Electromagnetic Theory (v2) #26 | Problem Solving #03: Gradient in Cartesian

Instructor: Dr. Süleyman Burak ÇELİK

Third problem-solving episode of the EMT (v2) series. The first differential operator of the course — the gradient — computed in Cartesian coordinates on a worked example: f(x, y, z) = x²y + z at Q = (1, 2, 3). Step-by-step partials give ∂f/∂x = 2xy = 4, ∂f/∂y = x² = 1, ∂f/∂z = 1, so ∇f|_Q = 4 x̂ + ŷ + ẑ with |∇f| = √18 = 3√2 ≈ 4.243. Visualization section shows level curves of x²y at z = 3 with the gradient arrow at Q perpendicular to the f = 5 curve, pointing toward f = 6 — the direction of steepest ascent. Reference: Sadiku, Ch.3 (gradient section).