14. Directional Derivatives and Gradients

c.2. Applications of Directional Derivatives

The temperature in a room is given by \(T=(275+\sin(\pi x)\cos(\pi y)+\sin(\pi z))^\circ K\). Find the rate of change of \(T\) at \((2,-1,2)\) in the direction toward \((3,-3,4)\). Assume \(x\), \(y\) and \(z\) are in meters and \(T\) is in \(^\circ\text{K}\).

We need to find the rate of change of the temperature in the direction of the vector from \(P=(2,-1,2)\) to \(Q=(3,-3,4)\). This is a directional derivative. The vector from \(P\) to \(Q\) is: \[ \overrightarrow{PQ}=\langle 1,-2,2\rangle \] Its length is \(\left|\overrightarrow{PQ}\right|=\sqrt{1+4+4}=3\) and the unit vector is: \[ \widehat{PQ} =\left\langle \dfrac{1}{3},\dfrac{-2}{3},\dfrac{2}{3}\right\rangle \] The gradient of the temperature is: \[\begin{aligned} \vec\nabla T &=\left\langle \dfrac{\partial T}{\partial x}, \dfrac{\partial T}{\partial y}, \dfrac{\partial T}{\partial z}\right\rangle \\ &=\langle \pi\cos(\pi x)\cos(\pi y), -\pi\sin(\pi x)\sin(\pi y), \pi\cos(\pi z)\rangle \end{aligned}\] At \(P=(2,-1,2)\) this is \[\begin{aligned} \left.\vec\nabla T\right|_P &=\langle \pi\cos(2\pi)\cos(-\pi), -\pi\sin(2\pi)\sin(-\pi), \pi\cos(2\pi)\rangle \\ &=\langle-\pi,0,\pi\rangle \end{aligned}\] So the directional derivative is: \[\begin{aligned} \nabla_{\widehat{PQ}}T &=\widehat{PQ}\cdot\left.\vec\nabla T\right|_P =\left\langle \dfrac{1}{3},\dfrac{-2}{3},\dfrac{2}{3}\right\rangle \cdot\langle-\pi,0,\pi\rangle \\ &=-\,\dfrac{\pi}{3}+0+\dfrac{2\pi}{3}=\dfrac{\pi}{3} \end{aligned}\] The derivative \(\dfrac{\partial T}{\partial x}\) has units \(\dfrac{^\circ\text{K}}{\text{m}}\). So those are the units of the gradient. The displacement vector, \(\overrightarrow{PQ}\), and its length \(|\overrightarrow{PQ}|\), have units of \(\text{m}\). So the direction vector, \(\widehat{PQ}\), is dimensionless. So the directional derivative, \(\widehat{PQ}\cdot\left.\vec\nabla T\right|_P\), has units \(\dfrac{^\circ\text{K}}{\text{m}}\). So the answer is \[ \nabla_{\widehat{PQ}}T=\dfrac{\pi}{3}\,\dfrac{^\circ\text{K}}{\text{m}} \]

It is interesting to note the difference in units between the answers to the example and the exercise. The directional derivative in the example gives rate of change with respect to distance while the derivative along a vector in the exercise gives the rate of change with respect to time.