13.6 Directional Derivatives and Gradient Vectors

We can already find the slope of a surface at the point (x0,y0,f(x0,y0)) in the y direction, since it is simply fy(x,y), the slope of the tangent to the curve which is the intersection of the surface and the plane x = x0. In a similar fashion, we can determine the slope of a surface in the x direction with fx(x,y). What if we are interested in the slope of the surface in a different direction? This is called a directional derivative. Another vector of interest at a point (x0,y0,f(x0,y0)) is the gradient vector.

Gradient Vector

Def: Gradient Vector

If the partial derivatives of a function f(x,y) are defined at a point P(x0,y0) in the domain of f, then the gradient of  f at P is the vector

Directional Derivative  

Def: Directional Derivative

Let z = f(x,y) be a surface in R3 and let P(x0,y0) be a point in the  domain of f. Further let u = cos(q)i + sin(q)j be a vector in the xy plane that describes a direction from the origin. Then the directional derivative of f in the direction of u is the dot product :

Note that what this really is, is the length of the projection of the gradient onto the direction vector, recalling

and so we can see that

Ex 1 Find the derivative in the direction of u = cos(p/4)i + sin(p/4)j of the function of two variables f(x,y) = y-x2 at the point (-1,0). Sketch the surface, the gradient, and u (!)

The points noted are (-1,0) in the xy plane and (-1,0,-1) on the surface. The slope of the line segment tangent to the surface at (-1,0,-1) is 3sqrt(2)/2. This is the directional derivative at the point, in the direction of u. The gradient there is 2i + j, and is drawn in purple. Hope that helps clear this up!

Below is a contour plot with the gradient vector at (-1,0) pointing in the direction of steepest ascent 2i + j.

Note that the gradient appears to be normal to the level curve at (-1,0) above. That is one of several notable properties of the directional derivative:

Properties of the Directional Derivative (Theorem 13.11,13.12 and more besides)

For the directional derivative of f(x,y), with q the angle between u and  :

  1. The directional derivative will be a maximum if cos(q) = 1. That is, it will take on its maxmum rate of change whenever u is in the same direction as the gradient (since q = 0, hence cos(q) = 1). Another way to say this: The gradient always points in the direction of steepest ascent on the surface.  Or say it like this: f increases most rapidly in the direction of the gradient vector.
  2. f decreases most rapidly in the opposite direction of the gradient.
  3. The derivative in the direction of -u is -Du(f).
  4. The relationship of the partial derivatives of f to the directional derivatives are fx =  derivative in i direction, fy = derivative in j direction.  
  5. Any direction normal to the gradient is a direction of 0 change in f.  
  6. The gradient vector at P is normal to the level curve in the xy plane at P.  

 Gradient Vector for functions of three variables

We know already that it is impossible to sketch a function of three variables as in w = f(x,y,z) since the domain is already 3 dimensional. The gradient gives a tool to help in our understanding of such functions, regarding increase and decrease in different directions. Next, I'll define a gradient in R3 that points in the direction of increasing value of the function of three variables.

Def: Gradient Vector for three variable functions.

If the partial derivatives of f(x,y,z) are defined at  P(x0,y0,z0),  then the gradient at P is the vector

where the partial derivatives are evaluated at the point P. If f(x,y,z) has continuous partial derivatives at P(x0,y0,z0), and u is a unit vector, then the directional derivative of f at P in the direction of u is the number

Note: All of the properties given above hold true for 3 variable functions.

Ex 2 Applications of directional derivatives: Thermodynamics

Antoine the ant is standing on a plate whose temperature at point P(x,y) is given by the formula f(x,y) = 40 - x2 - 4y2. His feet (all six of them!) are much too hot as he stands at (-3,2).

(a) In what direction should he move to get his feet cooler?

(b) What is the rate of change of the temperature in that direction?

(c) What is the rate of change of the temperature in the direction of i + j?

Antoine should move in the direction opposite this: -6i + 16j.  The length of this vector is the rate of change in that direction, so we get 

The temperature is dropping at about 17 degrees per unit change of position in the opposite direction of the gradient.  (Note: increasing 17 degrees in the gradient's direction)

In the direction of i + j, we have:

The temperature drops 5 sqrt(2) per unit hange of position in that direction.

Ex 3 Find the directions in which f increases and decreases most rapidly at the given point. Then find the derivatives of f in these directions. f(x,y,z) = sin(x) + sin(y) + sin(z) at the point (0,0,0)

Increasing most rapidly in the direction i + j + k, decreasing most rapidly in the direction - i - j - k. Derivatives in these directions are sqrt(3) and -sqrt(3) respectively , as it is the length of the gradient vector, since u (in gradient direction) can be found by

Assignment Pg 940 # 1,5,9,13,17,25,29,33,35,37,39-46,55,59,63