Problem 10 - Calculus for Biology and Medicine

Question

The tangent plane at the indicated point \(\left(x_{0}, y_{0}, z_{0}\right)\) exists. Find its equation. $$ f(x, y)=\ln \left(x^{2}+y^{2}\right) ;(1,1, \ln 2) $$

Step-by-Step Solution

Answer

The equation of the tangent plane is \(z = x + y + \ln 2 - 2\).

Step 1: Find Partial Derivatives

First, we find the partial derivatives of the function. For a function \(f(x, y) = \ln(x^2 + y^2)\), the partial derivatives are necessary to create the tangent plane. The partial derivative with respect to \(x\) is \(f_x = \frac{2x}{x^2 + y^2}\). The partial derivative with respect to \(y\) is \(f_y = \frac{2y}{x^2 + y^2}\).

Step 2: Evaluate Partial Derivatives at Point

Now, evaluate the partial derivatives at the point \((1,1)\). Substitute \(x = 1\) and \(y = 1\) into the partial derivatives. \(f_x(1,1) = \frac{2 \times 1}{1^2 + 1^2} = \frac{2}{2} = 1\). Similarly, \(f_y(1,1) = \frac{2 \times 1}{1^2 + 1^2} = \frac{2}{2} = 1\).

Step 3: Write Equation of Tangent Plane

Use the linearization or tangent plane formula: \(z = f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)\). Substituting the given point \((1,1,\ln 2)\) and the evaluated partial derivatives, the equation becomes: \(z = \ln 2 + 1(x-1) + 1(y-1)\).

Step 4: Simplify the Tangent Plane Equation

Simplify the equation from the previous step: \(z = \ln 2 + x - 1 + y - 1\). This further simplifies to \(z = x + y + \ln 2 - 2\).

Key Concepts

Partial Derivatives

In the realm of multivariable calculus, the concept of partial derivatives is quite pivotal. When dealing with functions of two or more variables, we often need to understand how the function behaves as each variable changes independently.
\(f(x, y) = \ln(x^2 + y^2)\) is a function of two variables, \(x\) and \(y\). These variables can be altered separately, leading us to our need for partial derivatives.

• The partial derivative with respect to \(x\), written as \(f_x(x, y)\), tells us how the function \(f\) changes as \(x\) changes, while \(y\) stays constant.
• Similarly, the partial derivative with respect to \(y\), \(f_y(x, y)\), indicates how \(f\) changes with a change in \(y\), while keeping \(x\) fixed.

To solve our problem, we found:

• \(f_x(x, y) = \frac{2x}{x^2 + y^2}\)
• \(f_y(x, y) = \frac{2y}{x^2 + y^2}\)

These expressions let us create a linear approximation or tangent plane to the surface defined by \(f(x, y)\). Understanding these derivatives is crucial for finding the precise equation for the tangent plane at any given point.

Multivariable Calculus

Multivariable calculus is an extension of single-variable calculus into more complex scenarios where functions involve multiple variables. This branch of mathematics allows us to explore more dynamic and real-world situations.
Imagine functions like \(f(x, y) = \ln(x^2 + y^2)\), which depend on several inputs.

• We delve into topics like gradients, where partial derivatives combine to guide us towards the steepest point of ascent.
• Level curves or surfaces are explored in-depth, showing sets of points where the function maintains constant elevation.

In our specific example, we worked with a surface in three-dimensional space and used partial derivatives to map a tangent plane at a point on this surface. This is where multivariable calculus shines, allowing us to simplify, analyze, and predict the behavior of such complex relationships.

Linearization

Linearization is a method of approximating a complex function using its linear counterpart. This is particularly useful when dealing with functions that are difficult to handle analytically across a broad range.
Linearization focuses on creating the tangent plane of a function at a specified point to approximate the function's behavior in that vicinity.

• For a function \(f(x, y)\), the linear approximation or tangent plane can often be written as \(z = f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)\).
• This approximation simplifies predictions of how the function behaves close to the point \((x_0, y_0)\).
• In our exercise, this meant forming the equation of the tangent plane: \(z = \ln 2 + x - 1 + y - 1\), which simplifies to \(z = x + y + \ln 2 - 2\).

By using linearization, we take a complex surface and create a simple planar model, which is much easier to use for certain calculations. This approximated solution is typically quite accurate for small deviations from the point of tangency.