X y z figure 750 exercise 7710 by calculating partial

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Differential Equations with Boundary-Value Problems
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Chapter 7 / Exercise 50
Differential Equations with Boundary-Value Problems
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X Y Z Figure 7.50 Exercise 7.7.10 By calculating partial derivatives, verify that ( X, Y ) = (0 , 0) is a critical point of the function Z = f ( X, Y ) = 0 . 5( X 2 - Y 2 ) . So how do we classify critical points as maxima, minima or saddle points? We will use the deep relationship between critical points of functions and equilibrium points of di ff erential equations. Given any function Z = f ( X, Y ) we can define a new vector field on ( X, Y ) space by X 0 = dX dt = @ f @ X and Y 0 = d Y dt = @ f @ Y (Recall that X 0 is the change of X with respect to time, dX dt .) This new vector field, derived from the function Z = f ( X, Y ) , is called the gradient vector field of f , called “ grad f ” and often written as “ r f ”. Exercise 7.7.11 Compute r f for the functions in Exercise 7.7.9 . What are the equilibrium points of this vector field? By definition, they are points where X 0 = 0 and Y 0 = 0 , that is, @ f @ X = 0 and @ f @ Y = 0 . But we just said that a critical point of the function f is a point where @ f @ X = 0 and @ f @ Y = 0 . Therefore, the critical points of f are exactly the equilibrium points of the vector field r f . Exercise 7.7.12 Verify that at the critical points you found in Exercise 7.7.9 , r f = 0 .
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Differential Equations with Boundary-Value Problems
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Chapter 7 / Exercise 50
Differential Equations with Boundary-Value Problems
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7.7. Optimization 379 If Z = f ( X, Y ) is a height function, we can define the gradient vector field r f as X 0 = @ f @ X Y 0 = @ f @ Y Critical points of f (maxima, minima, saddles) exactly correspond to equilibrium points (stable, purely unstable, saddle) of the gradient vector field r f . We will now make the key connection that will enable us to identify critical points as maxima, minima or saddles. First, let’s consider three simple height functions. We will plot the function f and project it down onto the X and Y axes, where we have calculated and plotted the vector field r f . The first example is a hill (Figure 7.51 left). The function is Z = f ( X, Y ) = 5 - X 2 2 - Y 2 4 The vector field r f is then X 0 = @ f @ X Y 0 = @ f @ Y So X 0 = - X Y 0 = - 0 . 5 Y This is obviously a linear vector field that has a stable equilibrium point at (0 , 0) . X Y Z f f Figure 7.51 The second example is a bowl (Figure 7.51 middle): Z = f ( X, Y ) = X 2 2 + Y 2 4 The vector field r f is then X 0 = @ f @ X = X Y 0 = @ f @ Y = 0 . 5 Y
380 Multivariable Systems which again is a linear di ff erential equation, with an unstable equilibrium point at (0 , 0) . The third example is a saddle (Figure 7.51 right): Z = f ( X, Y ) = 0 . 5( X 2 - Y 2 ) The vector field r f is then X 0 = @ f @ X = X Y 0 = @ f @ Y = - Y which again is a linear di ff erential equation, this time with a saddle point at (0 , 0) . So in this example, maxima of f correspond to stable equilibrium points (stable nodes) of r f . minima of f correspond to purely unstable equilibrium points (unstable nodes) of r f . saddle points of f correspond to saddle points of r f . This is true in general, due to the definition of the gradient vector field. Since Z = f ( X, Y ) , we know that the change in Z Δ Z = @ f @ X · Δ X + @ f @ Y · Δ Y But, from the definition of r f we know that 8 > > < > > : X 0 = @ f @ X = ) Δ X Δ t = @ f @ X = ) Δ X = @ f @ X · Δ t Y 0 = @ f @ Y = ) Δ Y Δ t = @ f @ Y = ) Δ Y = @ f @ Y · Δ t If we substitute these expressions for Δ X and Δ Y in the Δ Z equation, we get Δ Z = @ f @ X · Δ X + @ f @ Y · Δ Y = @ f @ X · @ f @ X · Δ t + @ f @ Y · @ f @ Y · Δ t = @ f @ X 2 · Δ t + @ f @ Y 2 · Δ t Since Δ t > 0 , df dX 2 > 0 , and df d Y 2 > 0 the whole Δ Z expression is positive. Therefore, Z will always increase following the gradient function r f .

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