Problem 25
Question
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 .
Step-by-Step Solution
Verified Answer
Draw circles with radii 0, 2, 2.8, 3.5, and 4 centered at the origin.
1Step 1: 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.
2Step 2: 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.
3Step 3: Calculate the level curves
Compute the level curves for \( k = 8, 6, 4, 2, \) and \( 0 \):\( f(x, y) = 8: \frac{1}{2} (x^2 + y^2) = 8 \Rightarrow x^2 + y^2 = 16 \) \( f(x, y) = 6: \frac{1}{2} (x^2 + y^2) = 6 \Rightarrow x^2 + y^2 = 12 \) \( f(x, y) = 4: \frac{1}{2} (x^2 + y^2) = 4 \Rightarrow x^2 + y^2 = 8 \) \( f(x, y) = 2: \frac{1}{2} (x^2 + y^2) = 2 \Rightarrow x^2 + y^2 = 4 \) \( f(x, y) = 0: \frac{1}{2} (x^2 + y^2) = 0 \Rightarrow x^2 + y^2 = 0 \)
4Step 4: 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 \( (0,0) \).
5Step 5: Label the level curves
Clearly label each circle with the corresponding function value \( k = 8, 6, 4, 2, \) and \( 0 \).
Key Concepts
level curvesfunction of two variablescontour mapradius calculationcircle equation
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.
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.
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.
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 \( x^2 + y^2 \):
A contour for \( f = 0 \) yields \( x^2 + y^2 = 0 \), or a single point at the origin.
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’s radius, determined by the function's value at different levels:
This relationship between \( x \) and \( y \) forms the foundation of sketching each level curve, visualizing numerous function values seamlessly.
Here, the variable \( r \) stands for the circle’s 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 = \( \sqrt{12} \) ≈ 3.5.
This relationship between \( x \) and \( y \) forms the foundation of sketching each level curve, visualizing numerous function values seamlessly.
Other exercises in this chapter
Problem 25
In Exercises 25 and 26 , find \(f_{x}(x, y)\) and \(f_{y}(x, y)\). \(f(x, y)=\int_{x}^{y} \ln \sin t d t\)
View solution Problem 25
In Exercises 24 through 29 , determine if the indicated limit exists. \(\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y^{2}}{x^{2}+y^{2}}\)
View solution Problem 26
Given \(u=e^{y} \cos x, x=2 t, y=t^{2}\). Find \(d^{2} u / d t^{2}\) in three ways: (a) by first expressing \(u\) in terms of \(t ;(\mathrm{b})\) by using the f
View solution Problem 26
If \(f\) and \(g\) are differentiable functions of \(x\) and \(y\) and \(u=f(x, y)\) and \(v=g(x, y)\), such that \(\partial u / \partial x=\partial v / \partia
View solution