Q 49 - Calculus

Question

In Exercises $43-52$, sketch the level curves $c=-3,-2,-1,0,1,2,3$ if they exist for the specified function.

$f\left(x,y\right)=\sin \left(x+y\right)$.

Step-by-Step Solution

Answer

The required level curves $c=-1,0,1$ for the function $f\left(x,y\right)=\sin \left(x+y\right)$

as shown as following :-

For $c=-3,-2,2,3$, the function does not exist, so there is no level curves for these points.

Step 1. Given Information

We have given the following function :-

$f\left(x,y\right)=\sin \left(x+y\right)$

We have to graph the level curves $c=-3,-2,-1,0,1,2,3$ for this function.

Step 2. Graph of level curves

For each value of $c$, the level curve is the graph of the equation $f(x,y)=c$.

We know that the sine function lies between $-1 and 1$. So there will be no level curves for $c=-3,-2,2,3$

That is we have to graph the following equations :-

$$\begin{gathered} \sin \left(x+y\right)=-1 \\ \sin \left(x+y\right)=0 \\ \sin \left(x+y\right)=1 \end{gathered}$$

All of these level curves are straight lines.

Then we have the following graph of these level curves :-