Problem 13 - 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). $$ f(x, y)=\cos \pi x \sin \pi y+\sin 2 \pi y, \mathbf{p}=\left(-1, \frac{1}{2}\right) $$

Step-by-Step Solution

Answer

Gradient: \((0, -2\pi)\), Tangent Plane: \(z = 1 + \pi - 2\pi y\)."

Step 1: Calculate Partial Derivatives

To find the gradient vector, we first need to calculate the partial derivatives of the function with respect to both variables, \(x\) and \(y\). For \(f(x, y) = \cos(\pi x) \sin(\pi y) + \sin(2\pi y)\):- The partial derivative with respect to \(x\) is: \(f_x(x, y) = \frac{d}{dx}[\cos(\pi x) \sin(\pi y)] = -\pi \sin(\pi x)\sin(\pi y)\).- The partial derivative with respect to \(y\) is: \(f_y(x, y) = \frac{d}{dy}[\cos(\pi x) \sin(\pi y) + \sin(2\pi y)] = \pi \cos(\pi x)\cos(\pi y) + 2\pi \cos(2\pi y)\).

Step 2: Evaluate the Gradients at Point p

Now, substitute the coordinates of point \(\mathbf{p} = (-1, \frac{1}{2})\) into the partial derivatives to find the gradient vector at this point.- For \(f_x\): \(f_x(-1, \frac{1}{2}) = -\pi \sin(-\pi) \sin\left(\frac{\pi}{2}\right) = 0\) because \(\sin(\pi) = 0\).- For \(f_y\): \(f_y(-1, \frac{1}{2}) = \pi \cos(-\pi)\cos\left(\frac{\pi}{2}\right) + 2\pi \cos(\pi) = 0 + 2\pi(-1) = -2\pi\).

Step 3: Construct the Gradient Vector

The gradient vector \(abla f\) at point \(\mathbf{p}\) is \(abla f(-1, \frac{1}{2}) = (0, -2\pi)\). This vector will guide us to find the tangent plane in the next step.

Step 4: Equation of the Tangent Plane

The equation for the tangent plane at a point \((x_0, y_0)\) is given by:\[ z = f(x_0, y_0) + abla f(x_0, y_0) \cdot \begin{bmatrix} (x - x_0) \ (y - y_0) \end{bmatrix}.\]Substitute \((x_0, y_0) = (-1, \frac{1}{2})\) and \(abla f(-1, \frac{1}{2}) = (0, -2\pi)\):- First, compute \(f(-1, \frac{1}{2}) = \cos(-\pi)\sin\left(\frac{\pi}{2}\right) + \sin(\pi) = 1\cdot 1 + 0 = 1\).- Plug into the plane equation: \[ z = 1 + 0(x + 1) - 2\pi(y - \frac{1}{2}) \].- Simplify the equation to \[ z = 1 - 2\pi y + \pi \].

Step 5: Simplify Tangent Plane Equation

Simplify the tangent plane equation further:- Combine like terms: \[ z = 1 + \pi - 2\pi y \].- Therefore, the equation of the tangent plane is: \[ z = 1 - 2\pi y + \pi = 1 + \pi - 2\pi y \].

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They involve taking the derivative of a function with respect to one variable, while keeping all other variables constant. This is crucial for understanding how changes in one variable affect the overall behavior of a function.
For functions in two variables like \(f(x, y)\), the partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), indicates the rate of change of \(f\) as \(x\) changes while \(y\) is held constant. Similarly, the partial derivative with respect to \(y\), denoted as \(f_y(x, y)\), shows the rate of change of \(f\) with changing \(y\) while \(x\) remains fixed.

• Notation: \(f_x(x, y)\) and \(f_y(x, y)\) are the common notations for partial derivatives, regardless of the function's complexity.
• Calculation: Use standard differentiation rules, treating other variables as constants.
• Purpose: Helps find the gradient vector and understand the function's behavior and slope.

In this exercise, the calculated partial derivatives are essential for obtaining the gradient vector needed for formulating the equation of the tangent plane.

Tangent Plane Equation

The equation of the tangent plane is a key element in multivariable calculus, providing a linear approximation of a surface at a given point. This equation helps to simplify the examination of complex surfaces by replacing the surface around a small region with a plane.
The general formula for the tangent plane to a surface \(z = f(x, y)\) at a point \((x_0, y_0, z_0)\) is:
\[ z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0). \]

• Point of tangency: \((x_0, y_0, z_0)\) is the point where the plane is tangent to the surface.
• Gradient vector: \((f_x(x_0, y_0), f_y(x_0, y_0))\) directs the slope and orientation of the tangent plane.
• Linear approximation: Provides a simple approach to analyze changes near the tangent point.

In our example, the tangent plane equation is derived using the gradient vector \((0, -2\pi)\) at the point \((-1, \frac{1}{2})\) and simplifies to \(z = 1 + \pi - 2\pi y\). This offers an insightful way to comprehend the function around that point.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of multiple variables. It allows us to understand and analyze scenarios where more than one variable influences the outcome, such as physical systems and geometrical surfaces.
Some key concepts include:

• Function of multiple variables: For example, \(f(x, y)\) can represent a surface where \(x\) and \(y\) are independent variables.
• Partial derivatives: As discussed, these are derivatives with respect to one variable at a time.
• Gradient vector: A vector composed of partial derivatives, pointing towards the direction of the steepest ascent of the function.
• Tangent planes and surfaces: Local linear approximations to better understand and manipulate 3D surfaces.

Multivariable calculus is essential in fields like physics, engineering, and economics, where multiple factors concurrently affect outcomes. Mastery of this subject provides powerful tools for modeling and solving complex real-world problems.