# Directional Derivative with Unit Vector

Suppose that $f(x,y)=x^2+xy-y^2$. How can I find the largest and the smallest possible values of the directional derivative $D_{u}f(1,2)$ in which $u$ is a unit vector?

• This will get downvoted unless you show some thought. Let me ask: IIf $u=(1/2,\sqrt 3/2)$ how do you compute $D_uf(1,2)?$ – zhw. Jul 9 '15 at 19:26
• @zhw: Why are you using $\sqrt 3$? don't mean to nitpick, but it may confuse a beginner. – Gary. Jul 9 '15 at 19:54
• There's nothing confusing about the unit vector $(1/2,\sqrt 3/2)$ as far as I can see. – zhw. Jul 9 '15 at 19:55
• OK, never mind. – Gary. Jul 9 '15 at 19:57

$||Df_u||=||< f_x,f_y>.<1,2>||=.....$ We normalize $<1,2>$ to $\frac{1}{\sqrt 5}<1,2>$. Then what formula do you know for the dot product $a.b$?

We know that $<a,b>. <c,d> =|<a,b>||<c,d>cos \theta$. And we are given $<a,b>=<f_x, f_y>$ and $<c,d>= <1,2>$ , in the formula:

$$||Df_u ||=|<f_x,f_y>,<1,2>= |<f_x,f_y>||<1,2>| cos \theta$$

Remember we have $|cos \theta \leq 1 |$. We also have that $<f_x, f_y>$ are known , once we know $f$. So the only thing that can affect the value of the formula above is the value of $|cos \theta |$. What value of $\theta$ maximizes the expression?

Since $|cos\theta|<1$, the value of the expression is maximized, as you said, when $\theta=0$, so that $\cos \theta =1$. Then the expression $||Df_u||$, given by $$||Df_u (1,2)||=|<f_x, f_y>(1,2)<u_1, u_2>cos \theta$$ is maximized by : $$<f_x,f_y>(1,2)$$ , which is the gradient, and $<f_x, f_y>(1,2)$ is evaluation of $<f_x, f_y>$ at the point $(1,2)$.

Notice that the smallest value of $||Df_u||=|<f_x, f_y>||u_1, u_2| cos \theta$ would be zero, and it would happen at $\theta =\pi/2$.

• Comments are not for extended discussion; this conversation has been moved to chat. – Daniel Fischer Jul 9 '15 at 21:40

By a theorem, if $f(x,y)$ is differentiable at a point $(a,b)$ and the gradient of $f(x,y)$ at $(a,b)$ ($\nabla f(a,b)$) is not equal to $(0,0)$, then the maximum value of $D_{\hat{u}}f(a,b)$ is $||\nabla f(a,b)||$, and occurs when $\hat{u}$ is in the direction of $\nabla f(a,b)$

Proof:

Since $f$ is differentiable at $(a,b)$ and $||\hat{u}|| = 1$, \begin{align*} D_{\hat{u}}f(a,b) &= \nabla f(a,b) \cdot \hat{u} \\ &= ||\nabla f(a,b)|| ||\hat{u}|| \cos \theta \\ &= ||\nabla f(a,b) || \cos \theta \end{align*} where $\theta$ is our angle between $\hat{u}$ and $\nabla f(a,b)$. Then $D_{\hat{u}}f(a,b)$ will take its largest value when $\cos \theta = 1$, and our result follows. $$\blacksquare$$

You can apply this theorem quite easily to your function to find the maximum directly, and use an idea in the theorem (orthogonality) to find your minimum value(s).

To find the maximum :

1. Calculate the gradient of your function $f$ at the point $(a,b)$.

2. Take the Euclidean norm $||\nabla f(a,b)||$

• I'm still unsure on how to use this. – Anne Jul 9 '15 at 19:31
• @Anne I wrote you some steps, see if you can apply them. – miradulo Jul 9 '15 at 19:36
• I like your theorem based answer. But can you show it in my case if you can? – Anne Jul 9 '15 at 19:51
• @Anne You now have all of the tools you need, I'm not particularly interested in using them for you. If you run into any specific problem along the way while attempting to answer your question yourself with this method, update your question with an edit and I'll be happy to help! – miradulo Jul 9 '15 at 19:53
• Can you just start it off for me? – Anne Jul 9 '15 at 19:53