Problem 14
Question
Sketch the level curve \(f(x, y)=c\). \(f(x, y)=6 x^{2} ; c=6,24\)
Step-by-Step Solution
Verified Answer
Level curves \( x = \pm 1 \) for \( c = 6 \) and \( x = \pm 2 \) for \( c = 24 \).
1Step 1: Identify the function
The given function is a product of the variable squared and its coefficient, specifically, the function is \( f(x, y) = 6x^2 \). This implies that the function depends solely on the variable \( x \), and does not vary with \( y \).
2Step 2: Set up equations for level curves
We need to find the level curves for different constant values \( c \). For \( c = 6 \) and \( c = 24 \), we set up the equations \( 6x^2 = 6 \) and \( 6x^2 = 24 \) respectively.
3Step 3: Solve the first level curve equation
For the level curve when \( c = 6 \), solve the equation \( 6x^2 = 6 \) to find \( x \).Divide both sides by 6:\[ x^2 = 1 \]Solve for \( x \):\[ x = \pm 1 \].This implies that the level curve for \( c = 6 \) is a pair of vertical lines \( x = 1 \) and \( x = -1 \).
4Step 4: Solve the second level curve equation
For the level curve when \( c = 24 \), solve the equation \( 6x^2 = 24 \).Divide both sides by 6:\[ x^2 = 4 \]Solve for \( x \):\[ x = \pm 2 \].Thus, the level curve for \( c = 24 \) is a pair of vertical lines \( x = 2 \) and \( x = -2 \).
5Step 5: Sketch the level curves on the coordinate plane
To sketch the level curves:1. Draw the y-axis as a vertical line and the x-axis as a horizontal line.2. Draw vertical lines at \( x = 1 \) and \( x = -1 \) for \( c = 6 \).3. Draw vertical lines at \( x = 2 \) and \( x = -2 \) for \( c = 24 \).4. Since the function \( f(x, y) = 6x^2 \) is independent of \( y \), the level curves are parallel lines unaffected by the value of \( y \).
Key Concepts
Graphing Level CurvesSingle-Variable FunctionsCoordinate Plane Sketching
Graphing Level Curves
Level curves are a great way to visualize functions of two variables. They show sets of points where the function takes the same constant value. For the function \( f(x, y) = 6x^2 \), we need to graph level curves for specific values of \( c \). The process starts by setting the function equal to a constant \( c \). Once you solve this equation, the level curves emerge.
- For \( c = 6 \), the equation becomes \( 6x^2 = 6 \). Solving for \( x \), you find \( x = \pm 1 \).
- For \( c = 24 \), the equation is \( 6x^2 = 24 \). Solving gives \( x = \pm 2 \).
Single-Variable Functions
The function in our exercise, \( f(x, y) = 6x^2 \), is an example of how specific calculations often revolve around just one variable, even in two-variable situations. Since there is no \( y \) variable in this function, it is effectively a single-variable function disguised as a two-variable one.
- This type of problem often simplifies complexity, focusing only on one key variable.
- Here, you can directly see that any change in \( y \) does not influence the function's value.
- The function boils down to finding where \( 6x^2 \) equals a constant.
Coordinate Plane Sketching
Sketching level curves on a coordinate plane involves identifying where those curves will lie and how they are positioned relating to the axes. Start with drawing your basic coordinate plane: - The \( x \)-axis is horizontal- The \( y \)-axis is verticalFor our task:
- Mark where \( x = 1 \) and \( x = -1 \) with vertical lines for the level curve with \( c = 6 \).
- Do the same for \( x = 2 \) and \( x = -2 \) for \( c = 24 \).
Other exercises in this chapter
Problem 14
Find the first partial derivatives of the function. $$ f(x, y, z)=x(\cos y) e^{z} $$
View solution Problem 14
Evaluate the limit. $$ \lim _{(x, y) \rightarrow(0,0)} \frac{x y}{\sqrt{x^{2}+y^{2}}} $$
View solution Problem 15
Find the points on the surface \(x^{2}-y z=1\) that are closest to the origin.
View solution Problem 15
Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ k(x, y)=e^{x} \
View solution