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Chapter 11: Problem 54

Describe the relationship of the gradient to the level curves of a surface given by \(z=f(x, y)\).

Short Answer

Expert verified
The gradient of a function at a point is always perpendicular to the level curve of the function at that point and points in the direction of the greatest increase of the function. The level curves represent places where the function value remains constant.

Step by step solution

01

Define level curves

The level curves of a function \(f(x, y)\) are the curves obtained by setting \(f(x, y)\) equal to a constant \(c\). Each of these curves corresponds to a particular height \(z\) on the surface of \(f\). The curves are formed by the points \( (x, y) \) where the function takes constant value \(c\).
02

Define the gradient

The gradient of a function \(f(x, y)\) at a point \( (x_0, y_0) \), denoted as \( \nabla f(x_0, y_0) \), is a vector in the xy-plane. This vector points in the direction of the greatest rate of increase of the function at the point \( (x_0, y_0) \), and its magnitude is equal to the rate of increase. The gradient is calculated as \( \nabla f = [f_x, f_y] \) where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively.
03

Discuss the relationship

For a function \( f(x,y) \), the gradient \( \nabla f \) at a point is always perpendicular to the level curve of the function at that point. This follows from the direction of greatest increase being orthogonal to the direction where the function value does not change, i.e., along the level curve. Moreover, the gradient points towards higher values of the function. If the function decreases along the direction of the level curve, then the gradient points in opposite direction.

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Most popular questions from this chapter

Define the gradient of a function of two variables. State the properties of the gradient.

In Exercises \(39-42,\) find \(\partial w / \partial r\) and \(\partial w / \partial \theta\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(r\) and \(\boldsymbol{\theta}\) before differentiating. \(w=x^{2}-2 x y+y^{2}, x=r+\theta, \quad y=r-\theta\)

Find a normal vector to the level curve \(f(x, y)=c\) at \(P.\) $$ \begin{array}{l} f(x, y)=6-2 x-3 y \\ c=6, \quad P(0,0) \end{array} $$

In Exercises 35-38, use the gradient to find a unit normal vector to the graph of the equation at the given point. Sketch your results $$ 4 x^{2}-y=6,(2,10) $$

Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \frac{\text { Function }}{f(x, y, z)=x e^{y z}} \frac{\text { Point }}{(2,0,-4)} $$
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