Di®erential Equation Notes Dan Singer Minnesota State University, Mankato [email protected] 1 Trajectories n n A trajectory in R is a function ® :[t0; t1] ! R of the form ®(t) = (®1(t); ®2(t); : : : ; ®n(t)): The graph of the trajectory is f®(t): t 2 [t0; t1]g: Example 1.1. Let ® : [0; 2¼] ! R2 be de¯ned by ®(t) = (cos t; sin t): The graph of ® is f(x; y) 2 R2 : x2 + y2 = 1g, the unit circle. 1 0.5 -1 -0.5 0.5 1 -0.5 -1 1 Example 1.2. Let ® : [0; 2¼] ! R2 be de¯ned by ®(t) = (8 cos t; 3 sin t): 2 x2 y2 The graph of ® is f(x; y) 2 R : 82 + 32 = 1g, the ellipse with major axis of length 16 and minor axis of length 6. 3 2 1 -7.5 -5 -2.5 2.5 5 7.5 -1 -2 -3 Example 1.3. Let ® : [0; 13¼] ! R3 be de¯ned by p ®(t) = (2 cos t; 2 sin t; t): The graph of ® is a helix of radius 1. As t increases the graph becomes increasingly compressed. -1 1 -0.5 0 0.5 0.5 0 1 -0.5 -1 6 4 2 0 2 2 Direction vectors Each point ®(t) of a trajectory ® can be regarded as a vector which begins at the origin and ends at ®(t). The displacement from position ®(t) to ®(t0) is the vector ®(t + h) ¡ ®(t). The average rate of change of position from time t to time t + h is 1 (®(t + h) ¡ ®(t)): h The instantaneous rate of change of the position vector at time t is the limit 1 ®0(t) = lim (®(t + h) ¡ ®(t)): h!0 h If ®(t) = (®1(t); ®2(t); : : : ; ®n(t)); then 0 0 0 0 ® (t) = (®1(t); ®2(t); : : : ; ®n(t)): We can interpret ®0(t) as the direction a particle is heading in at time t as it is traveling along the trajectory ®. If ®(t) = (x(t); y(t)) then the slope of y0(t) 0 the direction vector at time t is x0(t) , assuming x (t) 6= 0. Example 2.1. In Example 1.1, ®0(t) = (¡ sin t; cos t). At time t = ¼ , the p p 4 ¼ 2 2 particle is in position ®( 4 ) = ( 2 ; 2 ) and is heading in direction p p ¼ 2 2 ®0( ) = (¡ ; ) 4 2 2 with slope ¡1. Example 2.2. In Example 1.2, ®0(t) = (¡8 sin t; 3 cos t). At time t = ¼ , the p p 4 ¼ 2 2 particle is in position ®( 4 ) = (8 2 ; 3 2 ) and is heading in direction p p ¼ 2 2 ®0( ) = (¡8 ; 3 ) 4 2 2 3 with slope ¡ 8 . 3 Example 2.3. In Example 1.2, ®0(t) = (¡ sin t; cos p1 ). At time t = k¼, pt 1 · k · 12, the particle is in position ®(k¼) = (¡1; 0; k¼) and is heading in direction ®0(k¼) = (0; ¡1; p2 ). Notice that the direction vectors are k¼ becoming more horizontal as t increases. Example 2.4. In Examples 1.1 and 1.2, both trajectories travel exactly once around their graphs in the counter-clockwise direction. At what times are both particles traveling in the same direction? Answer: at those times t in which the direction vectors are parallel to each other, namely when (¡ sin t; cos t) = ¸(t)(¡8 sin t; 3 cos t) 1 for some ¸(t) 6= 0. When sin t 6= 0 then we must have ¸(t) = 8 , which 3 forces cos t = 8 cos t, which forces cos t = 0, which forces sin t = §1. This ¼ 3¼ corresponds to t = 2 and t = 2 . When sin t = 0 we must have cos t = §1 1 and ¸(t) = 3 . This corresponds to t = 0 and t = ¼. Exercise 1: Let P be a particle traveling clockwise around the circle x2 + y2 = 25. (a) Find the direction P is traveling in at the moment it passes through the point (¡3; ¡4). (b) Find both points along the circle at which the particle is traveling in a direction which is parallel to the line y = 2x. Hint: the line has direction vector (1; 2). x2 Exercise 2: Let P be a particle traveling clockwise around the ellipse 25 + y2 100 = 1. (a) Find the direction P is traveling in at the moment it passes through the point (3; 8). (b) Find both points along the ellipse at which the particle is traveling in a direction which is parallel to the line y = 2x. 4 3 Surfaces Let f : Rn ! R be given. A surface is the set of solutions to f(x1; x2; : : : ; xn) = 0: Example 3.1. Let f(x; y) = x2 + y2 ¡ 1. The surface associated with f is the unit circle (see Example 1.1). x2 y2 Example 3.2. Let f(x; y) = 82 + 32 ¡ 1. The surface associated with f is an ellipse (see Example 1.2). Example 3.3. Let f(x; y; t) = (x ¡ cos t2)2 + (y ¡ sin t2)2. The surface associated with f is f(cos t2; sin t2; t): t 2 Rg = p p f(cos t; sin t; t): t 2 Rg [ f(cos t; sin t; ¡ t): t 2 Rg: This is a helix. Compare with Example 1.3. 4 Partial derivatives and the chain rule 2 3 Let f(x; y; z) = y ¡ 2zx + 3 z . The partial derivatives of f are @f @f @f (x; y; z) = ¡2z; (x; y; z) = 1; (x; y; z) = ¡2x + 2z2: @x @y @z If we assume that x, y, and z are functions of t, then we can de¯ne 2 F (t) = f(x(t); y(t); z(t)) = y(t) ¡ 2z(t)x(t) + z(t)3: 3 The derivative is F 0(t) = y0(t) ¡ 2z0(t)x(t) ¡ 2z(t)x0(t) + 2z(t)2z0(t) = 5 ¡2z(t) ¢ x0(t) + 1 ¢ y0(t) + (¡2x(t) + 2z(t)2) ¢ x0(t) = · ¸ · ¸ · ¸ @f @f @f (x(t); y(t); z(t)) x0(t)+ (x(t); y(t); z(t)) y0(t)+ (x(t); y(t); z(t)) z0(t): @x @x @x In short, d f = f x0 + f y0 + f z0: dt x y z 5 Calculating direction vectors on a curve de- ¯ned as a surface Let ®(t) = (x(t); y(t)) be a trajectory which traces out the curve de¯ned by the surface f(x; y) = 0: Then the direction of the trajectory at time t is the vector (x0(t); y0(t)). On the other hand, we know that F (t) = f(x(t); y(t)) = 0 for all t, therefore 0 0 0 F (t) = fxx + fyy = 0 0 0 0 0 for all t, therefore (x ; y ) is perpendicular to (fx; fy) and (x ; y ) is parallel to (¡fy; fx). Example 5.1. Let f(x; y) = x2 + y2 ¡ 1. The direction of any trajectory along the surface generated by f is parallel to (¡fy; fx) = (¡2y; 2x) x with slope ¡ y at the point (x; y). In particular, the slope of the direction 2 2 vectorp p of a particle traveling around the unit circle x + y = 1 at the point 2 2 ( 2 ; 2 ) is ¡1. Compare this with Example 2.1. 6 x2 y2 Example 5.2. Let f(x; y) = 82 + 32 ¡ 1. The direction of any trajectory along the surface generated by f is parallel to 2 2 (¡f ; f ) = (¡ y; x) y x 9 64 ¡9x with slope 64y at the point (x; y). In particular, the slope of the direction 2 vector of a particle traveling around the ellipse x2 + y = 1 at the point p p 82 32 2 2 ¡72 ¡3 (8 2 ; 3 2 ) is 192 = 8 . Compare this with Example 2.2. 6 1-parameter families The function f : R3 ! R can be used to de¯ne a one-parameter family of curves. The curve corresponding to z = c is the graph of all (x; y) satisfying f(x; y; c) = 0: 2 3 Example 6.1. f(x; y; z) = y ¡ 2zx + 3 z . For c = 1 we get 2 y = 2x ¡ : 3 For c = 2 we get 16 y = 4x ¡ : 3 All the curves in this family are straight lines. If we plot the lines corre- sponding to c = 0:5k for k 2 f¡10; ¡9;:::; 9; 10g then we obtain 100 50 -5 5 10 15 20 -50 -100 7 There seems to be a curve which is tangent to all these lines. This is called the envelope of the 1-parameter family. The envelope is 100 50 5 10 15 20 -50 -100 7 Calculating the envelope of a 1-parameter family Given a 1-parameter family of curves de¯ned by f(x; y; z) = 0, we can cal- culate the envelope (if it exists) as follows: Let ®(t) = (x(t); y(t)) be a parameterized curve which is a candidate for the envelope. Properties ® must have: 1. It must pass through every curve de¯ned by the equation f(x; y; c) = 0 for each c. Therefore, for each t there must be a number c(t) such that f(x(t); y(t); c(t)) = 0. We can say that ® passes through the curve de¯ned by f(x; y; c(t)) = 0 at time t. 2. We will make the assumption that c(t) is a di®erentiable function of t and that the partial derivatives of f can be computed at all points of ®.