/*! 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 4 Find the gradient of the functio... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the gradient of the function at the given point. Then sketch the gradient together with the level curve that passes through the point. $$g(x, y)=\frac{x^{2}}{2}-\frac{y^{2}}{2}, \quad(\sqrt{2}, 1)$$

Short Answer

Expert verified
Gradient at \((\sqrt{2}, 1)\) is \((\sqrt{2}, -1)\); level curve is a hyperbola passing through \((\sqrt{2}, 1)\).

Step by step solution

01

Find the Partial Derivatives

To find the gradient, we first calculate the partial derivatives of the function \( g(x, y) = \frac{x^2}{2} - \frac{y^2}{2} \). The partial derivative with respect to \( x \) is \( \frac{\partial g}{\partial x} = x \), and the partial derivative with respect to \( y \) is \( \frac{\partial g}{\partial y} = -y \).
02

Evaluate the Gradient at the Given Point

The gradient \( abla g(x, y) \) is \( (x, -y) \). At the point \( (\sqrt{2}, 1) \), the gradient becomes \( (\sqrt{2}, -1) \).
03

Understand the Level Curve Equation

A level curve for the function \( g(x, y) \) is given by fixing \( g(x, y) = c \), where \( c \) is a constant. At the point \( (\sqrt{2}, 1) \), we have \( g(\sqrt{2}, 1) = \frac{(\sqrt{2})^2}{2} - \frac{1^2}{2} = 1 - \frac{1}{2} = \frac{1}{2} \). Thus the level curve is \( \frac{x^2}{2} - \frac{y^2}{2} = \frac{1}{2} \).
04

Sketch the Gradient and Level Curve

The gradient vector \( (\sqrt{2}, -1) \) at \( (\sqrt{2}, 1) \) points in the direction of greatest increase of \( g \). The level curve \( \frac{x^2}{2} - \frac{y^2}{2} = \frac{1}{2} \) is a hyperbola. Plot the point \( (\sqrt{2}, 1) \), draw the gradient vector from there, and sketch the hyperbola passing through that point following the general hyperbolic shape.

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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 crucial tool in multivariable calculus used to find the direction of the steepest ascent of a function. It is constructed from the partial derivatives of the function. For the function \( g(x, y) = \frac{x^2}{2} - \frac{y^2}{2} \), the gradient vector is composed of its partial derivatives:
  • The partial derivative with respect to \( x \) is \( \frac{\partial g}{\partial x} = x \).
  • The partial derivative with respect to \( y \) is \( \frac{\partial g}{\partial y} = -y \).
This gives us the gradient vector \( abla g(x, y) = (x, -y) \). At the point \((\sqrt{2}, 1)\), the vector evaluates to \((\sqrt{2}, -1)\). This vector points in the direction where the function \( g \) increases most rapidly.
Partial Derivatives
Partial derivatives represent the function's rate of change with respect to one of the variables, holding the other variables constant. In multivariable calculus, they allow us to understand how a function changes as we change just one variable at a time. Given the function \( g(x, y) = \frac{x^2}{2} - \frac{y^2}{2} \), we calculated:
  • The partial derivative with respect to \( x \) is \( \frac{\partial g}{\partial x} = x \), which tells us how the function changes as \( x \) changes with \( y \) fixed.
  • The partial derivative with respect to \( y \) is \( \frac{\partial g}{\partial y} = -y \), which shows how \( y \) affects the function when \( x \) is held constant.
These derivatives form the components of the gradient vector. Understanding partial derivatives helps us visualize how a multivariable function behaves in different settings.
Level Curve
A level curve is a curve along which the function has a constant value. For a function like \( g(x, y) = \frac{x^2}{2} - \frac{y^2}{2} \), you can find a level curve by setting \( g(x, y) = c \), where \( c \) is a constant. At the point \((\sqrt{2}, 1)\), we evaluated the function to get \( c = \frac{1}{2} \). Thus, the level curve is described by the equation \( \frac{x^2}{2} - \frac{y^2}{2} = \frac{1}{2} \).Level curves help us understand the terrain of a function. They provide a contour map that shows regions of equal value, much like elevation lines on a topographic map. For each level curve, the gradient vector is perpendicular, highlighting the direction of maximum increase away from the curve.
Hyperbola
A hyperbola is a type of conic section that appears as a set of open curves. When considering the level curve \( \frac{x^2}{2} - \frac{y^2}{2} = \frac{1}{2} \) from our function \( g(x, y) \), we recognize it as an equation for a hyperbola. Hyperbolas feature two branches that mirror each other across axes.The equation of the hyperbola in standard form is derived from the level curve. Hyperbolas are defined by their distinct shape, where differences in the squared terms signify the open nature of the curves. In multivariable calculus, hyperbolas can help visualize complex function relationships. The gradient vector's role in this context is as a normal to the hyperbola, illustrating the perpendicular nature of the gradient vector to the level curves.

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