Problem 181 - Calculus 3

Question

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for \(z\) in terms of \(x\) and \(y . )\) $$ z=x^{2}-2 x y+y^{2}, P(1,2,1) $$

Step-by-Step Solution

Answer

The tangent plane is \( z = -2x + 2y - 1 \).

Step 1: Identify the function and point

The function given is \( z = x^2 - 2xy + y^2 \). We are tasked with finding the equation of the tangent plane at the point \((x_0, y_0, z_0) = (1, 2, 1)\).

Step 2: Calculate partial derivatives

To find the tangent plane, we need the partial derivatives of \( z \) with respect to both \( x \) and \( y \). Compute \( \frac{\partial z}{\partial x} = 2x - 2y \) and \( \frac{\partial z}{\partial y} = -2x + 2y \).

Step 3: Evaluate partial derivatives at the point

Substitute \( x = 1 \) and \( y = 2 \) into the partial derivatives: \( \frac{\partial z}{\partial x}(1, 2) = 2(1) - 2(2) = -2 \) and \( \frac{\partial z}{\partial y}(1, 2) = -2(1) + 2(2) = 2 \).

Step 4: Write the equation of the tangent plane

The equation of the tangent plane is given by \( z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \). Substituting the point and partial derivatives, we get: \( z - 1 = -2(x - 1) + 2(y - 2) \).

Step 5: Simplify the equation of the tangent plane

Expand and simplify the equation: \( z - 1 = -2x + 2 + 2y - 4 \). Simplify to get \( z = -2x + 2y - 1 \).

Key Concepts

Partial Derivatives

Partial derivatives are a foundational concept in Calculus 3, where the focus is on functions of several variables. In our exercise, the function is given as \( z = x^2 - 2xy + y^2 \), a surface determined by two variables, \( x \) and \( y \). To understand how this surface changes with respect to each variable, we compute partial derivatives. These derivatives show how the surface slopes in the direction of each axis.

For \( z \), the partial derivative with respect to \( x \) is represented as \( \frac{\partial z}{\partial x} = 2x - 2y \). This measures the rate of change of \( z \) as \( x \) changes, with \( y \) held constant. Similarly, \( \frac{\partial z}{\partial y} = -2x + 2y \) indicates how \( z \) changes as \( y \) changes, holding \( x \) constant.

These partial derivatives are essential for finding the tangent plane because they tell us the direction and steepness of the slope of the tangent at any given point on the surface. By substituting the specific point \((1, 2, 1)\), we evaluate these derivatives to find the slope exactly at that point.

Calculus 3

In Calculus 3, we deal with multivariable calculus, which extends the principles of differentiation and integration to functions of two or more variables. In our problem, we're working with a function \( z = f(x, y) \), which forms a surface in a three-dimensional space.

Understanding how to compute and use partial derivatives is key in this realm of mathematics. These derivatives help us explore the nature of surfaces, curves, and beyond. Calculus 3 allows us to study the behavior of these entities with precision, enabling us to find tangent planes, the gradient, and multi-dimensional integrals.

By employing these concepts, we move from understanding simple slopes in single-variable calculus to more complex constructs, such as the tangent plane in our example. These skills are crucial for modeling and solving real-world problems in engineering, physics, and computer graphics, to name a few.

Equation of a Surface

The equation of a surface like \( z = x^2 - 2xy + y^2 \) represents a three-dimensional shape where every point is defined by the coordinates \((x, y, z)\). This surface is smooth and continuous, and for any given coordinates \((x, y)\), there's a corresponding \( z \) value that tells us the location of the surface in space.

To find the equation of the tangent plane to this surface at a particular point \((1, 2, 1)\), we start by understanding the concept of the tangent plane itself. A tangent plane is a flat surface that just "touches" the original surface at that specific point – it reflects the surface's immediate orientation and slope there.

Using the partial derivatives, as shown earlier, we calculate the slope of the tangent plane through the point. This leads to the equation \( z - 1 = -2(x - 1) + 2(y - 2) \). Upon simplifying, we get the linear equation \( z = -2x + 2y - 1 \), which provides a precise mathematical description of the plane at the touchpoint. This is how calculus forms a bridge between the geometrical shapes and their algebraic representations.