¡@

Directional Derivatives and the Gradient Vector

Download Maple Worksheet

Consider the surface S with equation z = f(x,y)and let z[0] = f(x[0],y[0]).  Then the point P(x[0],y[0],z[0])lies on S. What is the rate of change of zat (x[0], y[0]) in the direction of an arbitrary unit vector u = ( a, b) ?

Suppose the vertical plane that passes through P in the direction of u intersects S in a curve C . If C has a tangent line at P then the slope of the tangent line T to C at P is the rate of change of z in the direction of u , which is called the directional derivative of f at ( x[0], y[0]) in the direction of u , denoted by D[u]*f(x[0],y[0]) .

[Maple Plot]
3D view

 

Hence

D[u]*f(x[0],y[0]) = limit((f(x[0]+h*a,y[0]+h*b)-f(x... ,

if the limit exists.

¡@

Notice that for u = ( 1, 0), D[u]*f(x[0],y[0]) = f[x](x[0],y[0]) and for u = ( 0, 1), D[u]*f(x[0],y[0]) = f[y](x[0],y[0]).

 

On the other hand, if we define a function g of single variable t by

g(t) = f(x[0]+a*t,y[0]+b*t),

then

g ' (0) = limit((g(t)-g(0))/t,t = 0) = limit((f(x[0]+t*a,y[0]... = D[u]*f(x[0],y[0]).

If f is a differentiable function of x and y, by Chain Rule, we have

g ' ( t ) = f[x]*(x[0]+a*t, y[0]+t*b)*a+f[y]*(x[0]+a*t, y[0]+t*...,

Put t = 0, we get

D[u]*f(x[0],y[0]) = f[x]*(x[0], y[0])*a+f[y]*(x[0],....

¡@

So we have the following theorem :

Theorem

If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u = ( a, b) and

D[u]*f(x,y) = f[x]*(x, y)*a+f[y]*(x, y)*b

 

Note that the theorem is also true if f is a differentiable function of three variables.

In terms of vector functions, the curve C obtained by intersection of the vertical plane that passes through P in the direction of u with S can be described by the vector function

F (t) = ( x[0]+a*t, y[0]+b*t, f(x[0]+a*t,y[0]+b*t) )

and the tangent vector T of C at P is given by

F '(0) = (a, b, f[x]*(x[0], y[0])*a+f[y]*(x[0], y[0])*b ).

 

However,

the directional derivative D[u]*f(x[0],y[0]) in the direction of u = ( a, b) at ( x[0], y[0]) is the rate of change of zat  ( x[0], y[0] ) in the direction of u which is the third component of F '(0).

Let C[x], C[y]be the curves obtained by intersection of the vertical planes y = y[0] , x = x[0] , respectively. Then the tangent vectors of C[x] , C[y] at P are

T[1] = ( 1, 0, f[x](x[0],y[0]) ) and T[2] = ( 0, 1, f[y](x[0],y[0]) ), respectively.

The above theorem says that T = a*T[1]+b*T[2] , which also indicates that T , T[1] , T[2] lies on the same plane ( the tangent plane of S at P ).

[Maple Plot]

  3D view

 

If f is a function of xand y, then the gradient of f is the vector function defined by

(x, y) = (f[x](x,y), f[y](x,y))

 

So if f is differentiable, then D[u]*f(x,y) = . u for any unit vector u .

 

If angle between u and ( x, y ) is theta , then we have

 

D[u]*f(x,y) = | ( x, y ) | | u | cos(theta) [Maple OLE 2.0 Object] | ( x, y ) |.

 

If f is a function of three variables, then the gradient of f is defined by

 

(x, y, z) = (f[x](x,y,z), f[y](x,y,z), f[z](x,y,z))...

 

and it is also true that if f is differentiable, then D[u]f ( x ) = ( x ) [Maple OLE 2.0 Object] u for any unit vector u .

 

Therefore, we have

Theorem

If fis a function of two or three variables, then the maximum value of the directional derivative D[u]f( x ) is |(x)| and it occurs when u has the same direction as (x).

¡@

Let C is a level curve of f , suppose that C has the vector equation r (t) = (x(t), y(t)), that is, f(x(t),y(t)) = k for some constant k . If x, yand f are all differentiable, by Chain Rule, we get

F[x] d*x/(d*t) + F[y] d*y/(d*t) = 0 .

¡@

This says that the gradient vector ( x(t), y(t) ) is perpendicular to the tangent vector r ' ( t) to C .

f := proc (x, y) options operator, arrow; 1/100*((x...

 

[Maple Plot]

 

gradf := [proc (x, y) options operator, arrow; 1/25...
gradf := [proc (x, y) options operator, arrow; 1/25...

 

 

[Maple Plot]

 

The gradient of a function f satisfies the following properties :

1. points in the direction of maximum increase of the function.

2. | | is the rate of increase of f in the direction of maximum increase.

3. is perpendicular to each level curve of the function f .

4. | | is inversely proportional to the spacing between the level curves.

 

Suppose S is a surface with equation F(x,y,z) = k , that is, it is a level surface of a function F of three variables, let P(x[0],y[0],z[0]) be a point on S . Let C be a curve that lies on the surface S and passes through point P, say C is given by the vector function

r (t) = ( x(t), y(t), z(t) ) with r (t[0]) = ( x[0], y[0], z[0] ) .

Since C lies on S , we have

F(x(t),y(t),z(t)) = k .  

Suppose that x, y, z and F are all differentiable, by Chain Rule, we get

F[x] d*x/(d*t) + F[y] d*y/(d*t) + F[z] d*z/(d*t) = 0 ,

that is, . r ' ( t ) = 0 for all t.

¡@

In particular,

( x[0], y[0], z[0]) . r ' ( t[0] ) = 0.

This says that the gradient vector ( x[0], y[0], z[0]) is perpendicular to the tangent vector r ' ( t[0]) to any curve C on S that passes through P .

 

If ( x[0], y[0], z[0] ) is a nonzero vector, then it is a normal vector of the tangent plane to S at P . In this case, the equation of the tangent plane is given by

F[x](x[0],y[0],z[0])*(x-x[0])+F[y](x[0],y[0],z[0])*... .

 

The normal line to S at P , which is the line passing through P and perpendicular to the tangent plane, is given by the symmetric equations

(x-x[0])/F[x](x[0],y[0],z[0]) = (y-y[0])/F[y](x[0],... = (z-z[0])/F[z](x[0],y[0],z[0]) .

[Maple Plot]
3D view

¡@

Example Find the equation of the tangent plane and normal line at ( -2, 1, -3) to the ellipsoid

x^2/4+y^2+z^2/9 = 3

Note that the ellipsoid is a level surface of the function

F(x,y,z) = x^2/4+y^2+z^2/9

Therefore, we have

F[x](x,y,z) = x/2 , F[y](x,y,z) = 2*y , F[z](x,y,z) = 2*z/9

F[x](-2,1,-3) = -1 , F[y](-2,1,-3) = 2 , F[z](-2,1,-3) = -2/3

 

Hence the equation of the tangent plane is

-(x+2)+2*(y-1)-2/3*(z+3) = 0

and the symmetric equations of the normal line are

(x+2)/(-1) = (y-1)/2 = (z+3)/(-2/3)


Download Maple Worksheet