The Gradient and Directional Derivative

• Be able to compute a gradient vector, and use it to compute a directional derivative of a given function in a given direction.
• Be able to use the fact that the gradient of a function $f(x,y)$ is perpendicular (normal) to the level curves $f(x,y)=k$ and that it points in the direction in which $f(x,y)$ is increasing most rapidly.

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If $f(x,y)=x^2-xy$, then $\nabla f$ is $\left[\begin{array}{c}2x-y\\ -x\end{array}\right]$

Notice: $\nabla f$ is a vector-valued function, specifically one with a two-dimensional input and a two-dimensional output.

Directional Derivative

Say that the direction vector of interest is $\vec{l}$. Take the dot product between $\nabla F$ and the unit vector in the direction of $\vec{l}$, which should give:

$\frac{\partial F}{\partial \vec{l}}=\nabla F\cdot \frac{\vec{l}}{\Vert \vec{l}\Vert }$

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Webwork 5 Problem 8

$$\begin{array}{rl}\nabla F& =(\frac{1}{2},\frac{3}{4})\\ & \\ \vec{v}& =(-2,2)\\ & \\ D_{u}f& =\nabla F\cdot \frac{\vec{v}}{\Vert \vec{v}\Vert }\\ & =(\frac{1}{2},\frac{3}{4})\cdot \frac{(-2,2)}{\sqrt{8}}\\ & =(\frac{1}{2},\frac{3}{4})\cdot (-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\\ & =-\frac{1}{2\sqrt{2}}+\frac{3}{4\sqrt{2}}\\ & =\frac{1}{4\sqrt{2}}\end{array}$$

Max rate of change is $||\nabla F(P)||$

We find the (unit) direction vector in which the maximum rate of change occurs at P by normalizing the gradient vector $\frac{\nabla F}{||\nabla F||}$

Webwork 5 Problem 9 Here’s the question with proper LaTeX markdown:

Find the directional derivative of $f(x,y)=\sin (x+2y)$ at the point $(4,3)$ in the direction $\theta =\pi /4$.

The directional derivative is:

$$\begin{array}{rlr}{f}_x=\cos (x+2y)& & f_y=2\cos (x+2y)\\ \nabla f& =⟨\cos (x+2y),2\cos (x+2y)⟩& \\ \nabla f(4,3)& =⟨\cos (x+2y),2\cos (x+2y)⟩& \\ & =⟨\cos (4+6),2\cos (4+6)⟩& \\ & =⟨\cos (10),2\cos (10)⟩& \\ & & \\ \text{Directional derivative}& =\nabla f\cdot \vec{u}& \\ \theta =\frac{\pi }{4}⟹\vec{u}& =⟨\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}⟩& \\ & & \\ & =⟨\cos (10),2\cos (10)⟩\cdot ⟨\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}⟩& \\ & =\cos \left(10\right)\frac{\sqrt{2}}{2}+2\cos \left(10\right)\frac{\sqrt{2}}{2}& \\ & \approx -1.77993950433& \end{array}$$

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