Problem 55 - Calculus with Concepts in Calculus

Question

Let \(g\) be a differentiable function of one variable and let \(f(x, y)=x g(y / x)\). Show that every plane tangent to the graph of \(f\) passes through the origin.

Step-by-Step Solution

Answer

Every tangent plane to the graph of \(f\) passes through the origin because its equation simplifies to \(0 = 0\).

Step 1: Understanding the Problem

We need to show that every plane tangent to the graph of the function \( f(x, y) = x \, g(y/x) \) passes through the origin. This means that for any point \((x_0, y_0)\) on its surface, the tangent plane at that point should intersect the origin \((0,0,0)\).

Step 2: Calculating Partial Derivatives

Find the partial derivatives of \(f(x, y)\). Using the chain rule: the partial with respect to \(x\), \( f_x(x, y) = g(y/x) - \frac{y}{x} g'(y/x) \). For the partial with respect to \(y\), \( f_y(x, y) = \frac{1}{x} g'(y/x) \).

Step 3: Tangent Plane Equation

The equation of the tangent plane at a point \((x_0, y_0)\) on the surface is given by: \[ z - f(x_0, y_0) = f_x(x_0, y_0) (x - x_0) + f_y(x_0, y_0) (y - y_0) \].

Step 4: Substituting Back into Tangent Plane

Substitute \((x_0, y_0)\) into the function and derivatives: \(f(x_0, y_0) = x_0 g(y_0/x_0)\), \(f_x(x_0, y_0) = g(y_0/x_0) - \frac{y_0}{x_0} g'(y_0/x_0)\), \(f_y(x_0, y_0) = \frac{1}{x_0} g'(y_0/x_0)\).

Step 5: Setting the Plane Through the Origin

Substitute \((0, 0, 0)\) into the tangent plane equation: \[ 0 - x_0 g(y_0/x_0) = (g(y_0/x_0) - \frac{y_0}{x_0} g'(y_0/x_0))(0 - x_0) + \frac{1}{x_0} g'(y_0/x_0)(0 - y_0) \].

Step 6: Simplifying to Show Solution

Simplify the above equation:\[ -x_0 g(y_0/x_0) = -x_0 g(y_0/x_0) + \frac{y_0}{x_0} g'(y_0/x_0) x_0 - y_0 g'(y_0/x_0) \]. This simplifies to \[ 0 = 0 \]. Thus, the tangent plane goes through the origin.

Key Concepts

Tangent Plane

In multivariable calculus, a tangent plane is a flat surface that just "touches" a point on a curved surface. Think of it as a pillow gently resting on a bumpy hill. This plane represents the best linear approximation of the surface at that specific point. When dealing with functions of two variables, like in our problem with function \( f(x, y) = x g(y/x) \), the equation of a tangent plane at a given point \((x_0, y_0)\) is crucial to solving problems like these.
To find the tangent plane, we use partial derivatives. The general formula of the tangent plane is:

• \[ z - f(x_0, y_0) = f_x(x_0, y_0) (x - x_0) + f_y(x_0, y_0) (y - y_0) \]

This expression includes partial derivatives \( f_x \) and \( f_y \), which help us capture the changes in the function along the \( x \) and \( y \) axes respectively. In our specific case, showing that this tangent plane intersects the origin means verifying that plugging \((0,0,0)\) into the equation is valid and results in a true statement.

Partial Derivatives

Partial derivatives are a way to measure how a function changes as each of its variables change, while the other variables are held constant. They are a foundational concept in calculus and help us understand how multivariable functions behave in different dimensions.
For our function \( f(x, y) = x g(y/x) \), the partial derivative with respect to \( x \) is calculated using the chain rule. It gives us:

• \[ f_x(x, y) = g(y/x) - \frac{y}{x} g'(y/x) \]

This result helps us understand how changes in \( x \) affect the function while keeping \( y \) constant.
Similarly, the partial derivative with respect to \( y \) is:

• \[ f_y(x, y) = \frac{1}{x} g'(y/x) \]

These derivatives allow us to formulate the tangent plane equation later on because they give the slopes in \( x \) and \( y \) directions at a point on the surface.

Differentiable Functions

Differentiable functions are smooth and do not have any abrupt changes or sharp corners. This means we can find derivatives at every point across their domain. In multivariable calculus, this concept extends to finding partial derivatives for functions of several variables.
If a function is differentiable, we can approximate it linearly using its tangent plane. In our example with \( f(x, y) = x g(y/x) \), we're assured that the function is differentiable, so we can confidently find the tangent plane that contains partial derivatives. This allows us to study how the function changes in response to variations in each variable.
The idea of differentiability is crucial because it provides the groundwork for analyzing the behavior of functions. It guarantees that expressions like tangent plane equations are well-defined, ensuring we can pursue deeper investigations into the properties of functions in multivariable calculus.