Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

find the gradient vector of the given function at the given point \(\mathbf{p}\). Then find the equation of the tangent plane at \(\mathbf{p}\) \((\) see Example 1). $$ f(x, y)=x^{2} y-x y^{2}, \mathbf{p}=(-2,3) $$

Step-by-Step Solution

Answer

Gradient vector: \( \langle -21, 16 \rangle \); tangent plane: \( z = -21x + 16y - 8 \).

Step 1: Find the Partial Derivative with respect to x

To find the gradient vector, we first calculate the partial derivative of the function \( f(x, y) = x^2y - xy^2 \) with respect to \( x \). Differentiating, we get \( f_x(x, y) = 2xy - y^2 \).

Step 2: Find the Partial Derivative with respect to y

Next, compute the partial derivative with respect to \( y \). Differentiating the function \( f(x, y) = x^2y - xy^2 \) with respect to \( y \), we get \( f_y(x, y) = x^2 - 2xy \).

Step 3: Calculate the Gradient at Point \( \mathbf{p} = (-2, 3) \)

Evaluate the partial derivatives at the point \((-2, 3)\): \[ f_x(-2, 3) = 2(-2)(3) - 3^2 = -12 - 9 = -21 \] \[ f_y(-2, 3) = (-2)^2 - 2(-2)(3) = 4 + 12 = 16 \] Thus, the gradient vector at \( \mathbf{p} \) is \( abla f(-2, 3) = \langle -21, 16 \rangle \).

Step 4: Use Gradient to Form the Tangent Plane Equation

The equation of the tangent plane at a point \( \mathbf{p} = (a, b) \) is given by: \[ z - f(a, b) = f_x(a, b)(x - a) + f_y(a, b)(y - b) \]We already have \( f_x(-2, 3) = -21 \), \( f_y(-2, 3) = 16 \), and \( f(-2, 3) = (-2)^2(3) - (-2)(3)^2 = 12 \ - 18 = -6 \).So, the tangent plane is: \[ z + 6 = -21(x + 2) + 16(y - 3) \] Simplifying, we get the equation: \[ z = -21x + 16y - 8 \]

Key Concepts

Partial Derivative

When working with functions of multiple variables, like in our exercise with the function \( f(x, y) = x^2y - xy^2 \), we often need to find how the function changes as we alter one variable at a time. This is where the concept of a partial derivative comes in. Unlike a single-variable derivative where we look at the slope of a curve, a partial derivative focuses on changes in a specific direction—holding other variables constant.
To find the partial derivative with respect to \( x \), we consider \( y \) as a constant, and differentiate the expression as we would with a single-variable function. In our example:

• First partial derivative with respect to \( x \): \( f_x(x, y) = 2xy - y^2 \).
• Similarly, for \( y \), treat \( x \) as a constant to find \( f_y(x, y) = x^2 - 2xy \).

Understanding partial derivatives is crucial as they help us explore how a function behaves locally, and in our steps, they guided us to create the gradient vector.

Tangent Plane

In multivariable calculus, a tangent plane is a flat surface that just barely touches a point on a function's graph. Much like how the tangent line touches a curve in calculus with one variable, the tangent plane provides a linear approximation at a specific point.
Given the point \( \mathbf{p} = (-2, 3) \) and our earlier gradient calculations \( \langle -21, 16 \rangle \), we use these results to create the equation of the tangent plane:

• The equation is based on: \( z - f(a, b) = f_x(a, b)(x - a) + f_y(a, b)(y - b) \).
• With \( f(-2, 3) = -6 \), the equation becomes \( z + 6 = -21(x + 2) + 16(y - 3) \).
• This simplifies to \( z = -21x + 16y - 8 \).

The tangent plane is vital for understanding the local behavior of surfaces, as it approximates the surface near the point \( \mathbf{p} \).

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions with more than one variable. It's an essential field that opens up a wide array of applications, from physics to engineering and economics.
Key features include:

• Functions with more than one input (e.g., \( f(x, y) \))
• Partial derivatives to study the effect of changing one variable while keeping others constant
• Gradient vectors, which are crucial for finding direction and rate of maximum increase of a function
• Surface integrals and tangent planes to better understand the geometry and behavior of multidimensional functions

In our problem, using partial derivatives and forming a gradient vector, we explored the 3D representation provided by the function, going beyond simple curves to intricate surfaces.

Differentiation

Differentiation is the process of finding derivatives, which measure how a function changes. In single-variable calculus, this involves computing the slope of a curve. However, in multivariable calculus, the process becomes richer as there are multiple pathways to explore the change.
For a function like \( f(x, y) \), we look at the rate of change with respect to each variable separately, using partial derivatives:

• \( f_x(x, y) \) gives us the slope in the \( x \)-direction, keeping \( y \) constant.
• \( f_y(x, y) \) does the same for the \( y \)-direction.

After finding each partial derivative, they form a component of the gradient vector, which is foundational for deeper analysis. Differentiation remains a powerful tool, helping us derive meaningful insights about the behavior of intricate systems and surfaces.