For the function z = = f(x, y) = x² − xy² + y³, find the equations of the tangent lines in both the x-direction & the y-direction at the point (2, 1, 3).
For the function z = = f(x, y) = x² − xy² + y³, find the equations of the tangent lines in both the x-direction & the y-direction at the point (2, 1, 3).
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Topic: Finding Tangent Lines of a Multivariable Function**
**Objective:**
For the function \( z = f(x, y) = x^2 - xy^2 + y^3 \), find the equations of the tangent lines in both the x-direction and the y-direction at the point \( (2, 1, 3) \).
**Function and Point:**
- Function: \( z = f(x, y) = x^2 - xy^2 + y^3 \)
- Point: \( (2, 1, 3) \)
### Step-by-Step Solution:
1. **Find Partial Derivatives:**
- Partial derivative with respect to \( x \):
\[
f_x(x, y) = \frac{\partial}{\partial x} (x^2 - xy^2 + y^3) = 2x - y^2
\]
- Partial derivative with respect to \( y \):
\[
f_y(x, y) = \frac{\partial}{\partial y} (x^2 - xy^2 + y^3) = -2xy + 3y^2
\]
2. **Evaluate Partial Derivatives at Point \( (2, 1) \):**
- \( f_x(2, 1) \):
\[
f_x(2, 1) = 2(2) - 1^2 = 4 - 1 = 3
\]
- \( f_y(2, 1) \):
\[
f_y(2, 1) = -2(2)(1) + 3(1)^2 = -4 + 3 = -1
\]
3. **Equations of Tangent Lines:**
- Tangent line in x-direction:
\[
z - z_0 = f_x(x_0, y_0)(x - x_0)
\]
Given \( (x_0, y_0, z_0) = (2, 1, 3) \), the tangent line equation is:
\[
z - 3 = 3(x - 2) \implies z = 3x - 6 + 3 \implies z = 3x - 3
\]
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ce7c7b2-894f-4c8f-a982-e7518c82686a%2F50c26842-1a7e-4ad7-9d06-332fbd8c2b4e%2F7nygmrc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Finding Tangent Lines of a Multivariable Function**
**Objective:**
For the function \( z = f(x, y) = x^2 - xy^2 + y^3 \), find the equations of the tangent lines in both the x-direction and the y-direction at the point \( (2, 1, 3) \).
**Function and Point:**
- Function: \( z = f(x, y) = x^2 - xy^2 + y^3 \)
- Point: \( (2, 1, 3) \)
### Step-by-Step Solution:
1. **Find Partial Derivatives:**
- Partial derivative with respect to \( x \):
\[
f_x(x, y) = \frac{\partial}{\partial x} (x^2 - xy^2 + y^3) = 2x - y^2
\]
- Partial derivative with respect to \( y \):
\[
f_y(x, y) = \frac{\partial}{\partial y} (x^2 - xy^2 + y^3) = -2xy + 3y^2
\]
2. **Evaluate Partial Derivatives at Point \( (2, 1) \):**
- \( f_x(2, 1) \):
\[
f_x(2, 1) = 2(2) - 1^2 = 4 - 1 = 3
\]
- \( f_y(2, 1) \):
\[
f_y(2, 1) = -2(2)(1) + 3(1)^2 = -4 + 3 = -1
\]
3. **Equations of Tangent Lines:**
- Tangent line in x-direction:
\[
z - z_0 = f_x(x_0, y_0)(x - x_0)
\]
Given \( (x_0, y_0, z_0) = (2, 1, 3) \), the tangent line equation is:
\[
z - 3 = 3(x - 2) \implies z = 3x - 6 + 3 \implies z = 3x - 3
\]
-
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