/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Find the dual curve of \(x_{0}^{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the dual curve of \(x_{0}^{3}+x_{1}^{3}+x_{2}^{3}=0\).

Short Answer

Expert verified
The dual curve of \( x_{0}^{3} + x_{1}^{3} + x_{2}^{3} = 0 \) is \( y_{0}^{3} + y_{1}^{3} + y_{2}^{3} = 0 \).

Step by step solution

01

Identify the given curve

The equation of the given curve is: \[ x_{0}^{3} + x_{1}^{3} + x_{2}^{3} = 0 \]
02

Find the partial derivatives of the given curve

To find the dual curve, determine the partial derivatives of the given equation with respect to each variable: \[ \frac{\text{d}F}{\text{d}x_{0}} = 3x_{0}^{2}, \frac{\text{d}F}{\text{d}x_{1}} = 3x_{1}^{2}, \frac{\text{d}F}{\text{d}x_{2}} = 3x_{2}^{2} \]
03

Express the dual curve equation

The dual curve is found by setting the gradient vector equal to some vector in the dual space, i.e., a parameter set. This forms a system where the dual curve has variables that may be denoted \[ (y_{0}, y_{1}, y_{2}) \]. Using the partial derivatives, the dual curve equation can be written as: \[ x_{0}y_{0} = x_{1}y_{1} = x_{2}y_{2} \]
04

Rewrite partial derivatives in terms of dual variables

Using the relationship between the partial derivatives and the dual curve variables, substitute and solve for the coordinates of the dual curve: \[ y_{0}^{3} + y_{1}^{3} + y_{2}^{3} = 0 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Partial Derivatives
Partial derivatives are essential tools in calculus and algebraic geometry. They measure how a function changes as each variable changes while keeping the other variables constant.

In the given exercise, we start with the curve defined by \( x_{0}^{3} + x_{1}^{3} + x_{2}^{3} = 0 \). To move towards finding the dual curve, we need to compute the partial derivatives of this equation with respect to each variable:

\[\frac{\text{d}F}{\text{d}x_{0}} = 3x_{0}^{2}, \ \frac{\text{d}F}{\text{d}x_{1}} = 3x_{1}^{2}, \ \frac{\text{d}F}{\text{d}x_{2}} = 3x_{2}^{2} \]

Partial derivatives help inform how the function's slope behaves in different directions, crucial for finding a curve's gradient.
Dual Space
Dual space is a concept in linear algebra and algebraic geometry. It refers to the set of all linear functionals on a vector space, essentially flipping the roles of vectors and covectors.

In the context of finding the dual curve, we represent points in the dual space with variables \( y_{0}, y_{1}, y_{2} \). The gradient vector formed by the partial derivatives \( \left( 3x_{0}^{2}, 3x_{1}^{2}, 3x_{2}^{2} \right) \) must be proportional to some point in this dual set. This step allows us to translate the problem into dual coordinates.

The relationship between the original variables and the dual variables is given by:

\[ x_{0}y_{0} = x_{1}y_{1} = x_{2}y_{2} \]

This formula helps to map the original curve into its dual in the dual space.
Parameter Set
A parameter set is a collection of variables that describe a system or equation. For dual curves, this involves expressing the original curve's gradient in terms of the dual variables.

In our example, the dual variables are \( y_{0}, y_{1}, y_{2} \). We use the relationship derived from the partial derivatives, and the parameter set allows us to formulate the dual curve's corresponding equation in dual space terms.

Using the substitution from our derivatives:

\[ y_{0}^{3} + y_{1}^{3} + y_{2}^{3} = 0 \]

This equation is the representation of the dual curve in the parameter set \( y_{0}, y_{1}, y_{2} \).

By understanding partial derivatives, dual space, and parameter sets, you will be better equipped to tackle algebraic geometry problems related to dual curves.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X\) be an affine variety and \(K\) a finite extension of \(k(X)\). Prove that there exists an affine variety \(Y\) and a map \(f: Y \rightarrow X\) with the properties (1) \(f\) is finite; (2) \(Y\) is normal; (3) \(k(Y)=K\) with \(f^{*}: k(X) \hookrightarrow k(Y)=K\) the given inclusion. Prove that \(Y\) is uniquely determined by these properties. It is called the normalisation of \(X\) in \(K\).

Let \(V \subset \mathbb{A}^{3}\) be the quadratic cone defined by \(x y=z^{2}\); let \(X^{\prime} \rightarrow \mathbb{A}^{3}\) be the blowup of \(\mathbb{A}^{3}\) with centre in the origin, and \(V^{\prime}\) the closure of \(\sigma^{-1}(V \backslash 0)\) in \(X^{\prime}\). Prove that \(V^{\prime}\) is a nonsingular variety and that the inverse image of the origin under \(\sigma: V^{\prime} \rightarrow V\) is a nonsingular rational curve.

Prove that if \(X \subset \mathbb{P}^{n}\) is a nonsingular hypersurface and not a hyperplane, then as \(x\) runs through \(X\), the tangent hyperplanes \(\Theta_{x}\) form a hypersurface in the dual space \(\mathbb{P}^{n *}\)

For what values of \(a\) does the curve \(x_{0}^{3}+x_{1}^{3}+x_{2}^{3}+a\left(x_{0}+x_{1}+x_{2}\right)^{3}=0\) have a singular point? What are its singular points then? Is it reducible?

Suppose that a nonsingular irreducible affine \(n\)-dimensional variety \(X \subset \mathbb{A}^{n}\) is given by equations \(F_{1}=\cdots=F_{m}=0\), and that for every \(x=\left(x_{1}, \ldots, x_{n}\right) \in X\) the space defined by \(\sum\left(\partial F_{i} / \partial T_{j}\right)(x)\left(T_{j}-x_{j}\right)=0\) is \(n\)-dimensional. Prove that then \(\mathfrak{a}_{X}=\left(F_{1}, \ldots, F_{m}\right)\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.