MATH 251 Lecture 12

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Differential = Derivative = best linear approximation

Directional Derivative

Given a function $f(x,y)$ and a vector $\overrightarrow{v}$ (assuming $\overrightarrow{v}$ is a unit vector?).

$\left(D_{\overrightarrow{v}}f\right)(x,y)$ is the derivative of $f$ in the direction of $\overrightarrow{v}$.

The path is $(x+v_{x}t,y+v_{y}t)$

$\begin{array}{rl}{D}_{\overrightarrow{v}}f(x,y)& =\frac{\mathrm{d}}{\mathrm{d}t}f(x+v_{x}t,y+v_{y}t)\\ & =\frac{\partial f}{\partial x}\left(\frac{\mathrm{d}}{\mathrm{d}t}(x+v_{x}t)\right)+\frac{\partial f}{\partial y}\left(\frac{\mathrm{d}}{\mathrm{d}t}(y+v_{y}t)\right)\\ & =\mathrm{d}f(v_x,v_y)\\ & =\frac{\partial f}{\partial x}\left(v_x\right)+\frac{\partial f}{\partial y}\left(v_y\right)\\ & =\overrightarrow{\mathrm{\nabla }}f\cdot \hat{v}\end{array}$

Example

Find the directional derivative of $x^{2}y+y\sin x$ in the direction of $\overrightarrow{v}=⟨1,1⟩$ at $(\pi ,2)$.

Answer 1

$\begin{array}{rl}{\frac{\partial f}{\partial x}|}_{(\pi ,2)}& ={2xy+y\cos x|}_{(\pi ,2)}\\ & =4\pi -2\\ {\frac{\partial f}{\partial y}|}_{(\pi ,2)}& ={x^2+\sin x|}_{(\pi ,2)}\\ & ={\pi }^2\\ (4\pi -2)\left(1\right)+\left({\pi }^2\right)\left(1\right)& =4\pi -2+{\pi }^2\end{array}$

Answer 2

Convention: Direction always means unit direction. Replace $\overrightarrow{v}$ with $\hat{v}$. This is the way that the book uses.

$\begin{array}{rl}\hat{v}& =\frac{\overrightarrow{v}}{\Vert \overrightarrow{v}\Vert }=\frac{1}{\sqrt{2}}\\ (4\pi -2)\left(\frac{1}{\sqrt{2}}\right)+\frac{{\pi }^2}{\sqrt{2}}& =\frac{4\pi -2+{\pi }^2}{\sqrt{2}}\end{array}$

Gradient

A vector quantity

$\mathrm{\nabla }f(x,y)=⟨\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}⟩$

Valid in euclidean rectangular coordinates.

Facts:

1. It ponts in the direction of steepest ascent of $f$.
2. Points orthogonally to the level curves of $f$.

Example

Find the gradient of $f(x,y)={\sin }^{2}x{\mathrm{e}}^y$

$\begin{array}{rl}\mathrm{\nabla }f(x,y)& =2\sin x\cos x{\mathrm{e}}^y\hat{ı}+{\sin }^{}x{\mathrm{e}}^y\hat{ȷ}\\ \text{at }(\frac{3\pi }{2},1)& =⟨0,\mathrm{e}⟩\\ \text{Rate of Change}=\Vert \mathrm{\nabla }f\Vert & =\sqrt{0^2+{\mathrm{e}}^2}=\mathrm{e}\end{array}$