Problem 16 - Calculus Early Transcendentals

Question

\(11-16\) Explain why the function is differentiable at the given point. Then find the linearization \(L(x, y)\) of the function at that point. $$ f(x, y)=\sin (2 x+3 y), \quad(-3,2) $$

Step-by-Step Solution

Answer

The function is differentiable at \((-3, 2)\) with linearization \(L(x, y) = 2x + 3y\).

Step 1: Determine the differentiability criteria

A function of two variables is differentiable at a point if all of the partial derivatives exist and are continuous in a neighborhood around that point. In this case, consider the partial derivatives of the function with respect to both variables and check for continuity.

Step 2: Compute Partial Derivatives

Find the partial derivative of the function with respect to \(x\), denoted as \(f_x(x, y)\), and the partial derivative with respect to \(y\), denoted as \(f_y(x, y)\). For \(f(x, y) = \sin(2x + 3y)\), \[ f_x(x, y) = 2\cos(2x + 3y) \] \[ f_y(x, y) = 3\cos(2x + 3y) \]

Step 3: Evaluate the Partial Derivatives at the Point

Substitute \((-3, 2)\) into the partial derivatives obtained. This gives:\[ f_x(-3, 2) = 2\cos(2(-3) + 3(2)) = 2\cos(0) = 2 \] \[ f_y(-3, 2) = 3\cos(2(-3) + 3(2)) = 3\cos(0) = 3 \] Since both partial derivatives exist, are finite, and the sine function (and consequently the cosine function) is continuous everywhere, \(f(x, y)\) is differentiable at the point \((-3, 2)\).

Step 4: Compute the Linearization

The linearization of \(f(x, y)\) at a point \((x_0, y_0)\) is given by \[ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \] First, evaluate \(f(-3, 2)\):\[ f(-3, 2) = \sin(2(-3) + 3(2)) = \sin(0) = 0 \] Now apply the linearization formula:\[ L(x, y) = 0 + 2(x + 3) + 3(y - 2) \]

Step 5: Simplify the Linearization Expression

Simplify the expression obtained in the linearization formula.\[ L(x, y) = 2(x + 3) + 3(y - 2) = 2x + 6 + 3y - 6 = 2x + 3y \]Thus, the linearization at the point is \(L(x, y) = 2x + 3y\).

Key Concepts

Partial Derivatives

Let's explore the foundational concept of partial derivatives in the context of a multivariable function like \( f(x, y) = \sin(2x + 3y) \). In multivariable calculus, partial derivatives help us understand how a function changes as we change just one of its input variables while keeping the others constant.

For our function, to find the partial derivative with respect to \( x \) (denoted as \( f_x(x, y) \)), we treat \( y \) as a constant and differentiate \( f \) with respect to \( x \). The result is \( f_x(x, y) = 2\cos(2x + 3y) \). This shows how the function varies as \( x \) changes.

• Similarly, for the partial derivative with respect to \( y \) (\( f_y(x, y) \)), treat \( x \) as constant and differentiate with respect to \( y \). Thus, \( f_y(x, y) = 3\cos(2x + 3y) \).
• Evaluating these at the point \((-3, 2)\): \( f_x(-3, 2) = 2 \) and \( f_y(-3, 2) = 3 \). These derivatives inform us of the slope of the tangent in the \( x \) and \( y \) directions, respectively.

This calculation shows that both partial derivatives exist and are continuous, fulfilling one of the main conditions for differentiability of a function at a given point.

Linearization

Linearization is a valuable tool in approximating multivariable functions near a given point. It transforms non-linear functions into a linear form, making complex calculations more manageable. The linearization of a function at a specific point offers a linear approximation of the function's behavior near that point.
The formula for the linearization \( L(x, y) \) at a point \((x_0, y_0)\) is:

• \[ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \]

For the function \( f(x, y) = \sin(2x + 3y) \) with a linearization point \( (-3, 2) \), we've already found our partial derivatives \( f_x = 2 \) and \( f_y = 3 \). Evaluating the function at this point gives us \( f(-3, 2) = 0 \). Integrating these results into the linearization formula, we derive:

• \( L(x, y) = 0 + 2(x + 3) + 3(y - 2) \)
• This simplifies to \( L(x, y) = 2x + 3y \).

This linear equation, \( L(x, y) = 2x + 3y \), is an approximate representation of our original function around the point \((-3, 2)\). It efficiently describes the function's behavior near this point, emphasizing a straightforward relationship between \( x \) and \( y \).

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. It's the mathematics of curves, planes, and surfaces, allowing us to explore complex functions beyond the single-variable limitations. Understanding differentiability and linearization in this context aids in grasping changes in surface elevations or velocities in physics and engineering.

The function \( f(x, y) = \sin(2x + 3y) \) offers an insight into how these concepts play out when variables interact. The core elements include:

• Partial Derivatives: Key in understanding how the function changes as individual variables (\(x\) or \(y\)) change, setting the stage for analyzing gradients or directions of fastest change.
• Differentiability: To qualify as differentiable at a point, a function must have continuous partial derivatives in a neighborhood around that point. This ensures smooth transitions and no abrupt changes or cusps.
• Linearization: Streamlining complex, non-linear functions into linear models allows for easier computation and analysis, especially in practical applications such as computer graphics, economics, or physical simulations.

Through multivariable calculus, we deepen our exploration of mathematics, making essential applications possible in the technological and scientific framework of today's world.