I recently learned that when you have a real-valued function $f(x,y) = c$, you create a plane parallel to the $xy$-plane. Inside that plane we have a level set, such that the derivative of its position vector is perpendicular to the gradient at that point
(graph). Is there a case where this isn't true? Take for instance the following example found in my book (answer to a problem found in the book):
Let $x(t){\bf i} + y(t){\bf j}$ be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is $x'(t){\bf i} + y'(t){\bf i}$. Since we want the direction of motion to be $\nabla T(x, y)$, we have $x'(t){\bf i} + y'(t){\bf i} = \nabla T(x, y) =
−4x{\bf i}− 2y{\bf j}$.
I'm assuming the vector $x(t){\bf i} + y(t){\bf j}$ is the position vector of the curve at some time $t$, that is ${\bf r}(t)$. If that is true, then $x'(t){\bf i} + y'(t){\bf j}$ is the derivative of the position vector meaning it is ${\bf r}'(t)$ (the velocity vector). The relationship between ${\bf r}'(t)$ and the gradient is that they are perpendicular at any point. Yet, in this example we are saying they are equal to each other. What exactly is going on?
Book's problem:
The temperature at a point $(x,y)$ on a rectangular metal plate
is given by $T(x, y) = 100 – 2x^2 -y^2$. Find the path a heatseeking
particle will take, starting at (3, 4), as it moves in the
direction in which the temperature increases most rapidly.
Best Answer
The property that $\mathbf{r}'$ and $\nabla f$ are perpendicular is true when $\mathbf{r}'(t)$ is a level curve of $f$. If you picture the graph of $f$ as a hill, then the level curves are the paths of constant height, going around the hill; and the gradient $\nabla f$ is a vector field pointing up the hill, in the steepest direction.
The book's question is asking about the path of a particle with $\mathbf{r}'(t)=\nabla T$ - that is, one that moves always in the steepest direction. On our hill such a particle would move directly up towards the peak. From your question it seems you're assuming for some reason that $\mathbf{r}$ is a level curve of $T$, which is certainly not true in this case - in some sense it's as far from a level curve as you can get.