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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 \( f \):
\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc7db68c-7828-4a06-bc34-18cdd71bf49b%2F1e0cecf2-dba5-4dbc-ad74-d1ab199f95ae%2Fngwpha_processed.png&w=3840&q=75)
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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