For the function z = f(x, y) = x² − xy² + y³, find the equations of the tangent lines in both the c-direction & the y-direction at the point ( 2, 1, 3). 1.1 2.1 3.1 *Activity 3

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Chapter6: Rates Of Change
Section6.1: Velocity
Problem 2SBE: Sign of VelocityWhen directed distance is decreasing, is velocity positive or negative? What is the...
Question
### Finding Tangent Line Equations for Multivariable Functions

**Problem Statement:**
Given the function \( z = f(x, y) = x^2 - xy^2 + y^3 \), find the equations of the tangent lines in both the *x*-direction & the *y*-direction at the point \( (2, 1, 3) \).

**Visualization:**
The 3D plot on the right depicts the surface of the function \( z = f(x, y) = x^2 - xy^2 + y^3 \). The point \( (2, 1, 3) \) is marked on the surface. The plot gives a visual representation of how the surface behaves around the point of interest.

#### Steps to Find Tangent Lines:

1. **Calculate 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 the Point (2, 1, 3):**
   - Evaluate \( f_x(2, 1) \):
     \[
     f_x(2, 1) = 2(2) - (1)^2 = 4 - 1 = 3
     \]
   - Evaluate \( f_y(2, 1) \):
     \[
     f_y(2, 1) = -2(2)(1) + 3(1)^2 = -4 + 3 = -1
     \]

3. **Write Equations for Tangent Lines:**
   Using the point (2, 1, 3) and the slopes found from the partial derivatives:
   - Tangent line in the *x*-direction:
     \[
     z = 3 + 3(x - 2)
     \]
     Simplify:
     \[
     z = 3 + 3x - 6 \implies z = 3x
Transcribed Image Text:### Finding Tangent Line Equations for Multivariable Functions **Problem Statement:** Given the function \( z = f(x, y) = x^2 - xy^2 + y^3 \), find the equations of the tangent lines in both the *x*-direction & the *y*-direction at the point \( (2, 1, 3) \). **Visualization:** The 3D plot on the right depicts the surface of the function \( z = f(x, y) = x^2 - xy^2 + y^3 \). The point \( (2, 1, 3) \) is marked on the surface. The plot gives a visual representation of how the surface behaves around the point of interest. #### Steps to Find Tangent Lines: 1. **Calculate 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 the Point (2, 1, 3):** - Evaluate \( f_x(2, 1) \): \[ f_x(2, 1) = 2(2) - (1)^2 = 4 - 1 = 3 \] - Evaluate \( f_y(2, 1) \): \[ f_y(2, 1) = -2(2)(1) + 3(1)^2 = -4 + 3 = -1 \] 3. **Write Equations for Tangent Lines:** Using the point (2, 1, 3) and the slopes found from the partial derivatives: - Tangent line in the *x*-direction: \[ z = 3 + 3(x - 2) \] Simplify: \[ z = 3 + 3x - 6 \implies z = 3x
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