Question

In Exercises 49-56, 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

The level curves for the function \(z = x + y\) for values \(c=-1, 0, 2, 4\) are lines with a negative slope of -1, intercepting the y-axis at -1, 0, 2 and 4 respectively.

Step-by-step solution

Understand the Function

Given the function \(z = x + y\), the first thing to understand is that y represents the height for any given x. With this in mind, it’s clear that any increase or decrease in x will be mirrored by the y value.

Forming the Equations

The level curves are given by the equation \(x + y = c\) for different values of \(c\), which are: -1, 0, 2, and 4. So we produce four different equations each corresponding to a specific \(c\)-value: \(x + y = -1\), \(x + y = 0\), \(x + y = 2\), and \(x + y = 4\).

Describing the Level Curves

In case \(c = -1\), we can rearrange the equation to \(y = -1 - x\), which is a straight line with a slope of -1 and cuts the y-axis at -1. For \(c = 0\), the equation is \(y = -x\), which is a straight line with a slope of -1 and cuts the y-axis at 0. For \(c = 2\), the equation becomes \(y = 2 - x\), a straight line with a slope of -1 and y-intercept at 2. For \(c = 4\), the equation finally becomes \(y = 4 - x\), a straight line with slope -1 which intercepts the y-axis at 4.

Sketching the Level Curves

Plot each of these lines on a graph. All these lines must have a slope of -1 represented by a 45-degree angle to the x-axis and the y-intercept varies according to the provided \(c\)-values