Multivariate Calculus - Bruce E. Shapiro

80 LECTURE 11. GRADIENTS AND THE DIRECTIONAL DERIVATIVE
Hence
At the point (1,1,1) we have
200(xi + yj + zk)
∇T (x, y, z) = −
(10 + x 2 + y 2 + z 2 ) 2
∂
∂x
∣
−200x ∣∣∣(1,1,1)
10 + x 2 + y 2 + z 2 = −200
169
and similarly for the other two partial derivatives. Hence the direction of steepest
increase in temperature is
∇T (x, y, z)| (1,1,1)
= − 200 (i + j + k)
169
Theorem 11.5 The gradient vector is perpendicular to the level curves of a function.
To see why this theorem is true, recall the definition of a level curve: it is a
curve along which the value of z = f(x, y) is a constant. Thus if you move along
a tangent vector to a level curve, the function will not change, and the directional
derivative is zero. Since the directional derivative is the dot product of the gradient
and the direction of motion, which in this case is the tangent vector to the level
curve, this dot product is zero. But a dot product of two non-zero vectors can only
be zero if the two vectors are perpendicular to one another. Thus the gradient must
be perpendicular to the level curve.
Corollary 11.1 Let f(x, y, z) = 0 describe a surface in 3D space. Then the threedimensional
gradient vector ∇f is a normal vector to the surface.
This result follows because a surface f(x, y, z) = 0 is a level surface of a function
w = f(x, y, z). The value of f(x, y, z) does not change. So if you consider any tangent
vector, the directional derivative along the tangent vector is zero. Consequently
the tangent vector is perpendicular to the gradient.
Revised December 6, 2006. Math 250, Fall 2006