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Chapter 2: Problem 4

Show that the tangent planes of a surface given by \(z=x f(y / x), x \neq 0\), where \(f\) is a differentiable function, all pass through the origin \((0,0,0)\).

Short Answer

Expert verified
The tangent planes at any point on the surface all pass through (0,0,0).

Step by step solution

01

Find the partial derivatives of the surface equation

The surface is given by the equation \[ z = x f\left(\frac{y}{x}\right) \]Compute the partial derivatives with respect to x and y. Let \[ u = \frac{y}{x} \]Then the surface equation becomes\[ z = x f(u) \].
02

Compute Partial Derivative with respect to x

Differentiate \( z = x f(u) \) with respect to x, using the chain rule:\[ \frac{\partial z}{\partial x} = f\left(\frac{y}{x}\right) + x \cdot f'\left(\frac{y}{x}\right) \cdot \left( \frac{-y}{x^2}\right) \]Simplify to get:\[ \frac{\partial z}{\partial x} = f\left(\frac{y}{x}\right) - \frac{y}{x} f'\left(\frac{y}{x}\right) \].
03

Compute Partial Derivative with respect to y

Differentiate \( z = x f(u) \) with respect to y, using the chain rule:\[ \frac{\partial z}{\partial y} = x \cdot f' \left( \frac{y}{x} \right) \cdot \frac{1}{x} \]Simplify to get:\[ \frac{\partial z}{\partial y} = f' \left( \frac{y}{x} \right) \].
04

Write the equation of the tangent plane

The equation of the tangent plane to the surface at a point (\(x_0, y_0, z_0\)) is given by:\[ z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0) \].
05

Plug in the expressions for the partial derivatives

At the point (\(x_0, y_0\)) on the surface, substitute the expressions found for \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) into the tangent plane equation:\[ z - z_0 = \left[ f\left( \frac{y_0}{x_0} \right) - \frac{y_0}{x_0} f'\left( \frac{y_0}{x_0} \right) \right](x - x_0) + f'\left( \frac{y_0}{x_0} \right)(y - y_0) \].
06

Show the tangent plane passes through (0,0,0)

At the point (\(x_0, y_0, z_0\)), express \(z_0\) in the form \(z_0 = x_0 f\left( \frac{y_0}{x_0} \right) \). The tangent plane equation becomes:\[ z - x_0 f\left( \frac{y_0}{x_0} \right) = \left[ f\left( \frac{y_0}{x_0} \right) - \frac{y_0}{x_0} f'\left( \frac{y_0}{x_0} \right) \right](x - x_0) + f'\left( \frac{y_0}{x_0} \right)(y - y_0) \].Evaluating this at (0,0,0) confirms it satisfies the equation, hence the tangent plane passes through the origin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

partial derivatives
To tackle problems involving surfaces, it's important to understand partial derivatives.
These derivatives represent the rate of change of a function with respect to one variable while keeping the other variables constant.
For a function of two variables, say \( u = f(x, y) \), the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \) tell us how the function changes as we move along the x and y directions, respectively.
In geometrical terms, partial derivatives help to measure the slope of the function in the direction of each axis.
tangent plane equation
The tangent plane to a surface at a given point provides an approximation of the surface near that point.
The equation of the tangent plane is derived using the partial derivatives of the surface's defining function. For a surface defined by \( z = g(x, y) \), the tangent plane equation at a point \((x_0, y_0, z_0)\) is:
\[ z - z_0 = \frac{\partial g}{\partial x}(x - x_0) + \frac{\partial g}{\partial y}(y - y_0) \]
This equation uses the tangents to the surface along the x and y directions to construct a plane that closely approximates the surface at \( (x_0, y_0, z_0) \).
chain rule
The chain rule is a formula for computing the derivative of a function based on its composition with other functions.
In the context of partial derivatives, it's used when differentiating a function involving a composition of variables.
For instance, if \( z = x f(u) \) where \( u = \frac{y}{x} \), we use the chain rule to differentiate with respect to x or y.
It ensures that we account for the dependency of 'u' on 'x' and 'y' correctly during differentiation.
In essence, the chain rule allows breaking down a complex differentiation problem into simpler parts.
surface equation
Surfaces in three-dimensional space can be defined by equations relating their coordinates x, y, and z.
In our case, the surface is given by \( z = x f\left( \frac{y}{x} \right) \).
This equation describes how the z-coordinate varies with x and y coordinates through a function 'f'.
Understanding the specific form of the surface equation is crucial to apply operations like differentiation and finding tangents.
The specific transformation \( u = \frac{y}{x} \) simplifies the problem and is a common technique in differential geometry.
differentiable functions
A differentiable function is one that has a well-defined, continuous derivative at every point in its domain.
This property is essential because it ensures smooth behavior of the function, making it possible to apply calculus tools like partial derivatives and the chain rule.
In our surface equation, the function 'f' needs to be differentiable to correctly compute the tangent plane.
Differentiability implies that small changes in the input variables produce small changes in the function's output, maintaining coherence and predictability in geometric interpretations.

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Most popular questions from this chapter

Show that $$ \begin{aligned} \mathbf{x}(u, v)=(u \sin \alpha \cos v, u \sin \alpha \sin v, u \cos \alpha) & \\\ 0

Let \(S^{2}=\left\\{(x, y, z) \in R^{3} ; x^{2}+y^{2}+z^{2}=1\right\\}\) and \(H=\left\\{(x, y, z) \in R^{3}\right.\); \(\left.x^{2}+y^{2}-z^{2}=1\right\\}\). Denote by \(N=(0,0,1)\) and \(S=(0,0,-1)\) the north and south poles of \(S^{2}\), respectively, and let \(F: S^{2}-\\{N\\} \cup\\{S\\} \rightarrow H\) be defined as follows: For each \(p \in S^{2}-\\{N\\} \cup\\{S\\}\) let the perpendicular from \(p\) to the \(z\) axis meet \(0 z\) at \(q\). Consider the half-line \(l\) starting at \(q\) and containing \(p\). Then \(F(p)=l \cap H\) (Fig. 2-20). Prove that \(F\) is differentiable.

One way to define a system of coordinates for the sphere \(S^{2}\), given by \(x^{2}+y^{2}+(z-1)^{2}=1\), is to consider the so-called stereographic projection \(\pi: S^{2} \sim\\{N\\} \rightarrow R^{2}\) which carries a point \(p=(x, y, z)\) of the sphere \(S^{2}\) minus the north pole \(N=(0,0,2)\) onto the intersection of the \(x y\) plane with the straight line which connects \(N\) to \(p\) (Fig. 2-12). Let \((u, v)=\pi(x, y, z)\), where \((x, y, z) \in S^{2} \sim\\{N\\}\) and \((u, v) \in x y\) plane. a. Show that \(\pi^{-1}: R^{2} \rightarrow S^{2}\) is given by $$ \pi^{-1}\left\\{\begin{array}{l} x=\frac{4 u}{u^{2}+v^{2}+4} \\ y=\frac{4 v}{u^{2}+v^{2}+4} \\ z=\frac{2\left(u^{2}+v^{2}\right)}{u^{2}+v^{2}+4} \end{array}\right. $$ b. Show that it is possible, using stereographic projection, to cover the sphere with two coordinate neighborhoods.

(Gradient on Surfaces.) The gradient of a differentiable function \(f: S \rightarrow R\) is a differentiable map grad \(f: S \rightarrow R^{3}\) which assigns to each point \(p \in S\) a vector grad \(f(p) \in T_{p}(S) \subset R^{3}\) such that \(\langle\operatorname{grad} f(p), v\rangle_{p}=d f_{p}(v) \quad\) for all \(v \in T_{p}(S)\) Show that a. If \(E, F, G\) are the coefficients of the first fundamental form in a parametrization \(\mathbf{x}: U \subset R^{2} \rightarrow S\), then grad \(f\) on \(\mathbf{x}(U)\) is given by $$ \operatorname{grad} f=\frac{f_{u} G-f_{v} F}{E G-F^{2}} \mathbf{x}_{u}+\frac{f_{v} E-f_{u} F}{E G-F^{2}} \mathbf{x}_{v} $$ In particular, if \(S=R^{2}\) with coordinates \(x, y\), $$ \operatorname{grad} f=f_{x} e_{1}+f_{y} e_{2} $$ where \(\left\\{e_{1}, e_{2}\right\\}\) is the canonical basis of \(R_{2}\) (thus, the definition agrees with the usual definition of gradient in the plane). b. If you let \(p \in S\) be fixed and \(v\) vary in the unit circle \(|v|=1\) in \(T_{p}(s)\), then \(d f_{p}(v)\) is maximum if and only if \(v=\operatorname{grad} f /|\operatorname{grad} f|(\) thus, grad \(f(p)\) gives the direction of maximum variation of \(f\) at \(p)\). c. If grad \(f \neq 0\) at all points of the level curve \(C=\\{q \in S ; f(q)=\) const. \(\\},\) then \(C\) is a regular curve on \(S\) and grad \(f\) is normal to \(C\) at all points of \(C\).

Let \(S\) be a regular surface covered by coordinate neighborhoods \(V_{1}\) and \(V_{2}\). Assume that \(V_{1} \cap V_{2}\) has two connected components, \(W_{1}, W_{2}\), and that the Jacobian of the change of coordinates is positive in \(W_{1}\) and negative in \(W_{2}\). Prove that \(S\) is nonorientable.
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