Question

(b). $f = \ln(x^2 + y^2)$ \newline $\nabla f = \frac{2x}{x^2 + y^2} \hat{i} + \frac{2y}{x^2 + y^2} \hat{j}$ \newline $\nabla \cdot \nabla f = 2 \frac{\partial}{\partial x} (\frac{x}{x^2 + y^2}) + 2 \frac{\partial}{\partial y} (\frac{y}{x^2 + y^2})$ \newline $= \frac{2}{x^2 + y^2} - \frac{2x}{(x^2 + y^2)^2} (2x) + \frac{2}{x^2 + y^2} - \frac{2y}{(x^2 + y^2)^2} (2y)$ \newline $= \frac{4}{x^2 + y^2} - \frac{1}{(x^2 + y^2)^2} (4x^2 + 4y^2) = 0$

Name: (b). f = ln(x^2 + y^2) ∇ f = (2x)/(x^2 + y^2)î + (2y)/(x^2 + y^2)ĵ ∇·∇ f = 2 (∂)/(∂ x) ((x)/(x^2 + y^2)) + 2 (∂)/(∂ y) ((y)/(x^2 + y^2)) = (2)/(x^2 + y^2) - (2x)/((x^2 + y^2)^2) (2x) + (2)/(x^2 + y^2) - (2y)/((x^2 + y^2)^2) (2y) = (4)/(x^2 + y^2) - (1)/((x^2 + y^2)^2) (4x^2 + 4y^2) = 0
Uploaded: 2023-04-06T14:22:09-08:00
Duration: 2 min 42 s
Channel: Madhur L
Description: (b). f = ln(x^2 + y^2) ∇ f = (2x)/(x^2 + y^2)î + (2y)/(x^2 + y^2)ĵ ∇·∇ f = 2 (∂)/(∂ x) ((x)/(x^2 + y^2)) + 2 (∂)/(∂ y) ((y)/(x^2 + y^2)) = (2)/(x^2 + y^2) - (2x)/((x^2 + y^2)^2) (2x) + (2)/(x^2 + y^2) - (2y)/((x^2 + y^2)^2) (2y) = (4)/(x^2 + y^2) - (1)/((x^2 + y^2)^2) (4x^2 + 4y^2) = 0

          (b). $f = \ln(x^2 + y^2)$ \newline $\nabla f = \frac{2x}{x^2 + y^2} \hat{i} + \frac{2y}{x^2 + y^2} \hat{j}$ \newline $\nabla \cdot \nabla f = 2 \frac{\partial}{\partial x} (\frac{x}{x^2 + y^2}) + 2 \frac{\partial}{\partial y} (\frac{y}{x^2 + y^2})$ \newline $= \frac{2}{x^2 + y^2} - \frac{2x}{(x^2 + y^2)^2} (2x) + \frac{2}{x^2 + y^2} - \frac{2y}{(x^2 + y^2)^2} (2y)$ \newline $= \frac{4}{x^2 + y^2} - \frac{1}{(x^2 + y^2)^2} (4x^2 + 4y^2) = 0$

(b). f = ln(x^2 + y^2) ∇ f = (2x)/(x^2 + y^2)î + (2y)/(x^2 + y^2)ĵ ∇·∇ f = 2 (∂)/(∂ x) ((x)/(x^2 + y^2)) + 2 (∂)/(∂ y) ((y)/(x^2 + y^2)) = (2)/(x^2 + y^2) - (2x)/((x^2 + y^2)^2) (2x) + (2)/(x^2 + y^2) - (2y)/((x^2 + y^2)^2) (2y) = (4)/(x^2 + y^2) - (1)/((x^2 + y^2)^2) (4x^2 + 4y^2) = 0

Added by Allen W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

04/06/2023

Step 1

Step 1: The given function is f=ln(x^(2)+y^(2)). Show more…

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Q2. Calculate grad^(2)f using grad^(2)f=div(gradf) Check by direct differentiation (b) f=ln(x^(2)+y^(2)) I have uploaded the answer that I was given. Please explain the differentiation used especially to calculate the divergence.