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A directional derivative of $$ f(x,y)=x^2-3y^3 $$ at the point $P(2,1)$ in some direction might be:

a) $-9$

b) $-10$

c) $6$

d) $11$

e) $0$

I'd say it's $-9$ for sure, but what about $0$ (the direction would be $<0,0>$)? Are there any other proper answers?

TomDavies92
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    A direction is given by some unit vector, $(0, 0)$ is not a direction. Hint: consider the gradient, and how do you calculate directional derivative from gradient? – Secret Math Jun 15 '13 at 06:54
  • Which formulae did you use? Why would you say "-9"? – Avitus Jun 15 '13 at 06:58
  • It's $$ gradf*[cos\alpha, cos\beta] $$ $$ [cos\alpha, cos\beta] $$ is a directional vector so its directions are less or equal to 1 – TomDavies92 Jun 15 '13 at 06:59
  • @Avitus, I'd say t's -9 since when the direction is <0,1> it's -9. We consider any direction here. – TomDavies92 Jun 15 '13 at 07:00

1 Answers1

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$$D_{\vec u}f(\vec x)=\nabla f_{(2,1)}\frac{\vec u}{||\vec u||}\cdot=4u_1-9u_2\;\;\wedge\;\;u_1^2+u_2^2=1$$

so you get a non-linear system of equations

$$\begin{align*}\text{I}&\;\;4u_1-9u_2=t\\\text{II}&\;\;\;\;u_1^2+\;u_2^2=\,1\end{align*}$$

and from here we get

$$u_1^2+\left(\frac{4u_1-t}{9}\right)^2=1\implies 97u_1^2-8tu_1+(t^2-81)=0$$

The above quadratic's discriminant is

$$\Delta=-324(t^2-97)\ge 0\iff |t|\le\sqrt{97}$$

Thus, the system has a solution for any $\;t\in\Bbb R\;\;,\;\;|t|\le\sqrt{97}\;$ , so all of $\,(a), (c), (e)\;$ fulfill this condition .

DonAntonio
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