7fd984…: “Up the dark cat”

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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 $e^{2x-y-2} = e^{2-2} =1$ .

The result of this exercise is not an accident; as we will see, there is a relationship between the gradient and level curves, but we need more mathematical technology to establish the result more generally!