Michael Corral: Vector Calculus

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78 CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES 2.4 Directional Derivatives and the Gradient For a function z= f(x,y), we learned that the partial derivatives ∂f ∂f ∂x and ∂y represent the (instantaneous) rate of change of f in the positive x and y directions, respectively. What about other directions? It turns out that we can find the rate of change in any direction using a more general type of derivative called a directional derivative. Definition2.5. Let f(x,y)beareal-valuedfunctionwithdomain Din 2 ,andlet(a,b) be a point in D. Let v be a unit vector in 2 . Then the directional derivative of f at (a,b) in the direction of v, denoted by D v f(a,b), is defined as D v f(a,b)= lim h→0 f((a,b)+hv)− f(a,b) h (2.8) Notice in the definition that we seem to be treating the point (a,b) as a vector, since we are adding the vector hv to it. But this is just the usual idea of identifying vectors with their terminal points, which the reader should be used to by now. If we were to write the vector v as v=(v 1 ,v 2 ), then f(a+hv 1 ,b+hv 2 )− f(a,b) D v f(a,b)= lim . (2.9) h→0 h From this we can immediately recognize that the partial derivatives ∂f ∂f ∂x and ∂y are special cases of the directional derivative with v=i=(1,0) and v=j=(0,1), respectively. That is, ∂f ∂x = D if and ∂f ∂y = D jf. Since there are many vectors with the same direction,weuseaunitvectorinthedefinition,asthatrepresentsa“standard”vector for a given direction. If f(x,y) has continuous partial derivatives ∂f ∂f ∂x and ∂y (which will always be the case in this text), then there is a simple formula for the directional derivative: Theorem 2.2. Let f(x,y) be a real-valued function with domain D in 2 such that the partial derivatives ∂f ∂f ∂x and ∂y exist and are continuous in D. Let (a,b) be a point in D, and let v=(v 1 ,v 2 ) be a unit vector in 2 . Then ∂f D v f(a,b)=v 1 ∂x (a,b)+v ∂f 2 (a,b). (2.10) ∂y Proof: Note that if v=i=(1,0) then the above formula reduces to D v f(a,b)= ∂f ∂x (a,b), which we know is true since D i f= ∂f ∂x , as we noted earlier. Similarly, for v=j=(0,1) the formula reduces to D v f(a,b)= ∂f ∂y (a,b), which is true since D jf = ∂f ∂y . So since i=(1,0) and j=(0,1) are the only unit vectors in 2 with a zero component, then we
2.4 Directional Derivatives and the Gradient 79 need only show the formula holds for unit vectors v=(v 1 ,v 2 ) with v 1 0 and v 2 0. So fix such a vector v and fix a number h0. Then f(a+hv 1 ,b+hv 2 )− f(a,b)= f(a+hv 1 ,b+hv 2 )− f(a+hv 1 ,b)+ f(a+hv 1 ,b)− f(a,b). (2.11) Since h0 and v 2 0, then hv 2 0 and thus any number c between b and b+hv 2 can be written as c=b+αhv 2 for some number 0
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About the author: Michael Corral is
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iv Preface matter, anyone can make
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vi Contents 4.4 Surface Integrals a
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2 CHAPTER 1. VECTORS IN EUCLIDEAN S
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188 Bibliography Pogorelov, A.V., A
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190 Appendix A: Answers and Hints t
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Appendix B We will prove the right-
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Appendix C 3D Graphing with Gnuplot
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202 GNU Free Documentation License
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Index Symbols D....................
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212 Index iterated integral .......
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Michael Corral: Vector Calculus

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