I'm having some trouble solving this directional derivative problem.

$$ f(x,y) = \sin(2x + 5y), P(-15, 6), \mathbf{u} = {1 \over 2} (\sqrt{3}\mathbf{i-j}) $$

I know the theorem for the bivariate directional derivative: $$D_u f(x, y) = f_x(x, y)a + f_y(x, y)b $$

But that is with the unit vector in form $\mathbf{u}=\langle a,b\rangle$. I think perhaps I am not translating the unit vector into the proper format.

I got the answer: $$ \sqrt{3} \mathbf{i}\cos(2x+5y)+\cos(2x+5y){-5 \over 2}\mathbf{j} $$ But that is incorrect.

  • $\begingroup$ Notationally, $\langle a,b\rangle=a\mathbf{i}+b\mathbf{j}$. So in this case $\mathbf{u}=\frac{1}{2}(\sqrt{3}\mathbf{i}-\mathbf{j})=\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\rangle$ $\endgroup$ – charlestoncrabb Dec 8 '15 at 23:25
  • $\begingroup$ Also you need to plug the point $P$ into your cosines... $\endgroup$ – charlestoncrabb Dec 8 '15 at 23:26
  • $\begingroup$ @charlestoncrabb Thanks, got it. $\endgroup$ – d0rmLife Dec 8 '15 at 23:37

$$ \sqrt{3} - {5 \over 2 } $$

  • $\begingroup$ You ought to have explained that that the scalar product is not a vector... $\endgroup$ – zoli Dec 8 '15 at 23:38

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