Problem 38
Question
Find an equation of the plane tangent to the graph of the given function at the indicated point(s). $$ f(x, y)=\left\\{\begin{array}{ll} \frac{x^{3}-y^{3}}{x^{2}+y^{2}} & \text { for }(x, y) \neq(0,0) \\ 0 & \text { for }(x, y)=(0,0) \end{array} ; \quad(1,0,1)\right. $$
Step-by-Step Solution
Verified Answer
The tangent plane equation is \( z = 3x - 2 \).
1Step 1: Understand the Problem
We have a function of two variables, \( f(x, y) = \frac{x^3 - y^3}{x^2 + y^2} \) for \((x, y) eq (0, 0)\), and we need to find the equation of the plane tangent to this function at the point \((1, 0, 1)\).
2Step 2: Verify the Function at Given Point
We need to evaluate \( f(1, 0) \) and check if it is equal to the z-coordinate given, which is 1. Calculating, \( f(1, 0) = \frac{1^3 - 0^3}{1^2 + 0^2} = 1 \), which confirms it matches the z-coordinate.
3Step 3: Find Partial Derivatives
To find the tangent plane, calculate the partial derivatives of \( f(x, y) \). Compute:\[\frac{\partial f}{\partial x} = \frac{(3x^2(x^2+y^2) - (x^3-y^3)(2x))}{(x^2+y^2)^2},\]\[\frac{\partial f}{\partial y} = \frac{(3y^2(x^2+y^2) - (x^3-y^3)(2y))}{(x^2+y^2)^2}.\]
4Step 4: Evaluate Partial Derivatives at (1,0)
Substitute \( x = 1 \) and \( y = 0 \) into the partial derivatives:\[\frac{\partial f}{\partial x}(1,0) = \frac{3 \times 1^2 \times 1}{1^2} = 3, \]\[\frac{\partial f}{\partial y}(1,0) = \frac{0}{1^2} = 0.\]
5Step 5: Write the Equation of the Tangent Plane
Using the tangent plane formula at \((x_0, y_0, z_0) = (1, 0, 1)\):\[ z = f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0) \]Insert the values:\[ z = 1 + 3(x - 1) + 0(y - 0) \]Simplify to:\[ z = 3x - 2.\]
Key Concepts
Partial DerivativesMultivariable CalculusFunction of Two Variables
Partial Derivatives
Understanding partial derivatives is key to tackling problems involving functions of multiple variables. When we take a partial derivative, we differentiate with respect to one variable and treat all other variables as constants. This concept is similar to regular differentiation, but here, each variable can be held constant while focusing on the others.
In the exercise at hand, the function given was \(f(x, y) = \frac{x^3 - y^3}{x^2 + y^2}\) with the specified point \((1, 0, 1)\). To find the equation of the tangent plane, we first calculated the partial derivatives:
In the exercise at hand, the function given was \(f(x, y) = \frac{x^3 - y^3}{x^2 + y^2}\) with the specified point \((1, 0, 1)\). To find the equation of the tangent plane, we first calculated the partial derivatives:
- \(\frac{\partial f}{\partial x}\), the derivative concerning \(x\), measures how \(f(x, y)\) changes as \(x\) changes, while \(y\) is constant.
- \(\frac{\partial f}{\partial y}\), the derivative with respect to \(y\), shows the change in \(f(x, y)\) as \(y\) varies, keeping \(x\) constant.
Multivariable Calculus
Multivariable calculus extends single-variable calculus to functions of multiple variables, like \(f(x, y)\). This branch of mathematics deals with functions that have two or more variables, making it crucial for understanding real-world systems where outcomes depend on several factors.
The application of multivariable calculus in the tangent plane problem involves calculating derivatives concerning each variable (partial derivatives) and evaluating them at a specific point. These derivatives help us determine how the function behaves close to the point.
The process of determining a tangent plane equation uses this concept to approximate the graph of \(f(x, y)\) at a particular point \((1, 0, 1)\). This approximation allows one to visualize the surface in the immediate vicinity, helping you gain insights into its structure.
The application of multivariable calculus in the tangent plane problem involves calculating derivatives concerning each variable (partial derivatives) and evaluating them at a specific point. These derivatives help us determine how the function behaves close to the point.
The process of determining a tangent plane equation uses this concept to approximate the graph of \(f(x, y)\) at a particular point \((1, 0, 1)\). This approximation allows one to visualize the surface in the immediate vicinity, helping you gain insights into its structure.
Function of Two Variables
A function of two variables, such as \(f(x, y)\), provides a rule to assign a unique output \(z\) for each input pair \((x, y)\). These types of functions often represent scalar fields, where each point in a plane is assigned a scalar value, forming a surface in three-dimensional space.
In our exercise, \(f(x, y)\) is defined such that it behaves differently at the origin compared to all other points. Such approaches to defining functions account for discontinuities or special cases where functions may not typically be defined.
To find the tangent plane for \(f(x, y)\) at \((1, 0, 1)\), the method utilizes the specific behavior (slope changes) at that point, provided by the function's partial derivatives. This local linear approximation encapsulates the surface's inclination at that point, acting as the best linear approximation nearby.
In our exercise, \(f(x, y)\) is defined such that it behaves differently at the origin compared to all other points. Such approaches to defining functions account for discontinuities or special cases where functions may not typically be defined.
To find the tangent plane for \(f(x, y)\) at \((1, 0, 1)\), the method utilizes the specific behavior (slope changes) at that point, provided by the function's partial derivatives. This local linear approximation encapsulates the surface's inclination at that point, acting as the best linear approximation nearby.
Other exercises in this chapter
Problem 38
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