Problem 20 - APEX Calculus

Question

A function \(z=f(x, y)\) and a point \(P\) are given. Find the equation of the tangent plane to \(f\) at \(P .\) Note: these are the same functions as in Exercises \(5-8 .\) \(f(x, y)=x^{2}-2 x-y^{2}+4 y, P=(1,2)\)

Step-by-Step Solution

Answer

The equation of the tangent plane is \(z = 3\).

Step 1: Recall the formula for the tangent plane

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

Step 2: Calculate the partial derivatives

First, find \(f_x(x, y)\), the partial derivative of \(f\) with respect to \(x\):\[ f_x(x, y) = \frac{\partial}{\partial x}(x^2 - 2x - y^2 + 4y) = 2x - 2 \]Next, find \(f_y(x, y)\), the partial derivative of \(f\) with respect to \(y\):\[ f_y(x, y) = \frac{\partial}{\partial y}(x^2 - 2x - y^2 + 4y) = -2y + 4 \]

Step 3: Evaluate the partial derivatives at the point P

Substitute the point \(P = (1, 2)\) into the partial derivatives:- \(f_x(1, 2) = 2(1) - 2 = 0\)- \(f_y(1, 2) = -2(2) + 4 = 0\)

Step 4: Calculate \(f(1, 2)\)

Evaluate the original function at the point \((1, 2)\):\[ f(1, 2) = 1^2 - 2(1) - 2^2 + 4(2) = 1 - 2 - 4 + 8 = 3 \]

Step 5: Formulate the equation of the tangent plane

Using the results from the previous steps, substitute into the tangent plane formula:\[ z - 3 = 0(x - 1) + 0(y - 2) \]Simplifying, we get:\[ z = 3 \]

Key Concepts

Partial Derivatives

Partial derivatives are crucial in calculus, specifically when dealing with functions of more than one variable. Imagine a scenario where you have a surface defined by a function of two variables:

• Partial derivatives allow us to see how the function changes as we tweak one of the variables while keeping the other constant. In mathematical terms, a partial derivative of a function with respect to one of its variables measures the function's rate of change along that variable.
• For example, if you have a function \( f(x, y) = x^2 - 2x - y^2 + 4y \), finding \( f_x(x, y) \) means differentiating with respect to \( x \), treating \( y \) as constant, resulting in \( 2x - 2 \). Similarly, \( f_y(x, y) \) involves differentiating with respect to \( y \), producing \( -2y + 4 \).

This gradient information is integral when finding tangent planes, indicating how the surface behaves near a point of interest.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. It includes exploring rates of change and the relationships between changing variables through:

• Tangent planes, which help approximate surfaces near a particular point, can be found using partial derivatives to determine their slopes.
• For functions like \( z = f(x, y) \), the tangent plane equation at a point \((x_0, y_0)\) uses these derivatives: \( z - f(x_0, y_0) = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \).

Thus, multivariable calculus provides essential tools for applications across physics, engineering, and other disciplines, offering insights into how things change with respect to multiple interacting factors.

Differential Calculus

Differential calculus focuses on the concept of derivatives, which quantify how a function changes as its input varies. For functions of multiple variables,

• Understanding and calculating derivatives in multiple dimensions involves partial derivatives, which help describe changes along each variable axis.
• The process of finding the equation of a tangent plane involves evaluating these derivatives at a specific point and using them to derive the plane's equation.

In practice, differential calculus aids in optimizing functions, modeling dynamic systems, and understanding geometrical properties, transforming abstract mathematical concepts into tools for scientific and practical problem-solving.