MATH 251 Lecture 12

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Differential = Derivative = best linear approximation

Directional Derivative

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

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

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

${\begin{aligned}D_{\vec {v}}f\left(x,\,y\right)&={\frac {\mathrm {d} }{\mathrm {d} t}}\,f\left(x+v_{x}t,\,y+v_{y}t\right)\\&={\frac {\partial f}{\partial x}}\left({\frac {\mathrm {d} }{\mathrm {d} t}}\left(x+v_{x}t\right)\right)+{\frac {\partial f}{\partial y}}\left({\frac {\mathrm {d} }{\mathrm {d} t}}\left(y+v_{y}t\right)\right)\\&=\mathrm {d} f\left(v_{x},\,v_{y}\right)\\&={\frac {\partial f}{\partial x}}\left(v_{x}\right)+{\frac {\partial f}{\partial y}}\left(v_{y}\right)\\&={\vec {\nabla }}f\cdot {\hat {v}}\end{aligned}}$

Example

Find the directional derivative of $x^{2}y+y\sin {x}$ in the direction of ${\vec {v}}=\left\langle 1,\,1\right\rangle$ at $\left(\pi ,\,2\right)$.

Answer 1

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

Answer 2

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

${\begin{aligned}{\hat {v}}&={\frac {\vec {v}}{\left\|{\vec {v}}\right\|}}={\frac {1}{\sqrt {2}}}\\\left(4\pi -2\right)\left({\frac {1}{\sqrt {2}}}\right)+{\frac {\pi ^{2}}{\sqrt {2}}}&={\frac {4\pi -2+\pi ^{2}}{\sqrt {2}}}\end{aligned}}$

Gradient

A vector quantity

$\nabla f\left(x,\,y\right)=\left\langle {\frac {\partial f}{\partial x}},\,{\frac {\partial f}{\partial y}}\right\rangle$

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

${\begin{aligned}\nabla f(x,y)&=2\sin {x}\cos {x}\mathrm {e} ^{y}{\hat {\imath }}+\sin ^{2}{x}\mathrm {e} ^{y}{\hat {\jmath }}\\{\mbox{at }}\left({\frac {3\pi }{2}},\,1\right)&=\left\langle 0,\,\mathrm {e} \right\rangle \\{\mbox{Rate of Change}}=\left\|\nabla f\right\|&={\sqrt {0^{2}+\mathrm {e} ^{2}}}=\mathrm {e} \end{aligned}}$