We have f(x,y,z) at point $m$ the directional derivative is maximum at the direction of vector $u=(1,1,1)$ And is equal to $\sqrt{27}$

find for which a the directional derivative for f in point m in direction of vector (1,2,a) is 5?

Since the D'D at point $m$ is maximum at the direction of vector $u=(1,1,1)$ that means $\triangledown f(m)$ is equal to the direction of vector $u$ and the direction of $u$ is $\frac{u}{\left \| u \right \|}=\frac{1}{ \sqrt{3}}$ that is supposed to mean that gardient vector for $\triangledown f(m)=\alpha \vec{u}$ for some $\alpha\epsilon \mathbb{Z}$

Am I correct so far? and if that's so i dont know how to continue from here


If the directional derivative of $f$ is at a maximum in some direction then $\nabla f$ points in that direction.

What is $\nabla f$

$\nabla f = \mu (1,1,1)\\ \nabla f \cdot (\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}) = \sqrt {27}\\ \mu\sqrt 3 = \sqrt {27}\\ \mu = 3\\ \nabla f = (3,3,3)$

What is the directional derivative in the direction $(1,2,a)$?

$\nabla f\cdot \frac {(1,2,a)}{\|(1,2,a)\|} = \frac {9+3a}{\sqrt {5+a^2}}$

We have been told that this equals $5$

$\frac {9+3a}{\sqrt {5+a^2}} = 5\\ 9+3a = 5\sqrt {5+a^2}$

Square both sides and solve the quadratic for $a.$

$81 + 54a + 9a^2 = 125 + 25a^2\\ 16a^2 - 54a + 44 = 0\\ 2(8a - 11)(a-2)=0\\ a = \frac {11}{8}, 2$


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.