Problem 24 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The linearization is \( L(x, y) = 2x + y - 1 \).

Step 1: Recall the Formula for Linearization

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

Step 2: Evaluate the Function at the Given Point

Evaluate the function \( f(x, y) = x^2 e^y \) at the point \((1,0)\):\[ f(1,0) = (1)^2 e^0 = 1 \]

Step 3: Compute the Partial Derivative with Respect to x

Differentiate \( f(x, y) = x^2 e^y \) with respect to \( x \):\[ f_x = \frac{\partial}{\partial x}(x^2 e^y) = 2x e^y \]Evaluate it at \( (1, 0) \):\[ f_x(1, 0) = 2(1) e^0 = 2 \]

Step 4: Compute the Partial Derivative with Respect to y

Differentiate \( f(x, y) = x^2 e^y \) with respect to \( y \):\[ f_y = \frac{\partial}{\partial y}(x^2 e^y) = x^2 e^y \]Evaluate it at \( (1, 0) \):\[ f_y(1, 0) = (1)^2 e^0 = 1 \]

Step 5: Construct the Linearization

Substitute all calculated values into the linearization formula:\[ L(x, y) = 1 + 2(x - 1) + 1(y - 0) \]Simplify:\[ L(x, y) = 2x + y - 1 \]

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are important tools that help us understand the nature of functions with more than one variable. When we take a partial derivative, we differentiate with respect to one variable while treating other variables as constants.
This is particularly useful in the context of functions like our exercise's function, where we have variables \(x\) and \(y\).

• The partial derivative with respect to \(x\), denoted \(f_x\), gives us the rate of change of the function as \(x\) changes, holding \(y\) constant.
• Similarly, the partial derivative with respect to \(y\), denoted \(f_y\), shows how the function changes as \(y\) changes, with \(x\) held constant.

For our given function \(f(x, y) = x^2 e^y\):

• Finding \(f_x\) involves differentiating \(x^2 e^y\) in terms of \(x\), resulting in \(2x e^y\).
• Calculating \(f_y\), we differentiate in terms of \(y\), resulting in \(x^2 e^y\).

These partial derivatives are crucial for constructing the linearization of a function, as seen in the solution steps.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus, such as differentiation, into functions of several variables rather than just one.
It allows us to analyze and approximate the behavior of complex systems that depend on multiple factors. In our exercise, we see a function \(f(x, y) = x^2 e^y\) that depends on two variables, \(x\) and \(y\). Understanding the behavior of such a function requires tools from multivariable calculus, including partial derivatives and linearization.
One of the key aspects of multivariable calculus is the ability to approximate functions using linearization. This is similar to using tangent lines in single-variable calculus.
The linearization formula allows us to create a tangent plane that approximates the function near a specific point. This is particularly useful when evaluating functions in contexts where exact calculations are difficult or unnecessary.

Differentiation

Differentiation, a fundamental technique in calculus, involves finding the rate at which things change. For multivariable calculus, we deal with partial derivatives as a form of differentiation, examining how individual changes in variables affect a function.
In the context of our exercise, differentiation helped us calculate the partial derivatives \(f_x\) and \(f_y\). Differentiating a function gives us insights into its behavior, showing exactly how changes in one variable influence the overall output.

• When differentiating \(f(x, y) = x^2 e^y\) with respect to \(x\), we obtained \(f_x = 2x e^y\).
• Upon differentiating with respect to \(y\), we calculated \(f_y = x^2 e^y\).

This differentiation tells us:

• For \(f_x\), how the function's value changes as \(x\) moves.
• For \(f_y\), how the function's value changes with \(y\).

These derivatives play a critical role in linearization, used to determine the best linear approximation of a function near a specific point, as demonstrated at the given point \((1,0)\). This approximation is extremely useful in various fields such as economics, physics, and engineering, where understanding and predicting change is key.