<<

9.7 of a . Directional

Some of the vector fields in applications can be obtained from scalar fields. This is very advantageous because scalar fields can be handled more easily.

The gradient of a given scalar f (,xyz ,) denoted by grad f or f (read nabla f ) is the vector function,

f ff grad ff i j  k . (9.7.1) x yz

Here x,,yz are Cartesian coordinates in a domain in 3-space in which f is defined and differentiable. (The symbol  is also called - pg. 283 Hildebrand.)

Example: If f (,xyz ,) 2 y3 4 xz 3 x then gradf  (4zy 3)i 62 j  4xk .

The  (read nabla or del) is defined by,

 i j  k . (9.7.1*) x yz

Gradients are useful in several ways, notably in giving the rate of change of f (,xyz ,) in any direction in space, in obtaining surface vectors, and in deriving vector fields from scalar fields.

9.7.1 Directional derivative

 From we know that ,, give the rates of change of f (,xyz ,) in the direction of x yz the coordinate axes. It seems natural to extend this and ask for the rate of change of f in an arbitrary direction in space.

The directional derivative Dfb or df /ds of a function f (,,)xyz at a point P in the direction of a vector b is,

df f() Q f () P Dfb lim . (9.7.2) dss0 s

Here Q is a variable point on the straight line L in the direction of b , and s is the distance between P and Q. Also s  0 if Q lies in the direction of b (see Fig. 213), s  0 if Q lies in the direction of b .

May 30, 2011 9.7-1 Next, use Cartesian xyz -coordinates and for b a . Then the line L is given by

ri()sxsyszs () ()j ()kp 0 s b, (9.7.3)

where p0 is the position vector of P. Equation (9.7.2) now shows that Db f df/ ds is the derivative of the function f (xs ( ), ys ( ), zs ( )) with respect to the s of L. Hence, assuming that f has continuous partial and applying the

df f f f Df x  y  z, (9.7.4) b ds x y z where primes denotes derivatives with respect to s (which are taken at s  0 ). But here differentiating (9.7.3) gives ri()sxyzj kb . Hence (9.7.4) is simply the inner product of grad f and b ,

df Dfb  grad f. (9.7.5) b ds

(Note that in eqn (9.7.5) b 1.)

Example 1: Gradient. Directional derivative Find the directional derivative of f (,xyz ,) 2 x22 3 y z2 at P :(2,1,3) in the direction of ai2 k.

Solution: gradf  4xyzi 6j 2 k , gives at P the vector gradfP ( ) 8i  6j 6k . 1 baˆ (2 i  k). 5 14 DfP() (ik 2)(86 i  jk 6) . b 55 The minus sign indicates that at P the function f is decreasing in the direction of a .

9.7.2 Gradient is a vector

Since grad f is defined in terms of components depending on the Cartesian coordinates, to prove that grad f is actually a vector, we must show that grad f has a length and direction independent f ff of the choice of those coordinates. In contrast i 2 jk also looks like a vector but does x yz not have a length and direction independent of the choice of Cartesian coordinates. grad f points in the direction of maximum increase of f .

Theorem 1: Vector character of gradient. Maximum increase Let f ()(,,)Pfxyz be a scalar function having continuous first partial derivatives in some domain B in space. Then grad f exists in B and is a vector (i.e., its length and direction are independent of the particular choice of Cartesian coordinates). If gradfP ( )  0 at some point P, it has the direction of maximum increase of f at P .

May 30, 2011 9.7-2 Proof: From eqn (9.7.5),

Dfb bb grad f grad f cos  grad f cos , (9.7.6)

where  is the angle between b and grad f . Now f is a scalar function. Hence its value at a point P depends on P but not on the particular choice of coordinates. The

same holds for the arc length s of the line L (Fig. 213), hence also for Dfb . Now

(9.7.6) shows that Dfb is maximum when   0 and then Dfb  grad f. It follows that the length and direction of grad f are independent of the choice of coordinates. Since   0 if and only if b has the direction of grad f , the latter is the direction of maximum increase of f at P , provided grad f  0 at P .

9.7.3 Gradient as surface normal vector

Let S be a surface represented by f (,,)x y z c const , where f is differentiable. Such a surface is called a level surface of f , and for different c we get different level surfaces. Let C be a curve on S through a point P of S . As a curve in space, C has a representation ri()txtytzt() ()j ()k . For C to lie on the surface S , the components of r()t must satisfy f (,xy,)z c, i.e.,

f ((),(),())xt yt zt c. (9.7.7)

A vector of C is ri()txtytzt ()  ()j  ()k (pg. 392 EK). The tangent vectors of all curves on S passing through P will generally form a plane, called the tangent plane of S at P . (Exceptions occur at edges or cusps of S , for instance, for the cone in Fig. 215 at the apex.) The normal of this plane (the straight line through P perpendicular to the tangent plane) is called the surface normal to S at P . Differentiating eqn (9.7.7) with respect to t ,

fff xyzgrad f r   0 . xyz

Hence, grad f is orthogonal to all the vectors r in the tangent plane, so that it is a normal vector of S at P .

Theorem 2: Gradient as surface normal vector Let f be a differentiable scalar function in space. Let f (,,)x y z c const represent a surface S . Then if the gradient of f at a point P of S is not the zero vector, it is a normal vector of S at P .

May 30, 2011 9.7-3 Example 2: Gradient as surface normal vector. Cone Find a unit normal vector n of the cone of revolution zxy224(2 ) at the point P :(1,0,2).

Solution: The cone is the level surface f  0 of f (,xyz ,) 4( x22 y ) z2. Thus 1 gradf  8xyzi 8j 2 k , gradfP () 8ik 4 . nik (2 ) . 5 The vector n points downward since it has a negative z -component. The other unit normal vector of the cone at P is n .

9.7.4 Vector fields that are of scalar fields ("Potentials")

Some vector fields have the advantage that they can be obtained from scalar fields, which can be handled more easily. Such a vector field is given by a vector function v()P which is obtained as the gradient of a scalar function,

v()Pf grad()P.

The function f ()P is called a potential function or a potential of v()P . Such a v()P and the corresponding vector field are called conservative because in such a vector field, energy is conserved; i.e., no energy is lost (or gained) in displacing a body (or a charge in the case of an electric field) from a point P to another point in the field and back to P . (See Sec. 10.2.)

Conservative fields play a central role in physics and engineering. A basic application concerns the gravitational force (Ex. 3 in Sec. 9.4) and we show that it has a potential which satisfies Laplace's equation, the most important partial in physics and its applications.

May 30, 2011 9.7-4 Theorem 3: Gravitational field. Laplace's equation The force of attraction,

cc pr [(xx  ) i  ( yy  )j  ( zz  )k ], (9.7.8) rr33000

between two particles at points Px0000:( ,yz , ) and Pxyz:(,,) (as given by Newton's law of gravitation) has the potential f (,xyz ,) c/ r, where r (0) is the distance

between P0 and P . c Thus p gradf grad . This potential f is a solution of Laplace's equation, r

222fff 2 f   0 . (9.7.9) xyz222

(2 f (read nabla squared f , or del squared f ) is called the Laplacian of f .)

2221/2 Proof: The distance is rxxyyzz[()(000  )(  )]. ccc   pipp123j pki grad f   j  k x ryrzr  

dr1  cx() x0 dr1  cy() y0  pc1  3 , pc2  3 , dr r x r dr r y r

dr1  cz() z0 pc3  3 dr r z r  p given by eqn (9.7.8) is the gradient of the scalar function (,,)/f xyz c r.

Consider

2 ff cx() x 1 3() x x 2  00c . 233  5 xxxx  r r r

Similarly

2222 ff113(yy00 ) 3(zz )  235cc  , 235   . yrr zrr  222fff  0 . Hence 2 f  0 . xyz222

2 f is also denoted by  f . The differential operator

222 2   , (9.7.11) x22yz2 read "nabla squared" or "del squared" or delta, is called the . It can be shown that the field of force produced by any distribution of masses is given by a vector function that is the gradient of a scalar function f and f satisfies eqn. (9.7.9) in any region that is free of matter.

May 30, 2011 9.7-5 The Laplace equation is very important because there are other laws in physics that are of the same form as Newton's law of gravitation. For instance in elastostatics the force of attraction (or repulsion) between two particles of opposite (or like) charge Q1 and Q2 is,

k pr . Coulomb's law (9.7.12) r3

The Laplace equation will be discussed in detail in Ch. 12 and Ch. 18.

A method for finding out whether a given vector field has a potential will be explained in Sec. 9.9.

May 30, 2011 9.7-6