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Directional Derivatives, Gradient of f and the Min-Max

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  • 0:03 Higher Dimensional Living
  • 0:42 The Directional Derivative
  • 1:25 Calculating…
  • 3:28 Gradient
  • 3:53 Minimum and Maximum
  • 6:05 Lesson Summary
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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Using partial derivatives and a direction vector, we can find the directional derivative. In this lesson, we clarify the meaning of the directional derivative and relate it to gradients, minimums, and maximums.

Higher Dimensional Living

Meet my friend, Bugs McFly. He lives happily in a flat two-dimensional space. But what he thinks is two dimensions is actually the surface of a three-dimensional object he can't see. When he moves in any direction, he is forced to stay on the surface of this object so it always feels like two-dimensions. We can do some math with Bugs to help him gain some insight for this third dimension.

In calculus, derivatives give us rates of change, and Bugs is interested in movement in a particular direction. The mathematical concept we will use is called the directional derivative. The sign of the directional derivative tells us whether we are moving up or down.

The Directional Derivative

Let's walk through an example together:

A point (the green dot) is placed on the surface of a three-dimensional curved object:


A_point_(the_green_dot)_is_placed_on_the_surface


Looking down from the z-axis, a new x-y sits on the point.


A_new_x-y_sits_on_the_point


The vector u tells us which direction we should move the point in, while staying on the surface of the curve.


Move_in_the_direction_of_the_vector_but_stay_on_the_surface


The curved surface is some function, f(x, y). We calculate the directional derivative of f(x, y) along a direction vector. In this example, the directional derivative is positive, indicating an upwards (positive z) movement.

What if the point started at a different location?

Again, we should begin by locating the x-y on the point.


Again,_locate_the_x-y_on_the_point


Do you see how the directional derivative is now pointing down? That means in this case, it is negative!


The_directional_derivative_will_be_negative


Calculating the Directional Derivative

Now for some math fun! The steps for finding the directional derivative:

Step 1: Normalize the direction vector

In our example, the direction vector is:


u_arrow=i+j


To normalize, we divide the vector by its length. The length of the unit vector is:


length_of_u=sqrt(1^2+1^2)=sqrt(2)


The normalized directional vector is indicated by a u with a ''hat'' on top of it:


u_hat=i/sqrt(2)+j/sqrt(2)


Step 2: Identify the components of the normalized direction vector

Focusing on just the components, we often write the normalized direction vector as:


u_hat=a,b


Thus, in our example, a = 1/√(2) and b = 1/√(2).

Step 3: Compute the partial derivatives fx and fy

The function in our example is:


f(x,y)=-x^2/4-y^2/2+x/2+y


Find fx, by taking the derivative of f(x, y) with respect to x while keeping y constant:


f_x=-x/2+1/2


Likewise, finding fy means differentiating with respect to y with x constant:


f_y=-y+1


Step 4: Substitute into the equation for the directional derivative

The symbol for the directional derivative is Du.

The equation for the directional derivative of f(x, y) is:


D_u_f(x,y)=af_x+bf_y


We have all the parts for the right-hand-side. Thus,


D_u_f(x,y)=(1/sqrt(2))(-x/2+1/2)+(1/sqrt(2))(-y+1)


Step 5: Evaluate the directional derivative at a specific point (x, y)

The first point we looked at resulted in a positive directional derivative. The point was (-2, 0, -2). Thus, x = -2 and y = 0. Substituting, we get:


D_u_f(-2,0)=(1/sqrt(2))(-(-2)/2+1/2)+(1/sqrt(2))(-(0)+1)=2.8284


As expected, we get a positive number.

For the second point on the surface of f(x, y), the directional derivative was pointing down and thus, negative. The second point was (0, 3, -1.5), i.e. , x = 0 and y = 3: Substituting x and y in, we get:


D_u_f(0,3)=(1/sqrt(2))(-(0)/2+1/2)+(1/sqrt(2))(-(3)+1)=-.7071


This time, we get a negative number!

Gradient

Closely related to the directional derivative is the gradient, ∇f(x, y), which also uses partial derivatives:


nabla_f(x,y)=f_x_i+f_y_b


If we take the dot product of the gradient, ∇f(x, y), with the normalized direction vector we get:


nabla_f(x,y).u=af_x+bf_y


which is our definition of the directional derivative. The relationship between the gradient and the directional derivative is:


D_u_f(x,y)=nabla_f(x,y).u


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