Gradient: deﬁnition and properties Deﬁnition of the gradient ∂w ∂w If w = f (x, y), then and are the rates of change of w in the i and j directions. ∂x ∂y It will be quite useful to put these two derivatives together in a vector called the gradient of w. � � ∂w ∂w , grad w = . ∂x ∂y We will also use the symbol �w to denote the gradient. (You read this as ’gradient of w’ or ’grad w’.) Of course, if we specify a point P0 = (x0 , y0 ), we can evaluate the gradient at that point. We will use several notations for this � � � � ∂w �� ∂w �� grad w(x0 , y0 ) = � w|P0 = � w|o = , . ∂x �o ∂y �o Note well the following: (as we look more deeply into properties of the gradient these can be points of confusion). 1. The gradient takes a scalar function f (x, y) and produces a vector �f . 2. The vector �f (x, y) lies in the plane. For functions w = f (x, y, z) we have the gradient � � ∂w ∂w ∂w grad w = �w = , , . ∂x ∂y ∂z That is, the gradient takes a scalar function of three variables and produces a three dimen sional vector. The gradient has many geometric properties. In the next session we will prove that for w = f (x, y) the gradient is perpendicular to the level curves f (x, y) = c. We can show this by direct computation in the following example. Example 1: Compute the gradient of w = (x2 + y 2 )/3 and show that the gradient at (x0 , y0 ) = (1, 2) is perpendicular to the level curve through that point. y� Answer: The gradient is easily computed � 2 �w = �2x/3, 2y/3� = �x, y�. 3 At (1, 2) we get �w(1, 2) = 23 �1, 2�. The level curve through (1, 2) is 2 2 (x + y )/3 = 5/3, x2 + y 2 √ which is identical to = 5. That is, it is a circle of radius 5 centered at the origin. Since the gradient at (1,2) is a multiple of �1, 2�, it points radially outward and hence is perpendicular to the circle. Below is a ﬁgure showing the gradient ﬁeld and the level curves. � � � � � � � � � � � � � � � � � � � x� � � z = (x2 + y 2 )/3 Example 2: Consider the graph of y = ex . Find a vector perpendicular to the tangent to y = ex at the point (1, e). Old method: Find the slope take the negative reciprocal and make the vector. New method: This graph is the level curve of w = y − ex = 0. �w = �−ex , 1� ⇒ (at x = 1) �w(1, e) = �−e, 1� is perpendicular to the tangent vector to the graph, v = �1, e�. Higher dimensions Similarly, for w = f (x, y, z) we get level surfaces f (x, y, z) = c. The gradient is perpendic ular to the level surfaces. Example 3: Find the tangent plane to the surface x2 + 2y 2 + 3z 2 = 6 at the point P = (1, 1, 1). Answer: Introduce a new variable w = x2 + 2y 2 + 3z 2 . Our surface is the level surface w = 6. Saying the gradient is perpendicular to the surface means exactly the same thing as saying it is normal to the tangent plane. Computing �w = �2x, 4y, 6z� ⇒ �w|P = �2, 4, 6�. Using point normal form we get the equation of the tangent plane is 2(x − 1) + 4(y − 1) + 6(z − 1) = 0, or 2x + 4y + 6z = 12.