Question

Let \(f\) be a differentiable function and consider the surface \(z=x f(y / x)\). Show that the tangent plane at any point \(P\left(x_{0}, y_{0}, z_{0}\right)\) on the surface passes through the origin.

Short answer

Firstly, represent the function \(z=x f(y / x)\) in terms of two variables. Then find out the partial derivatives \(\partial z / \partial x\) and \(\partial z / \partial y\). Use these derivatives to form the equation of the tangent plane. Lastly, substitute the origin point into the equation to verify it does pass through the origin(0,0,0).

Step-by-step solution

Find the parametric representation of the function

Given the function \(z=x f(y / x)\), substitute \(y/x\) by \(u\), that means \(f(y / x) = f(u)\) and \(y = ux\). Now the function is \(z = xf(u)\) with constraints \(x = x\) and \(y = ux\). With these, the function \( z \) is a function of two variables \( x \) and \( u \).

Find the partial derivatives of function

The surface is differentiable, so calculate the partial derivatives \(\partial z/\partial x\) and \(\partial z/\partial u\). Using the product rule for differentiation, \(\partial z / \partial x = f(u) + x \cdot f'(u) \cdot (-u)\) and \(\partial z / \partial u = x \cdot f'(u)\). Substituting \(u = y/x\), these can be rewritten as \(\partial z / \partial x = f(y/x) - y \cdot f'(y/x)\) and \(\partial z / \partial y = x \cdot f'(y/x)\). Remember that \(\partial z / \partial x\) is the z-coordinate change rate in x direction, \(\partial z / \partial y\) is in the y direction.

Find the equation of tangent plane

The equation of the tangent plane at point \(P\left(x_{0}, y_{0}, z_{0}\right)\) is given by \[z - z_{0} = \frac{\partial z}{\partial x}\left(x_{0}, y_{0}\right) (x - x_{0}) + \frac{\partial z}{\partial y}\left(x_{0}, y_{0}\right) (y- y_{0})\]. Substitute the values from earlier into this equation.

Show the plane passes origin

Substitute the origin point \((0,0,0)\) into the tangent plane equation from step 3. If the equality holds, it means the tangent plane does pass through the origin.

Key concepts

Partial Derivatives

Partial derivatives play a vital role in the study of multivariable calculus, as they measure the rate of change of a function with respect to one variable while holding other variables constant. In the context of the exercise, you encounter a surface defined by a multivariable function, specifically, \( z = x f(y / x) \). To understand how this surface bends and curves, you need to understand how changes in \( x \) and \( y \) affect \( z \), and this is where partial derivatives come into play.

To compute the partial derivative of \( z \) with respect to \( x \), we symbolize it as \( \partial z / \partial x \), and similarly, \( \partial z / \partial y \) represents the rate of change of \( z \) with respect to \( y \). These partial derivatives are the core components needed to determine the tangent plane's orientation at any given point on the surface.

Parametric Representation

Parametric representation is a method of defining geometric objects using parameters. For functions of several variables, it can provide an alternative and often more intuitive way of visualizing and working with a surface or curve. In our problem, we applied parametric representation by substituting \( y/x \) with \( u \), allowing us to rewrite the function as \( z = xf(u) \), effectively setting up the function in terms of two independent variables, \( x \) and \( u \).

This approach simplifies the process of differentiating the function because it reduces the complex relationship between \( x \) and \( y \) into a more manageable form. This method is instrumental when dealing with surfaces in three dimensions as it aids in visualizing how the surface extends in space.

Product Rule for Differentiation

The product rule is a fundamental theorem in calculus used when differentiating the product of two functions. It states that the derivative of a product of two functions is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.

In mathematical notation, for functions \( u(x) \) and \( v(x) \), the product rule is given as \( \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) \). This rule is especially useful in our exercise where the function \( z \) is defined by the product of \( x \) and \( f(u) \), making it essential for calculating the partial derivatives that are required to determine the tangent plane equation.

Tangent Plane Equation

The equation of a tangent plane to a surface at a specific point provides a linear approximation of the surface near that point. This plane is defined in such a way that it just touches the surface at the point of tangency and has the same slope as the surface does at that point. The general equation for the tangent plane at point \( P(x_0, y_0, z_0) \) for a function \( z \) with two variables is:
\[ z - z_0 = \frac{\partial z}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial z}{\partial y}(x_0, y_0)(y - y_0) \].

This equation incorporates the partial derivatives representing the surface's slopes along the \( x \) and \( y \) directions at the point \( (x_0, y_0) \), further emphasizing the significance of understanding and computing partial derivatives accurately. In the step-by-step solution provided, we applied this equation to prove that the tangent plane at any given point on the given surface passes through the origin.