Problem 43
Question
Sketch the following by finding the level curves. Verify the graph using technology. $$f(x, y)=2-\sqrt{x^{2}+y^{2}}$$
Step-by-Step Solution
Verified Answer
The level curves of \( f(x, y) = 2 - \sqrt{x^2 + y^2} \) are circles centered at the origin with decreasing radii as \( c \) increases towards 2.
1Step 1: Understand the Function
The function given is \( f(x, y) = 2 - \sqrt{x^2 + y^2} \). It represents a surface in three dimensions where each point \((x, y)\) on the surface has the height \(z\) given by the formula.
2Step 2: Set Up Level Curves
Level curves are created by setting the function equal to a constant \( c \). So we have \( 2 - \sqrt{x^2 + y^2} = c \). This simplifies to \( \sqrt{x^2 + y^2} = 2 - c \).
3Step 3: Solve for Level Curves
Solving for \(x\) and \(y\), we find that \( x^2 + y^2 = (2-c)^2 \). These are equations of circles centered at the origin \((0,0)\) with radius \(|2-c|\).
4Step 4: Analyze Validity of \(c\)
Since \( \sqrt{x^2 + y^2} \) is non-negative, \(2 - c\) must also be non-negative. Thus, \( 2 \geq c \).
5Step 5: Sketch the Circles
For each valid constant \( c \) from the range \( -\infty < c \leq 2 \), draw a circle of radius \( |2 - c| \) centered at \((0,0)\). As \( c \) increases towards 2, the circle's radius decreases towards 0.
6Step 6: Verify Using Technology
Use graphing software such as Desmos or GeoGebra to enter the equation \( f(x, y) = 2 - \sqrt{x^2 + y^2} \) and observe the level curves by setting \( f(x, y) = c \) for different values of \( c \). Confirm the graph shows concentric circles with diminishing radius as predicted.
Key Concepts
Multivariable CalculusEquations of CirclesThree-Dimensional Surfaces
Multivariable Calculus
In multivariable calculus, we study functions with two or more variables, like our function \( f(x, y) = 2 - \sqrt{x^2 + y^2} \). Unlike single-variable calculus, multivariable calculus deals with surfaces and curves in higher dimensions. This adds a beautiful complexity to our mathematical toolkit, because rather than just focusing on points along a line, we explore
- Points on surfaces in three-dimensional space
- Curves that form from intersecting these surfaces
- Ways of visualizing gradients and directions along these surfaces
Equations of Circles
The original exercise shows that the function \( f(x, y) = 2 - \sqrt{x^2 + y^2} \) leads to level curves that represent circles. These are equations of the form \( x^2 + y^2 = r^2 \), where \( r \) is the radius. To better understand how this connects to our function, notice that by setting \( f(x, y) = c \) and simplifying, we ended up with \( \sqrt{x^2 + y^2} = 2 - c \), which is rearranged to \( x^2 + y^2 = (2-c)^2 \).
Here are the key points:
Here are the key points:
- The center of the circle is always at \((0,0)\), known as the origin.
- The radius of the circle is determined by the expression \(|2-c|\).
- As \( c \) changes, the size of the circles change, reflecting different heights of the surface.
Three-Dimensional Surfaces
Three-dimensional surfaces can initially seem overwhelming, but they're simply an extension of two-dimensional graphs. When dealing with an equation like \( f(x, y) = 2 - \sqrt{x^2 + y^2} \), you are picturing a surface floating in three-dimensional space where every point on the surface corresponds to a pair of \( x \) and \( y \) values and a resulting \( z \) value.
Key aspects to keep in mind:
Key aspects to keep in mind:
- Each pair \((x, y)\) corresponds to a specific height \(z\) on the surface.
- Level curves, like the circles we plotted, are the intersections of the surface with horizontal planes of constant \(z\).
- Visualizing these surfaces can reveal insights on slopes and variations across different regions.
Other exercises in this chapter
Problem 40
For the following exercises, plot a graph of the function. $$z=x^{2}+y^{2}$$
View solution Problem 42
Sketch the following by finding the level curves. Verify the graph using technology. $$f(x, y)=\sqrt{4-x^{2}-y^{2}}$$
View solution Problem 44
Sketch the following by finding the level curves. Verify the graph using technology. $$z=1+e^{-x^{2}-y^{2}}$$
View solution Problem 45
Sketch the following by finding the level curves. Verify the graph using technology. $$z=\cos \sqrt{x^{2}+y^{2}}$$
View solution