The following exercise foreshadows an important fact that we will study again later.

Suppose that \(F(x,y) =5e^{2x-y-2}\).

The equation of the level curve associated to \(F(x,y)=5\) is \(\answer {2x-y} = 2\). Letting \(x(t)=t\), we find that a parametric description of the level curve to be

\[ \vec {c}(t) = \vector {t, \answer {2t-2}}. \]

A vector parallel to the level curve is \(\vec {v} = \vector {\answer {1},\answer {2}}\).

The level curve should be a line; by setting \(F(x,y)=5\), we find that
\begin{align*} 5e^{2x-y-2} &=5 \\ e^{2x-y-2} &= \answer {1} \\ 2x-y-2 &= \answer {0} \end{align*}

In the last line, we take the natural logarithm of both sides.

Now, to find a vector parallel to a line, we can start by giving a parametric description of it. Then, we can bring it into the form

\[ \vecl (t) = \vec {p} + t\vec {v} \]
where \(\vec {v}\) and \(\vec {P}\) are constant vectors. As before, the vector \(\vec {v}\) will be parallel to the line.
By a routine computation, we find that \(\grad {F}(x,y) = \vector {\answer {10e^{2x-y-2}}, \answer {-5e^{2x-y-2}} }\).

An interesting observation can be made though. At any point on the level curve associated to \(z=5\), we found \(2x-y=2\), so at any point along the level curve, \(\grad {F}(x,y) = \vector {\answer {10}, \answer {-5} }\).

Note that \(\grad {F} \dotp \vec {v} = \answer {0}\) at any point along the level curve.

If \(2x-y=2\), then \(e^{2x-y-2} = e^{2-2} =1\).