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Chapter 19: Problem 25

In Exercises 21 through 26 , draw a sketch of a contour map of the function \(f\) showing the level curves of \(f\) at the given numbers. The function \(f\) for which \(f(x, y)=\frac{1}{2}\left(x^{2}+y^{2}\right)\) at \(8,6,4,2\), and 0 .

Short Answer

Expert verified
Draw circles with radii 0, 2, 2.8, 3.5, and 4 centered at the origin.

Step by step solution

01

Identify the function

The given function is \( f(x, y) = \frac{1}{2} (x^2 + y^2) \). This is similar to the equation of a circle in polar coordinates.
02

Set up the level curves

For various values of \( k \), where \( k = f(x,y) \), we need to find the corresponding level curves. Set \( f(x, y) = k \) for the given values.
03

Calculate the level curves

Compute the level curves for \( k = 8, 6, 4, 2, \) and \( 0 \):\( f(x, y) = 8: \frac{1}{2} (x^2 + y^2) = 8 \Rightarrow x^2 + y^2 = 16 \) \( f(x, y) = 6: \frac{1}{2} (x^2 + y^2) = 6 \Rightarrow x^2 + y^2 = 12 \) \( f(x, y) = 4: \frac{1}{2} (x^2 + y^2) = 4 \Rightarrow x^2 + y^2 = 8 \) \( f(x, y) = 2: \frac{1}{2} (x^2 + y^2) = 2 \Rightarrow x^2 + y^2 = 4 \) \( f(x, y) = 0: \frac{1}{2} (x^2 + y^2) = 0 \Rightarrow x^2 + y^2 = 0 \)
04

Draw the contour map

Sketch circles corresponding to each level curve. The radius for each circle can be found by taking the square root of the terms:- For \( f(x, y) = 8 \), draw a circle with radius \( \sqrt{16} = 4 \).- For \( f(x, y) = 6 \), draw a circle with radius \( \sqrt{12} \approx 3.5 \).- For \( f(x, y) = 4 \), draw a circle with radius \( \sqrt{8} \approx 2.8 \).- For \( f(x, y) = 2 \), draw a circle with radius \( \sqrt{4} = 2 \).- For \( f(x, y) = 0 \), a single point at \( (0,0) \).
05

Label the level curves

Clearly label each circle with the corresponding function value \( k = 8, 6, 4, 2, \) and \( 0 \).

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Key Concepts

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

level curves
Level curves are crucial in understanding contour maps. A level curve is a set of points where a function of two variables, like our function \( f(x, y) = \frac{1}{2}(x^2 + y^2) \), takes on a constant value.

For instance, if you take a specific value like \( k = 4 \), setting \( f(x, y) = 4 \), you get the equation \( \frac{1}{2}(x^2 + y^2) = 4 \). Solving for \( x \) and \( y \), you get the circle equation \( x^2 + y^2 = 8 \).

Each level curve can be seen as an 'isoline' representing different heights on a surface. The collection of these curves gives us a contour map.
function of two variables
Functions of two variables, such as \( f(x, y) \), generate values based on combinations of both \( x \) and \( y \).

In our example, the function is \( \frac{1}{2}(x^2 + y^2) \). This function shows how \( f \) varies with different \( x \) and \( y \) coordinates.

Typically, these functions are visualized using 3D plots or - more commonly - contour maps, where each level curve represents a constant value of the function at different points \( (x, y) \).

This approach helps simplify the visualization of the function's behavior across a plane.
contour map
A contour map is a graphical representation showing level curves of a function.

For the given function, our contour map consists of circles centered at the origin \( (0,0) \). Each circle represents a constant value of \( f(x, y) \).

In this case, we have computed level curves for \( f = 8, 6, 4, 2, \text{and} 0 \). Each of these values produces a circle with a distinct radius: for example, when \( f = 8 \), the resulting circle has a radius of 4, derived from the equation \( x^2 + y^2 = 16 \).

This visual representation using circles effectively showcases the function's characteristic.
radius calculation
Calculating the radius of each level curve requires solving the circle equation \( x^2 + y^2 = \text{constant} \).

For the given function \( f(x, y) = \frac{1}{2}(x^2 + y^2) \), compute by isolating \( x^2 + y^2 \):
  • For \( f(x, y) = 8 \), \( \frac{1}{2}(x^2 + y^2) = 8 \) results in \( x^2 + y^2 = 16 \), yielding a radius \( \text{radius} = \sqrt{16}\ = 4 \).
  • For \( f(x, y) = 6 \), \( x^2 + y^2 = 12 \), radius is \( \sqrt{12}\text{or} \approx 3.5 \).
  • For \( f(x, y) = 4 \), \( x^2 + y^2 = 8 \), radius is \( \sqrt{8}\text{or} \approx 2.8 \).
  • For \( f(x, y) = 2 \), \( x^2 + y^2 = 4 \), radius is \( 2 \).


A contour for \( f = 0 \) yields \( x^2 + y^2 = 0 \), or a single point at the origin.
circle equation
The equation of a circle \( x^2 + y^2 = r^2 \) is fundamental when sketching contour maps for functions like \( f(x, y) = \frac{1}{2}(x^2 + y^2) \).

Here, the variable \( r \) stands for the circle鈥檚 radius, determined by the function's value at different levels:
  • At level \( f = 8 \), our equation forms \( x^2 + y^2 = 16 \), r = 4.
  • At level \( f = 6 \), it becomes \( x^2 + y^2 = 12 \), r = \( \sqrt{12} \) 鈮� 3.5.


This relationship between \( x \) and \( y \) forms the foundation of sketching each level curve, visualizing numerous function values seamlessly.

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Most popular questions from this chapter

In Exercises 1 through 8 , do each of the following: (a) Find \(D_{11} f(x, y)\); (b) find \(D_{22} f(x, y) ;\) (c) show that \(D_{12} f(x, y)=D_{21} f(x, y) .\) $$ f(x, y)=\sin ^{-1} \frac{3 y}{x^{2}} $$

In Exercises 5 through 8 , prove that \(f\) is differentiable at all points in its domain by doing each of the following: (a) Find \(\Delta f\left(x_{0}, y_{0}\right)\) for the given function; (b) find an \(\epsilon_{1}\) and an \(\epsilon_{2}\) so that Eq. (3) holds; (c) show that the \(\epsilon_{1}\) and the \(\epsilon_{2}\) found in part (b) both approach zero as \((\Delta x, \Delta y) \rightarrow(0,0)\). $$ f(x, y)=x^{2} y-2 x y $$

In Exercises 18 through 21, prove that the function is discontinuous at the origin. Then determine if the discontinuity is removable or essential. If the discontinuity is removable, define \(f(0,0)\) so that the discontinuity is removed. $$ f(x, y)=\frac{\sqrt{x y}}{x+y} $$

Water is flowing into a tank in the form of a right-circular cylinder at the rate of \(\frac{4}{5} \pi \mathrm{ft}^{3} / \mathrm{min}\). The tank is stretching in such a way that even though it remains cylindrical, its radius is increasing at the rate of \(0.002 \mathrm{ft} / \mathrm{min}\). How fast is the surface of the water rising when the radius is \(2 \mathrm{ft}\) and the volume of water in the tank is \(20 \pi \mathrm{ft}^{3}\) ?

If \(f\) is a differentiable function of two variables \(u\) and \(v\), let \(u=x-y\) and \(v=y-x\) and prove that \(z=f(x-y, y-x)\) satisfies the equation \(\partial z / \partial x+\partial z / \partial y=0\).
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