Problem 24
Question
Sketch the level curves of the function corresponding to each value of \(z\). \(f(x, y)=e^{x}-y ; z=-2,-1,0,1,2\)
Step-by-Step Solution
Verified Answer
The level curves for the function \(f(x, y) = e^x - y\) with given z values are:
1. For \(z=-2\), the level curve equation is \(y = e^x + 2\)
2. For \(z=-1\), the level curve equation is \(y = e^x + 1\)
3. For \(z=0\), the level curve equation is \(y = e^x\)
4. For \(z=1\), the level curve equation is \(y = e^x - 1\)
5. For \(z=2\), the level curve equation is \(y = e^x - 2\)
Plot these equations on the xy-plane to visualize the level curves.
1Step 1: Set up equations for each z value
We will set up the equations by substituting \(f(x, y)\) with the given z values. This will give us 5 different equations for the level curves.
For \(z=-2\),
\[ -2=e^{x}-y \]
For \(z=-1\),
\[ -1 = e^x - y \]
For \(z=0\),
\[ 0 = e^x - y \]
For \(z=1\),
\[ 1 = e^x - y \]
For \(z=2\),
\[ 2 = e^x - y \]
We can now solve these equations for y.
2Step 2: Solve each equation for y
Solve for y in all the equations obtained in step 1.
For \(z=-2\),
\[y = e^x + 2 \]
For \(z=-1\),
\[y = e^x + 1 \]
For \(z=0\),
\[y = e^x \]
For \(z=1\),
\[y = e^x - 1 \]
For \(z=2\),
\[y = e^x - 2 \]
These are the equations of the level curves for the given function.
3Step 3: Sketch the level curves
To sketch the level curves, we will plot the equations obtained in step 2 on the xy-plane. For each level curve, keep in mind the points in which they intersect the x-axis (where y = 0) and y-axis (where x = 0). This will help in obtaining a better sketch of the level curves.
1. For \(z=-2\), the level curve equation is \(y = e^x + 2\)
2. For \(z=-1\), the level curve equation is \(y = e^x + 1\)
3. For \(z=0 \), the level curve equation is \(y = e^x\)
4. For \(z=1 \), the level curve equation is \(y = e^x - 1\)
5. For \(z=2 \), the level curve equation is \(y = e^x - 2\)
Note: Although we cannot sketch the level curves in this text-based format, you are encouraged to plot these on graphing software or on graph paper to better visualize the curves and their relative features.
Key Concepts
Exponential FunctionEquation SolvingGraphing Techniques
Exponential Function
The exponential function is a mathematical expression that involves the constant \( e \), approximately equal to 2.718. It commonly appears in formulas like \( f(x, y) = e^x - y \). By understanding this concept, one can explore how small changes in \( x \) affect the output in exponential growth or decay. The beauty of the exponential function lies in its property of increasing very rapidly across inputs. This is particularly useful in modeling phenomena like population growth or interest calculations.
Exponential functions also play a crucial role in understanding level curves, which are graphical representations of the function. When plotting the level curves of an exponential function, one can observe how the curves shift up or down depending on the constant \( z \). This shift highlights the geometric nature of exponential growth.
Exponential functions also play a crucial role in understanding level curves, which are graphical representations of the function. When plotting the level curves of an exponential function, one can observe how the curves shift up or down depending on the constant \( z \). This shift highlights the geometric nature of exponential growth.
Equation Solving
Solving equations involving exponential functions involves isolating variables to understand their relationships. In the context of level curves, we typically set the function equal to a constant value \( z \), like in our original exercise.
- First, substitute \( f(x, y) \) with \( z \), giving equations such as \( -2 = e^x - y \).
- Next, rearrange the equation to solve for \( y \), yielding expressions like \( y = e^x + 2 \).
Graphing Techniques
Graphing techniques help bring equations to life by visualizing functions on a plane. When graphing level curves like \( y = e^x + 2 \), understanding the intersection points is crucial.
- Plot points where the curve intersects the x-axis by setting \( y = 0 \).
- Determine where the curve crosses the y-axis by setting \( x = 0 \).
Other exercises in this chapter
Problem 24
C\&G Imports imports two brands of white wine, one from Germany and the other from Italy. The German wine costs $$\$ 4 /$$ bottle, and the Italian wine costs $$
View solution Problem 24
Find the first partial derivatives of the function. \(f(x, y, z)=x e^{y / z}\)
View solution Problem 25
Evaluate the first partial derivatives of the function at the given point. \(f(x, y)=x^{2} y+x y^{2} ;(1,2)\)
View solution Problem 25
Find an equation of the level curve of \(f(x, y)=\sqrt{x^{2}+y^{2}}\) that contains the point \((3,4)\).
View solution