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Chapter 13: Problem 33

Describe the level curves of the function. Sketch the level curves for the given c-values. $$ z=x+y \quad c=-1,0,2,4 $$

Short Answer

Expert verified
The level curves are lines with equations \(y = -x - 1\) for \(c = -1\), \(y = -x\) for \(c = 0\), \(y = -x + 2\) for \(c = 2\) and \(y = -x + 4\) for \(c = 4\).

Step by step solution

01

Understand the problem

We are given the function \(z = x + y\) and are asked to sketch the level curves for \(c = -1, 0, 2, 4\). This means we need to find all the (x, y) pairs such that the equation \(x + y = c\) is satisfied. Each of these pairs will produce a curve on the x-y plane, and these curves are our level curves.
02

Find the level curves

Let's substitute the given values of c into our function to attain the equation of the level curves. For \(c = -1\), the equation becomes \(x + y = -1\), or equivalently, \(y = -x -1\). For \(c = 0\), the equation becomes \(x + y = 0\), or equivalently, \(y = -x\). For \(c = 2\), the equation becomes \(x + y = 2\), or equivalently, \(y = -x + 2\). Lastly, for \(c = 4\), the equation becomes \(x + y = 4\), or equivalently, \(y = -x + 4\).
03

Sketch the level curves

Now let's sketch these equations. You can notice the equations are all straight lines with negative slope. The equation \(y = -x -1\) will be a line crossing the y-axis at -1. \(y = -x\) will be a line crossing the origin. \(y = -x + 2\) will be a line crossing y-axis at 2, and \(y = -x + 4\) a line crossing y-axis at 4. The level curves should be labeled with their respective \(c\) values.

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Key Concepts

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

Function sketching
When tackling function sketching, the goal is to visually represent a mathematical relationship. For the function given here, \( z = x + y \), sketching involves plotting level curves for specified \( c \) values. Each level curve represents a set of solutions where the function equals a constant \( c \). To sketch effectively:
  • Identify the type of function: In our case, it's a linear function due to the addition of \( x \) and \( y \) without any powers or products.
  • Substitute different \( c \) values: These are given as -1, 0, 2, and 4 in our task.
  • Plot the results on the coordinate plane, interpreting each equation as a line, because the resulting curves are linear.
Sketching helps visualize how the function behaves across different sets of values, offering insights into the relationships between \( x \), \( y \), and the output \( z \). Always label each line with its corresponding \( c \) for clarity.
Linear equations
Linear equations are foundational in mathematics. They consist of a constant term and the product of constants with variables, resulting in expressions like \( x + y = c \). Here, the equation is treated as a guideline for sketching level curves. Each equation:
  • Represents a straight line on a coordinate plane.
  • Has a slope and an intercept. In \( y = -x + c \), the slope is -1 (indicating a diagonal line sloping downwards), and \( c \) is the y-intercept.
  • Highlights simplicity in relationships. When changes in \( x \) to match changes in \( y \) are equal in magnitude but opposite in direction, the result is a direct, readable relation seen in straight lines.
Understanding these concepts smoothens the learning process when visualizing and sketching larger functions or more complex equations.
Coordinate plane
The coordinate plane is a central tool for graphing functions like \( z = x + y \). For this function, we represent solutions on a two-dimensional plane, where:
  • The horizontal axis (x-axis) and the vertical axis (y-axis) intersect at the origin (0,0).
  • Points on the plane correspond to pairs \((x, y)\) where a given equation holds true for specific values of \( c \).
  • Lines like \( y = -x - 1 \), \( y = -x \), \( y = -x + 2 \), and \( y = -x + 4 \) will visibly intercept the y-axis at defined points derived from \( c \).
The layout of the coordinate plane allows one to easily interpret these intercepts and examine how the corresponding values alter the positions of level curves. It鈥檚 a powerful way to see mathematics come to life in a spatial format.

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Most popular questions from this chapter

Exercises 55 and 56, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \int_{2}^{5} \int_{1}^{6} x d y d x=\int_{1}^{6} \int_{2}^{5} x d x d y $$

Find the average value of \(f(x, y)\) over the region \(R\). $$ \begin{aligned} &f(x, y)=e^{x+y}\\\ &R: \text { triangle with vertices }(0,0),(0,1),(1,1) \end{aligned} $$

Exercises 55 and 56, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \int_{-1}^{1} \int_{-2}^{2} y d y d x=\int_{-1}^{1} \int_{-2}^{2} y d x d y $$

Evaluate the double integral. $$ \int_{1}^{2} \int_{0}^{4}\left(3 x^{2}-2 y^{2}+1\right) d x d y $$

Evaluate the partial integral. $$ \int_{0}^{x}(2 x-y) d y $$
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