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Chapter 13: Problem 50

Sketch the level curve of \(f(x, y)\) that passes through \(P\) and draw the gradient vector at \(P\). $$ f(x, y)=x^{2}-y^{2} ; \quad P(2,-1) $$

Short Answer

Expert verified
The level curve is \( x^2 - y^2 = 3 \); gradient at \( P(2,-1) \) is \( (4, 2) \).

Step by step solution

01

Understand the Definition of a Level Curve

A level curve for a function of two variables, such as \( f(x, y) = x^2 - y^2 \), is a curve where the function has a constant value. To find the level curve that passes through point \( P(2, -1) \), we need to determine this constant value at \( P \).
02

Calculate the Level Curve's Constant Value

Substitute \( x = 2 \) and \( y = -1 \) into the function \( f(x, y) = x^2 - y^2 \). Calculate:\[ f(2, -1) = 2^2 - (-1)^2 = 4 - 1 = 3 \]Thus, the level curve passing through \( P \) is where \( f(x, y) = 3 \).
03

Write the Equation for the Level Curve

Since the constant from Step 2 is 3, the level curve is described by the equation:\[ x^2 - y^2 = 3 \]
04

Draw the Level Curve

Sketch the hyperbola given by the equation \( x^2 - y^2 = 3 \). This is a standard form of a hyperbola centered at the origin, opening along the x-axis.
05

Find the Gradient Vector at \( P \)

The gradient vector of a function \( f(x, y) \) is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For our function, compute:\[ \frac{\partial f}{\partial x} = 2x \] and \[ \frac{\partial f}{\partial y} = -2y \]
06

Calculate the Gradient Vector at \( P \)

Substitute \( x = 2 \) and \( y = -1 \) into the gradient expressions:\[ abla f(2, -1) = (2(2), -2(-1)) = (4, 2) \]The gradient vector at \( P(2, -1) \) is \( (4, 2) \).
07

Draw the Gradient Vector at \( P \)

On the sketch, draw the vector \( (4, 2) \) at the point \( (2, -1) \). The gradient vector points in the direction of the greatest increase of the function and is perpendicular to the level curve.

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Key Concepts

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

Gradient Vector
The gradient vector is a fundamental concept in multivariable calculus, representing the direction and rate of steepest ascent of a function. For a function of two variables, such as \( f(x, y) \), the gradient vector is denoted as \( abla f \).
  • The gradient vector components are found using partial derivatives with respect to each variable.
  • For \( f(x, y) = x^2 - y^2 \), the gradient vector is \( abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \).
  • The partial derivatives are calculated as follows: \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = -2y \).
For any point \( P(x, y) \), the gradient vector can be computed by substituting its coordinates into these partial derivative expressions. At point \( P(2, -1) \), the gradient vector becomes \( (4, 2) \), indicating the direction of maximum increase of the function.
Partial Derivatives
Partial derivatives are used to measure the rate at which a function changes with respect to one of its variables while keeping other variables constant. They are the building blocks for gradient vectors.
  • The partial derivative \( \frac{\partial f}{\partial x} \) represents how the function changes as \( x \) changes, holding \( y \) constant.
  • Similarly, \( \frac{\partial f}{\partial y} \) indicates how the function varies with changes in \( y \) while \( x \) remains unchanged.
To find these partial derivatives for a specific function, apply standard differentiation rules. For example, with \( f(x, y) = x^2 - y^2 \), differentiating with respect to \( x \) gives \( 2x \). Differentiating with respect to \( y \) gives \( -2y \). These derivatives provide insights into how the surface described by \( f(x, y) \) is oriented and how steep its slopes are.
Hyperbola
A hyperbola is a type of conic section that is characterized by its unique shape, usually featuring two distinct curves or branches. It is defined mathematically by an equation that resembles a difference of squares.
  • The general form of a hyperbola's equation is \( x^2/a^2 - y^2/b^2 = 1 \).
  • In our specific case, the level curve through the point \( P(2, -1) \) is described by \( x^2 - y^2 = 3 \).
To sketch this hyperbola, note that it is centered at the origin and opens along the x-axis. The branches extend infinitely, and the curve represents points where the function takes a constant value, maintaining the equation \( x^2 - y^2 = 3 \). This unique structure makes hyperbolas a fascinating study in geometry and calculus.
Function of Two Variables
A function of two variables involves input values \( x \) and \( y \), often denoted as \( f(x, y) \). Such functions plot a surface in three-dimensional space, with the output value representing the height above the \( xy \)-plane.
  • The expression \( f(x, y) = x^2 - y^2 \) provides a rule that assigns a single value to every pair of \( (x, y) \).
  • Level curves, or isoclines, are specific curves on this surface where the function has the same value.
  • They are found by setting \( f(x, y) = c \), where \( c \) is some constant, leading to equations like \( x^2 - y^2 = 3 \).
By analyzing such functions, one can gain insights into the geometry and behavior of surfaces. These concepts are essential in fields ranging from physics to economics, where real-world phenomena are often modeled with multivariable functions.

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