10 Directional Derivative
Lets say you are on a hillside and you want to know in what direction does the hill slope downward most steeply from this point? (This would be the direction you would slide if you lost your footing) - this would be the direction straight down.
Lets say we move a small distance \(\Delta s\) the vertical change will be \(\Delta z\) (positive, negative, or zero) meaning \(\frac{d z}{d s}\) depends upon duration (it is a directional derivative).
The direction of steepest slope is the direction in which \(\frac{d z}{d s}\) has its largest value.
Lets take a scalar field \(\phi(x, y, z)\). To find the directional derivative of \(\phi\) at a particular point in a particular direction we need to find \(\frac{d \phi}{d s}\) the rate of change of \(\phi\) with distance at a given point \(x_{0}, y_{0}, z_{0}\) and in a given direction (\(\mathbf{s}\)).
Lets define \(\mathbf{u}=\hat{\imath} a+\hat{j} b+\hat{k} c\) as a unit vector in a given direction (\(\mathbf{s}\) in this case).
\[ \begin{aligned} & (x, y, z)-\left(x_{0}, y_{0}, z_0\right)=\mathbf{u} s=(\hat{i} a+\hat{j} b+\hat{k}c) s \\ & x=x_{0}+a s \\ & y=y_{0}+b s \\ & z=z_{0}+c s \\ \end{aligned} \] parametric equations - \(x, y, z\) are equations of a single variable.
Lets write the full derivative: \[ \begin{aligned} & \frac{d \phi}{d s}=\frac{\partial \phi}{\partial x} \cdot \frac{d x}{d s}+\frac{\partial \phi}{\partial y} \cdot \frac{d y}{d s}+\frac{\partial \phi}{\partial z} \cdot \frac{d z}{d s} \\ & \frac{d \phi}{d s}=\frac{\partial \phi}{\partial x} \cdot a+\frac{\partial \phi}{\partial y} \cdot b+\frac{\partial \phi}{\partial z} \cdot c \\ & \nabla \phi=\frac{\partial \phi}{\partial x} \hat{i}+\frac{\partial \phi}{\partial y} \hat{j}+\frac{\partial \phi}{\partial z} \hat{k} \\ & \nabla \phi \cdot \mathbf{u}=\frac{\partial \phi}{\partial x} a+\frac{\partial \phi}{\partial y} b+\frac{\partial \phi}{\partial z} c \end{aligned} \] Thus the directional derivative is \(\frac{d \phi}{d s} = \nabla \phi \cdot \mathbf{u}\).
Find the directional derivative of \(\phi=x^{2} y+x\) at \((1,2,-1)\) in the direction \(\mathbf{A}=2 \hat{i}+2 \hat{\jmath}+\hat{k}\)
\[ \begin{aligned} & \mathbf{u}=\frac{\mathbf{A}}{|\mathbf{A}|} \\ & |\mathbf{A}|=\sqrt{4+4+1}=3 \\ & \mathbf{u}=\frac{1}{3}(2 \hat{\imath}-2 \hat{\jmath}+\hat{k}) \end{aligned} \] \[ \nabla \phi=(2 x y+z) \hat{\imath}+x^{2} \hat{\jmath}+x \hat{k} \] at point \((1,2,-1)\) \[ \begin{aligned} \nabla \phi(1,2,-1) & =(2 \cdot 1 \cdot 2-1) \hat{\imath}+1 \hat{\jmath}+1 \hat{k} \\ & =3 \hat{\imath}+\hat{\jmath}+\hat{k} \\ \frac{d \phi}{d s}=\nabla \phi \cdot \mathbf{u} & =2-2 / 3+1 / 3=5 / 3 \end{aligned} \]
\(\frac{d \phi}{d s}=|\nabla \phi||\underline{u}| \cos \theta=|\nabla \phi| \cos \theta\) where \(\theta\) is angle between \(\nabla \phi\) and \(\mathbf{u}\).
Maximum \(\frac{d \phi}{d s}\) or \(|\nabla \phi|\) is if \(\theta=0\) largest decrease occurs at \(\theta=180\) or \(-|\nabla \phi|\).
Lets consider the temperature in a room. The temperature follows \(T=x^{2}-y^{2}+x y z+273\). In which direction is the temp increasing most rapidly at \((-1,2,3)\)?
\[ \begin{aligned} & \nabla T=(2 x+y z) \hat{i}+(-2 y+x z) \hat{\jmath}+x y \hat{k} \\ & \nabla T(-1,2,3)=(-2+6) \hat{i}+(-4-3) \hat{\jmath}+(-2) \hat{k} \\ & =4 \hat{i}-7 \hat{\jmath}-2 \hat{k} \end{aligned} \]
and max rate of \(\Delta\) in direction of this vectors. \(\frac{dT}{ds} = |\Delta T| = \sqrt{69}\). \(-\nabla T\) is rate of max decrease \(\frac{d T}{d s}=-\sqrt{69}\). Heat flows in \(-\Delta T\) direction.
10.1 Normal Derivative
Now say \(\phi\) is constant at a point \(P\left(x_{0}, y_{0}, z_{0}\right)\) and the direction vector \(\mathbf{u}\) is tangent to \(\frac{d \phi}{d s}\). If \(\phi\) is constant then \(\frac{d \phi}{d s}=0\). This also means \(\nabla \phi \cdot \mathbf{u}=0\). \[ \begin{aligned} & |\nabla \phi||\mathbf{u}|\cos \theta=0 \\ & |\nabla \phi| \cos \theta=0 \end{aligned} \] \[ \theta=90^{\circ} \] In this case, \(\nabla \phi\) is perpendicular to the surface. Since \(|\nabla \phi|\) is the value of the directional derivative in the direction normal often called the normal derivative \[ |\nabla \phi|=\frac{d \phi}{d n}. \]