Fundamental Theorems

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Fundamental Theorems Chapter 5 Fundamental Theorems 5.1 Revisiting the 1D Fundamental Theorem of Calculus Key Ideas. Alternate take on the derivative of a function f : R R. • ! – The function f determines a one-dimensional vector field V~ = f = f@ = fi. h i x – Let Br =[p r, p + r] be the interval of “radius” r centered at point p . ⇤ − ⇤ ⇤ – The boundary of Br consists of two points p + r and p r.The“outwardunitnormal” ⇤ ⇤ − to Br is given by @x at x = p + r Nˆ = ⇤ ( @x at x = p r. − ⇤ − – The outflux of V~ across the boundary of Br is V~ Nˆ = f(p + r) f(p r). · ⇤ − ⇤ − boundaryX points f(p + r) f(p r) – The outflux per unit size is ⇤ − ⇤ − 2r f(p + r) f(p r) – Linear approximation gives us ⇤ − ⇤ − = f 0(p )+error, where error 0. 2r ⇤ ! Thus the derivative of f at p is the limit of the “e↵ect per unit size” of the function. • ⇤ The Fundamental Theorem of Calculus can be obtained by summing over small intervals. • Our plan is now to find notions of derivatives, and corresponding fundamental theorems, for • various quantities. 73 74 CHAPTER 5. FUNDAMENTAL THEOREMS 5.2 Fundamental Theorem for Gradients Key Ideas. Consider a function f : Rn R and a path P (t)=(x(t),y(t),...) defined for a t b. Let • ! A = P (a) and B = P (b). We have f P :[a, b] R. By the 1D Fundamental Theorem of Calculus we have ◦ ! b (f P )0(t) dt = f(A) f(B). ◦ − Za From the Chain Rule we have • @f dx @x(P (t)) dt @f dy (f P )0(t)=[Df(P (t))]P 0(t)=0 @y (P (t))1 0 dt 1 ◦ · . B . C B . C B . C B . C @ A @ A Recall that the gradient of a function f: • @f @f grad(f)= , ,... @x @y ⌧ Geometric idea of gradient: The gradient of a function measures the total change of a function. • If Aˆ is any unit vector, then grad(f) Aˆ tells us the extent to which f changes in that direction. (This is the same as the directional derivative.)· The gradient vector points in the direction of largest increase, with magnitude equal to the rate of change in that direction. Di↵erentiation rules: • – Scaling rule: grad(cf)=c grad(f) if c is a constant – Addition rule: grad(f + g) = grad(f) + grad(g) – Product rule: grad fg = grad(f)g + f grad(g) Fundamental Theorem of Calculus for gradients:Supposef : Rn R is smooth1 and • that C is a parametrized path from point A to point B. Then ! grad(f) Tdsˆ = f(B) f(A). · − ZC Exercises. 1. Compute the gradient for the following functions: (a) f(x, y)=x2 + y2 (b) f(x, y)=x2 y2 − (c) f(x, y)=ln(x2 + y2) 1In this course smooth means that the object is continuous and that derivatrives of the object are also continuous. 5.2. FUNDAMENTAL THEOREM FOR GRADIENTS 75 (d) f(x, y, z)= x2 + y2 + z2 2. Can you find a functionp f so that. (a) . grad(f)= 2, 3 ? h i (b) . grad(f)= x, y ? h i (c) . grad(f)= y, x ? h i (d) . grad(f)= y, x ? h − i 3. Let C be the path traveling counter-clockwise along the unit circle from (1, 0) to ( 1, 0). Let L be the path traveling along the x axis from (1, 0) to ( 1, 0). − − (a) Find parametrizations for C and L. (b) Compute x, y Tdsˆ by direct methods. C h i· (c) Compute R x, y Tdsˆ by direct methods. Lh i· (d) Find a function f such that grad(f)= x, y . R h i (e) Compute x, y Tdsˆ using the Fundamental Theorem for Gradients. C h i· (f) Compute R x, y Tdsˆ using the Fundamental Theorem for Gradients. Lh i· (g) Do your computationsR agree? Explain why/why not. 4. Let C be the path traveling counter-clockwise along the unit circle from (1, 0) to ( 1, 0). Let − C be the path traveling clockwise along the unit circle from (1, 0) to ( 1, 0). − e(a) Find parametrizations of C and C. (b) Compute y, x Tdsˆ by direct methods. C h − i· e (c) Compute R y, x Tdsˆ by direct methods. C h − i· (d) Do your computations agree? Explain why/why not. R e 76 CHAPTER 5. FUNDAMENTAL THEOREMS 5.3 Divergence of a vector field Key Ideas. Geometric idea of divergence: The divergence of vector field V~ at the point P measures the • expansion of V~ at P ⇤ ⇤ Definition of divergence using limits: Let Br be the solid ball of radius r centered at point P • ⇤ and Sr be the boundary of Br. 1 1 div V~ (P )=lim V~ NdAˆ and/or div V~ (P )=lim V~ Ndsˆ ⇤ r 0 B · ⇤ r 0 B · ! | r| ZZSr ! | r| ZSr Computation in Cartesian coordinates: Suppose V~ = V ,V ,... Then • h x y i @V @V div V~ = x + y + .... @x @y Di↵erentiation rules: • – Scaling rule: div(cV~ )=c div V~ if c is a constant – Addition rule: div(V~ + W~ )=divV~ +divW~ – Product rule: div(fV~ ) = grad(f) V~ + f div V~ if f is a function · Exercises. 1. For each vector field do the following: make a sketch of the vector field by hand, describe in words what the vector field does, compute the divergence of the vector field, comment on how your computation fits with your earlier description. (a) V~ = x, y h i (b) V~ = y, x h i (c) V~ = 3, 2 h − i (d) V~ = x , y h x2+y2 x2+y2 i (e) V~ = x y, x + y h − i (f) V~ = x + y, x y h − i (g) V~ = y2,y2 h i 2. Suppose f(x, y)=x2 y2 and V~ = x + y, x +2y . − h i (a) Compute fV~ and then compute div(fV~ ). (b) Use the product rule to compute div(fV~ ). (c) Prove the product rule formula for divergence. 5.3. DIVERGENCE OF A VECTOR FIELD 77 3. The Laplacian of a function f is defined by div(grad f). Compute the Laplacian of the following functions. (a) f(x, y)=2x 3y − (b) f(x, y)=x2 + y2 (c) f(x, y)=x2 y2 − (d) f(x, y) = cos(x)sin(y) 1 2 2 (e) f(x, y)= 2 ln(x + y ) 1 (f) f(x, y, z)= x2 + y2 + z2 What property ofp the function does the Laplacian seem to measure? 4. Consider the vector field given in spherical coordinates by 1 x y z V~ = @ = , , r2 r (x2 + y2 + z2)3/2 (x2 + y2 + z2)3/2 (x2 + y2 + z2)3/2 D E and concentric spheres Sr of radius r centered at the origin. (a) Compute the outward flux V~ NdAˆ using the definition of the flux integral; Sr · (b) Find the flux rate RR V~ NdAˆ lim Sr · r 0 Volume(B ) ! RR r where Br denotes the ball enclosed by Sr. (c) Compute the divergence of the vector field V~ . (d) What is the outward flux rate of V~ at the origin? What about anywhere other than the origin? 5. (Optional Challenge Problem) In this problem you consider the possibility of defining diver- gence of a vector field V~ (x, y) using a square, rather than a ball. Fill in the boxes in the argument below. (a) For simplicity we work in two dimensions. Write V~ (x, y)= Vx(x, y),Vy(x, y) .Fixa point (x ,y ). For small x and y we have h i ⇤ ⇤ @V @V V~ (x, y) V~ (x ,y )+(x x ) x (x ,y ), y (x ,y ) ⇡ ⇤ ⇤ − ⇤ @x ⇤ ⇤ @x ⇤ ⇤ ⌧ +(y y ) , − ⇤ * + 2 (b) Let Br denote the interior of the square in R having corners at (x + r, y + r), (x ⇤ ⇤ ⇤ − r, y + r), (x r, y r), (x + r, y r). Let Sr be the boundary of Br, oriented counter-clockwise.⇤ ⇤ − ⇤ − ⇤ ⇤ − 78 CHAPTER 5. FUNDAMENTAL THEOREMS Draw a picture showing Br and Sr. The area of Br is (c) The outward unit normal Nˆ along Sr is given by 1, 0 when x = x + r, h i ⇤ 8 when y = y + r, > ⇤ > ˆ > N = > > when , <> > > when , > > > (d) The integral V~ Ndsˆ can:> be approximated by Sr · r R r − V~ (x + r, y + t) 1, 0 dt + 0, 1 dt t= r ⇤ ⇤ ·h i t=r ·h i Z − Z r − + dt + . Zt=r (e) Using the approximation from the first part of this problem we we find that r @V @V V~ Ndsˆ V~ (x ,y )+r x (x ,y )+t x (x ,y ) dt S · ⇡ t= r ⇤ ⇤ @x ⇤ ⇤ @y ⇤ ⇤ Z r Z − ✓ ◆ r − + V~ (x ,y )+t + r dt 0 ⇤ ⇤ 1 Zt=r B C @ A r − V~ (x ,y )+r t dt − 0 ⇤ ⇤ − 1 Zt=r B C @ A r dt − t= r Z − (f) The previous integral simplifies to @V V~ Ndsˆ x (x ,y )+ . · ⇡ 0 @x ⇤ ⇤ 1 ZSr B C @ A 5.3. DIVERGENCE OF A VECTOR FIELD 79 Thus V~ Ndsˆ @V lim Sr · = x (x ,y )+ r 0 @x ⇤ ⇤ ! R and we are done! 80 CHAPTER 5. FUNDAMENTAL THEOREMS 5.4 Divergence Theorems Key Ideas. Two-dimensional Divergence Theorem. Let D be a bounded domain in R2, let C be the • boundary of D, and let Nˆ be the outward-pointing unit normal along C. Then for any smooth vector field V~ defined in D we have div VdA~ = V~ Nds.ˆ · ZZD ZC The two-dimensional divergence theorem is equivalent to Green’s Theorem.
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