Problem 22 - Calculus for Biology and Medicine

Question

Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right).\) $$ f(x, y)=e^{3 x+2 y} ;(1,2) $$

Step-by-Step Solution

Answer

The linearization of \(f(x, y)\) at (1, 2) is \(e^7(3x + 2y - 4)\).

Step 1: Recall the formula for linearization

The linearization of a multivariable function \(f(x,y)\) at a point \((x_0, y_0)\) is given by the formula \(L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\), where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\) respectively.

Step 2: Compute \(f_x(x, y)\)

Firstly, find the partial derivative of \(f(x,y)\) with respect to \(x\). The function is \(f(x,y) = e^{3x + 2y}\). Differentiate it with respect to \(x\): \[f_x(x, y) = \frac{\partial}{\partial x}(e^{3x + 2y}) = 3e^{3x + 2y}\].

Step 3: Compute \(f_y(x, y)\)

Next, find the partial derivative of \(f(x,y)\) with respect to \(y\). The function is \(f(x,y) = e^{3x + 2y}\). Differentiate it with respect to \(y\): \[f_y(x, y) = \frac{\partial}{\partial y}(e^{3x + 2y}) = 2e^{3x + 2y}\].

Step 4: Evaluate \(f(x, y)\) at \((x_0, y_0)\)

Now, substitute \((x, y) = (1, 2)\) into the original function to get \(f(x_0, y_0)\): \[f(1, 2) = e^{3(1) + 2(2)} = e^7\].

Step 5: Evaluate \(f_x(x_0, y_0)\) and \(f_y(x_0, y_0)\)

Substitute \((x, y) = (1, 2)\) into the expressions for the partial derivatives:\[f_x(1, 2) = 3e^7\]\[f_y(1, 2) = 2e^7\].

Step 6: Write the linearization function

Substitute \(f(1, 2)\), \(f_x(1, 2)\), and \(f_y(1, 2)\) into the linearization formula:\[L(x, y) = e^7 + 3e^7(x - 1) + 2e^7(y - 2)\].

Step 7: Simplify the linearization equation

Simplify the expression of \(L(x, y)\):\[L(x, y) = e^7 + 3e^7(x - 1) + 2e^7(y - 2) = e^7(1 + 3(x - 1) + 2(y - 2))\].Simplify the expression within the parentheses:\[= e^7(3x + 2y - 4)\].

Key Concepts

Partial Derivatives

When working with functions that have multiple variables, partial derivatives allow us to understand how changes in one variable affect the output of the function while keeping other variables constant. For a function of two variables, such as \( f(x, y) = e^{3x + 2y} \), there are two primary partial derivatives:

• The partial derivative of \( f \) with respect to \( x \) (denoted as \( f_x \)) shows how the function changes as \( x \) changes, while \( y \) remains fixed.
• The partial derivative of \( f \) with respect to \( y \) (denoted as \( f_y \)) shows how the function behaves as \( y \) varies, with \( x \) held constant.

In the example of \( f(x, y) = e^{3x + 2y} \), the process of finding \( f_x \) and \( f_y \) involves differentiating the exponential function, considering one variable at a time:

• Differentiate with respect to \( x \): \[ f_x(x, y) = \frac{\partial}{\partial x}(e^{3x + 2y}) = 3e^{3x + 2y} \]
• Differentiate with respect to \( y \): \[ f_y(x, y) = \frac{\partial}{\partial y}(e^{3x + 2y}) = 2e^{3x + 2y} \]

Understanding partial derivatives is crucial in fields ranging from economics to engineering, as they help describe how multidimensional systems respond to individual changes.

Multivariable Function

A multivariable function is a function that has more than one input variable and provides a single output. These functions are essential when modeling real-world phenomena where multiple factors simultaneously affect a system or process. For instance, in our exercise, the function \( f(x, y) = e^{3x + 2y} \) depends on two variables, \( x \) and \( y \). Each term in the exponent represents a linear combination of the variables, which can describe relations such as growth rates or changes in the context they model.
In such functions, we are usually interested in understanding:

• How the function behaves as each variable changes independently (captured by partial derivatives).
• The overall shape or surface that the function describes in space.

Multivariable functions can vary in nature and complexity, ranging from simple linear functions to intricate non-linear models in higher dimensions. Analyzing these functions includes studying their behavior through derivatives, visualizing their surfaces, and determining points of extremum or inflection.

Calculus

Calculus is a branch of mathematics focusing on rates of change and the accumulation of quantities. When dealing with multivariable calculus, we explore not only the rates of change concerning variables but also their interdependence and interaction. In the context of the exercise, multivariable calculus allows us to linearize a function at a particular point. The idea is to approximate the function \( f(x, y) \) near the point \((x_0, y_0)\) with a linear function. This linearization uses the concept of derivatives to provide the best linear approximation around that point:

• The formula \( L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \) stems from the tangent plane concept.
• The partial derivatives \( f_x \) and \( f_y \) give the slope of this plane in their respective directions.

The fundamental principle of calculus underpinning linearization is the ability to use derivative information to create a simpler representation of a complex surface locally. This concept is as vital for high-level theoretical constructs as it is for applied sciences, where calculating predictions based on localized models is common practice.