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Sketch the level curves f(r,y) = c of the function f(x,y) = ln(r² + y²) for c = -1,c = 0,c = 1,c = 2. Draw the gradient vectors of f on the level curve f(x, y) = 1. %3D %3D

Sketch the level curves f(r,y) = c of the function f(x,y) = ln(r² + y²) for c = -1,c = 0,c = 1,c = 2. Draw the gradient vectors of f on the level curve f(x, y) = 1. %3D %3D

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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**Sketch the Level Curves and Gradient Vectors** To illustrate the concept of level curves and gradient vectors, consider the function: \[ f(x, y) = \ln(x^2 + y^2) \] You are tasked to: 1. Sketch the level curves \( f(x, y) = c \) for the following values: - \( c = -1 \) - \( c = 0 \) - \( c = 1 \) - \( c = 2 \) 2. Draw the gradient vectors of \( f \) on the level curve where \( f(x, y) = 1 \). Level curves are the contours at which the function \( f(x, y) \) remains constant, giving you a comprehensive view of the function's topology. Each value of \( c \) will generate a different level curve for \( f \). To determine these curves: \[ f(x, y) = \ln(x^2 + y^2) = c \] Exponentiating both sides: \[ x^2 + y^2 = e^c \] This is the equation of a circle with radius \( \sqrt{e^c} \): - For \( c = -1 \): \[ x^2 + y^2 = e^{-1} \approx 0.368 \] Radius \( = \sqrt{0.368} \approx 0.607 \) - For \( c = 0 \): \[ x^2 + y^2 = e^0 = 1 \] Radius \( = \sqrt{1} = 1 \) - For \( c = 1 \): \[ x^2 + y^2 = e^1 \approx 2.718 \] Radius \( = \sqrt{2.718} \approx 1.649 \) - For \( c = 2 \): \[ x^2 + y^2 = e^2 \approx 7.389 \] Radius \( = \sqrt{7.389} \approx 2.718 \) Next, to find the gradient vectors of \( f \) on the level curve \( f(x, y) = 1 \): The gradient of \( f \): \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\
Transcribed Image Text:**Sketch the Level Curves and Gradient Vectors** To illustrate the concept of level curves and gradient vectors, consider the function: \[ f(x, y) = \ln(x^2 + y^2) \] You are tasked to: 1. Sketch the level curves \( f(x, y) = c \) for the following values: - \( c = -1 \) - \( c = 0 \) - \( c = 1 \) - \( c = 2 \) 2. Draw the gradient vectors of \( f \) on the level curve where \( f(x, y) = 1 \). Level curves are the contours at which the function \( f(x, y) \) remains constant, giving you a comprehensive view of the function's topology. Each value of \( c \) will generate a different level curve for \( f \). To determine these curves: \[ f(x, y) = \ln(x^2 + y^2) = c \] Exponentiating both sides: \[ x^2 + y^2 = e^c \] This is the equation of a circle with radius \( \sqrt{e^c} \): - For \( c = -1 \): \[ x^2 + y^2 = e^{-1} \approx 0.368 \] Radius \( = \sqrt{0.368} \approx 0.607 \) - For \( c = 0 \): \[ x^2 + y^2 = e^0 = 1 \] Radius \( = \sqrt{1} = 1 \) - For \( c = 1 \): \[ x^2 + y^2 = e^1 \approx 2.718 \] Radius \( = \sqrt{2.718} \approx 1.649 \) - For \( c = 2 \): \[ x^2 + y^2 = e^2 \approx 7.389 \] Radius \( = \sqrt{7.389} \approx 2.718 \) Next, to find the gradient vectors of \( f \) on the level curve \( f(x, y) = 1 \): The gradient of \( f \): \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\
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