Partial Derivatives, Gradients & Directional Derivatives
“Gradient vector grad(f), directional derivative D_u f = grad(f) . u, tangent hyperplanes, and Jacobian matrices.”
1.Multivariable Geometry, Gradients & Directional Derivatives
§ 2.1.For a scalar field f: R^n -> R, the Gradient Vector grad(f) packages all first-order partial derivatives:
grad(f) = [ df/dx_1, df/dx_2, ..., df/dx_n ]^T.
Geometric Meaning of the Gradient:
1. grad(f) is strictly orthogonal to the level curve (or level surface) f(x) = c at every point.
2. grad(f) points in the direction of fastest increase.
3. Its magnitude ||grad(f)|| is the maximum possible directional derivative.
The Directional Derivative:
For unit direction u (||u|| = 1), D_u f(x) = lim_{h -> 0} [ f(x + h u) - f(x) ] / h = grad(f(x)) . u.
By Cauchy-Schwarz: |D_u f| <= ||grad(f)|| * ||u|| = ||grad(f)||.
2. Worked Practice Problem: 3D Directional Derivative & Normal Vector
Normalize the given 3D direction vector before computing the inner product.
Problem: Find the directional derivative of f(x, y, z) = x^2 y z + 2 y^2 z at point P(1, 1, 2) in the direction of vector v = [ 1, 2, 2 ]^T.
Step 1: Compute partial derivatives of f:
df/dx = 2xy*z
df/dy = x^2 z + 4y*z
df/dz = x^2 y + 2y^2
Step 2: Evaluate gradient vector at P(1, 1, 2):
df/dx = 2(1)(1)(2) = 4.
df/dy = (1)^2(2) + 4(1)(2) = 2 + 8 = 10.
df/dz = (1)^2(1) + 2(1)^2 = 1 + 2 = 3.
grad(f)(P) = [ 4, 10, 3 ]^T.
Step 3: Normalize direction vector v:
||v|| = sqrt( 1^2 + 2^2 + 2^2 ) = sqrt( 1 + 4 + 4 ) = sqrt(9) = 3.
Unit vector u = [ 1/3, 2/3, 2/3 ]^T.
Step 4: Compute dot product D_u f = grad(f) . u:
D_u f = (4)(1/3) + (10)(2/3) + (3)(2/3)
D_u f = (4 / 3) + (20 / 3) + (6 / 3) = 30 / 3 = 10.
Conclusion: The directional derivative at P in the direction of v is D_u f = 10.
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