4.6 Directional derivatives and the gradient  (Page 3/8)

The gradient has some important properties. We have already seen one formula that uses the gradient: the formula for the directional derivative. Recall from The Dot Product that if the angle between two vectors $a$ and $b$ is $\phi ,$ then $a·\text{b}=\Vert a\Vert \Vert \text{b}\Vert \text{cos}\phi .$ Therefore, if the angle between $\nabla f\left(x_0,y_0\right)$ and $u=\left(\text{cos}\theta \right)i+\left(\text{sin}\theta \right)j$ is $\phi ,$ we have

The $\Vert u\Vert$ disappears because $u$ is a unit vector. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at $\left(x_0,y_0\right)$ multiplied by $\text{cos}\phi .$ Recall that $\text{cos}\phi$ ranges from $\mathrm{-1}$ to $1.$ If $\phi =0,$ then $\text{cos}\phi =1$ and $\nabla f\left(x_0,y_0\right)$ and $u$ both point in the same direction. If $\phi =\pi ,$ then $\text{cos}\phi =\mathrm{-1}$ and $\nabla f\left(x_0,y_0\right)$ and $u$ point in opposite directions. In the first case, the value of $D_{\text{u}}f\left(x_0,y_0\right)$ is maximized; in the second case, the value of $D_{\text{u}}f\left(x_0,y_0\right)$ is minimized. If $\nabla f\left(x_0,y_0\right)=0,$ then $D_{u}f\left(x_0,y_0\right)=\nabla f\left(x_0,y_0\right)·u=0$ for any vector $u.$ These three cases are outlined in the following theorem.

Finding a maximum directional derivative

Find the direction for which the directional derivative of $f\left(x,y\right)=3x^2-4xy+2y^2$ at $\left(\mathrm{-2},3\right)$ is a maximum. What is the maximum value?

The maximum value of the directional derivative occurs when $\nabla f$ and the unit vector point in the same direction. Therefore, we start by calculating $\nabla f\left(x,y\right)\text{:}$

$\begin{array}{ccc}\hfill f_x\left(x,y\right)& =\hfill & 6x-4y\text{and}{f}_y\left(x,y\right)=\mathrm{-4}x+4y,\text{so}\hfill \\ \hfill \nabla f\left(x,y\right)& =\hfill & f_x\left(x,y\right)i+f_y\left(x,y\right)j=\left(6x-4y\right)i+\left(\mathrm{-4}x+4y\right)j.\hfill \end{array}$

Next, we evaluate the gradient at $\left(\mathrm{-2},3\right)\text{:}$

$\nabla f\left(\mathrm{-2},3\right)=\left(6\left(\mathrm{-2}\right)-4\left(3\right)\right)i+\left(\mathrm{-4}\left(\mathrm{-2}\right)+4\left(3\right)\right)j=\mathrm{-24}i+20j.$

We need to find a unit vector that points in the same direction as $\nabla f\left(\mathrm{-2},3\right),$ so the next step is to divide $\nabla f\left(\mathrm{-2},3\right)$ by its magnitude, which is $\sqrt{{\left(\mathrm{-24}\right)}^2+{\left(20\right)}^2}=\sqrt{976}=4\sqrt{61}.$ Therefore,

$\frac{\nabla f\left(\mathrm{-2},3\right)}{\Vert \nabla f\left(\mathrm{-2},3\right)\Vert }=\frac{\mathrm{-24}}{4\sqrt{61}}i+\frac{20}{4\sqrt{61}}j=\frac{\mathrm{-6}\sqrt{61}}{61}i+\frac{5\sqrt{61}}{61}j.$

This is the unit vector that points in the same direction as $\nabla f\left(\mathrm{-2},3\right).$ To find the angle corresponding to this unit vector, we solve the equations

$\text{cos}\theta =\frac{\mathrm{-6}\sqrt{61}}{61}\text{and}\text{sin}\theta =\frac{5\sqrt{61}}{61}$

for $\theta .$ Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore, $\theta =\pi -\text{arcsin}\left(\left(5\sqrt{61}\right)\text{/}61\right)\approx 2.45\text{rad.}$

The maximum value of the directional derivative at $\left(\mathrm{-2},3\right)$ is $\Vert \nabla f\left(\mathrm{-2},3\right)\Vert =4\sqrt{61}$ (see the following figure).

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