noon, going 2 miles the first hour and 1 mile the second hour. What time did it start snowing ? We assume that the plow clears snow at a constant rate of k cubic miles per hour. Let t be the time in hours after noon, x(t) the depths in miles of the snow at time t, and y(t) the distance the plow has moved in t hours. The dy/dt is the velocity of the plow and the assumption (that plow clears snow at a constant rate) gives wx dy = k, dt where w is the width of the plow. Each side of this equation simply represents the volume of snow plowed in one hour. Now let t0 be the number of hours before noon when it started snowing and let s > 0 be the constant rate in miles per hour at which x increases. It means that dx/dt = s and therefore, for t > −t0 , x = s(t + t0 ). (Note, that in our notation, t = 0 is the time the snoplow started working and t = −t0 is the time it started 4 snowing.) The differential equation becomes k 1 dy . = dt ws t + t0 Integrating we obtain y= k [ln(t + t0 ) + c] , ws where c is a constant of integration. Now, when t = 0, y = 0 so c = − ln t0 and   t k ln 1 + . y= ws t0 Finally, from the fact that when t = 1, y = 2 and when t = 2, y = 3 (Do you know why?), we obtain  2  3 2 1 1+ = 1+ . t0 t0 Expanding and simplifying gives t20 + t0 − 1 = 0. The solutions are t0 = t0 = √ −1+ 5 2 √ −1± 5 . 2 Since t0 > 0, we have ≈ 0.618 hours ≈ 37 minutes. Thus it started snowing at about 11 : 23 in the morning. Nice problem, isn’t it !!! Problem 5. Suppose a hole is drilled through the center of the earth and a body of mass m is dropped into the hole. Discuss the possible motion of the mass. Construct a mathematical model that discribes the motion. Notation: Let the distance from the center of the earth to the mass at any time t be denoted by r, let M denote the mass of earth, let Mr denote the mass of that portion of the earth witin the sphere of radius r, and let δ denote the constant density of the earth. A possible motion for the mass is that it oscillates back and forth around the center of the earth. The gravitational force on mass m is F = −kMr m/r 2 , where r > 0 and k is the gravitational constant. Since Mr = 4πδr 3 /3 and M = 4πδR3 /3, we have Mr = r 3 M/R3 and F = −k mM Mr m = −k 3 r. 2 r R Now from Newton’s second law, F = ma = d2 r/dt2 , thus we have m mM d2 r = −k 3 r 2 dt R or d2 r M = −k 3 r. 2 dt R Remark 2. We will see later on that the solution to this differential equations is a periodic function of t.