Problem 35 - Calculus with Concepts in Calculus

Question

Find an equation of the plane tangent to the graph of the given function at the indicated point(s). $$ f(x, y)=(2+x-y)^{2} ;(3,-1,36) $$

Step-by-Step Solution

Answer

The equation of the tangent plane is \( z = 12x - 12y - 12 \).

Step 1: Identify the Gradient

To find the tangent plane, we'll first compute the gradient of the function at the given point. The gradient of a function \( f(x, y) \) is given by the vector of the partial derivatives \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 2: Compute Partial Derivatives

Differentiate \( f(x, y) = (2 + x - y)^2 \) with respect to \( x \) and \( y \). For \( x \), use the derivative \( \frac{\partial f}{\partial x} = 2(2 + x - y) \cdot 1 = 2(2 + x - y) \). For \( y \), use the derivative \( \frac{\partial f}{\partial y} = 2(2 + x - y) \cdot (-1) = -2(2 + x - y) \).

Step 3: Evaluate the Gradient at the Given Point

Plug the given point \( (3, -1) \) into the gradient. Compute \( \frac{\partial f}{\partial x} |_{(3,-1)} = 2(2 + 3 - (-1)) = 2 \times 6 = 12 \) and \( \frac{\partial f}{\partial y} |_{(3,-1)} = -2(2 + 3 - (-1)) = -2 \times 6 = -12 \). Thus, the gradient at \( (3, -1) \) is \( (12, -12) \).

Step 4: Find the Tangent Plane Equation

The equation of the tangent plane is given by \( z = f(3, -1) + \frac{\partial f}{\partial x} (x - 3) + \frac{\partial f}{\partial y} (y + 1) \). Since \( f(3, -1) = 36 \), substitute the values to get \( z = 36 + 12(x - 3) - 12(y + 1) \).

Step 5: Simplify the Tangent Plane Equation

Simplify the equation: \( z = 36 + 12x - 36 - 12y - 12 = 12x - 12y - 12 \). The tangent plane equation is \( z = 12x - 12y - 12 \).

Key Concepts

Partial Derivatives

Partial derivatives are key tools in calculus used to investigate functions of several variables. When a function is dependent on more than one variable, we can understand its behavior by finding how it changes with respect to each variable individually. This is done using partial derivatives.

For a function \( f(x, y) = (2 + x - y)^2 \), to find its rate of change concerning \( x \) and \( y \), we calculate \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). These represent the function's partial derivatives.

• \( \frac{\partial f}{\partial x} = 2(2 + x - y) \) indicates how \( f \) changes with a little adjustment in \( x \) while keeping \( y \) constant.
• \( \frac{\partial f}{\partial y} = -2(2 + x - y) \) shows the change in \( f \) when \( y \) varies and \( x \) stays fixed.

Understanding these derivatives helps us grasp how functions sculpt the surfaces we analyze.

Gradient

The gradient is an essential vector in calculus representing the multi-component rate of change of a function. It combines all partial derivatives into a single vector that points in the direction of the steepest ascent of the function.

For a function \( f(x, y) \), the gradient can be written as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For the function \( f(x, y) = (2 + x - y)^2 \), the gradient vector is \( (2(2 + x - y), -2(2 + x - y)) \).

• At the point \( (3, -1) \), the gradient evaluates to \( (12, -12) \).
• The gradient vector signifies how the function increases or decreases at that specific point.

Applying the gradient helps formulate the equation of a plane tangent to the function's surface at a given point.

Tangent Plane Equation

A tangent plane is a flat surface that just touches a curved surface at a point. The equation of a tangent plane provides a linear approximation of the function around this point.

The general form for a tangent plane equation to \( f(x, y) \) at point \((a, b, f(a, b))\) is:

\[ z = f(a, b) + \frac{\partial f}{\partial x}(x - a) + \frac{\partial f}{\partial y}(y - b) \]

Using the particular point \( (3, -1) \), where \( f(3, -1) = 36 \),the tangent plane equation becomes:

\[ z = 36 + 12(x - 3) - 12(y + 1) \]

After simplification, this equation is \( z = 12x - 12y - 12 \).

This formula illustrates the linear surface that approximates the curve just at that specific point.

Calculus Problem

Solving calculus problems typically involves understanding a variety of mathematical concepts and techniques. In the case of determining the tangent plane of a function, it requires a mix of differentiation and algebra.

• First, identify key points and evaluate their positions on the function’s surface using partial derivatives.
• Next, compute the gradient to ascertain the direction and magnitude of change.
• Finally, construct the tangent plane equation using derived formulas.

Problems like these demonstrate how calculus provides tools to model and predict behavior of real-world systems, offering linear approximations of complex surfaces at given points.