As you work through the exercises in Homework 2, you’ll come across this exercise in Section 14.6:

## 23. Temperature change along a circle

Suppose that the Celsius temperature at a point $(x,y)$ in the $xy$-plane is $T(x,y)=x\sin(2y)$ and that distance in the $xy$-plane is measured in meters. A particle is moving *clockwise* around the circle of radius 1 m centered at the origin at the constant rate of 2 m/sec.

**a.** How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point $P(1/2, \sqrt{3}/2)$?

**b.** How fast is the temperature experienced by the particle changing in degrees Celsius per second at $P$?

The first question is asking for a directional derivative, a concept with which we’re now familiar. We know that it can be computed by computing the dot product of $\nabla T(1/2,\sqrt{3}/2)$ with the unit vector ${\bf u}$ pointing in the direction of motion. How can we find ${\bf u}$ in this problem? We could sketch a diagram, in which case we would see that ${\bf u}$ is the unit vector tangent to the circle and hence perpendicular to the radial vector, as shown in the image below:

The position vector of the point $P$ is the vector $\begin{bmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{bmatrix}$, from which we can immediately deduce that ${\bf u} = \begin{bmatrix} \frac{\sqrt{3}}{2} \\ -\frac{1}{2}\end{bmatrix}$. Alternatively, we could find a parametric description of the path of the motion, from which we could compute ${\bf u}$ as the unit tangent vector. Recall that if ${\bf v}(t)$ is the particle’s velocity vector at time $t$, then the unit tangent vector is simply the direction of the velocity vector, i.e., $\frac{{\bf v}(t)}{\|{\bf v}(t)\|}$. (We will be talking a lot more about parametric curves in Chapter 16.)

In any case, note that the directional derivative does **not** take the particle’s speed into account. We have stressed that the directional derivative is always computed using a *unit* vector, for the very purpose of discounting any effect of speed. In this problem, the directional derivative measures the rate of change of $T$ with respect to distance moved, and in particular has units of degrees Celsius per meter. If we moved in direction ${\bf u}$ at the speed of one meter per second, then we would expect the temperature to increase at the rate of the directional derivative (now in degrees Celsius per second).

To take speed into account, we have two options:

*Option 1:* Multiply our directional derivative by the speed of motion. This is the fastest solution if we have already computed the directional derivative.

*Option 2**: *Compute the dot product of $\nabla T$ with the velocity vector. This is the fastest solution if we have not yet computed the directional derivative.

Observe that the two options really are equivalent. In the first option, we first dot the gradient of $T$ with the vector ${\bf u} = \frac{{\bf v}}{\|{\bf v}\|}$, and then multiply our answer by the speed, which is $\|{\bf v}\|$. The net effect of that two-step calculation is the same as simply dotting the gradient by ${\bf v}$.

## Moral of the Story

To measure the rate of change of a function $f$ as a particle moves through the domain, compute the dot product of the gradient, $\nabla f$, with the velocity vector, ${\bf v}$.