0
$\begingroup$

At a certain point $P$ the function $f(x,y)$ has a direction derivative $\sqrt{2}$ in the direction of $\hat i-\hat j$ and $ 3\sqrt{2}$ in the direction of $\hat i+\hat j$. What is the maximum directional derivative of $f$ at the point $P$

From the first directional derivative we got the equation $\frac{f_x}{\sqrt{2}} -\frac{f_y }{\sqrt{2}} =\sqrt{2}\rightarrow f_x -f_y=2$ From the second $f_x+f_y=6$ hence $f_x=4 , f_y=2$ now we want to maximize $(4,2).\frac{\hat v}{|| \hat v||}$ how to continue? Is there any easier approach?

$\endgroup$
1
$\begingroup$

Yes, the directional derivative is maximal in the direction pointing along the gradient, i.e., $$ \hat v = \frac{1}{\|\nabla f\|}\nabla f\,.$$ This is a general result of multivariable calculus. Since you know the partial derivatives, it is easy to compute $\nabla f$ and $\hat v$.

$\endgroup$
  • $\begingroup$ $(4i+2j).\frac{4i+2j}{\sqrt{20}}$ is the answer? Thanks i got it $\endgroup$ – AmerYR Mar 9 '17 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.