1
$\begingroup$

Find values $a, b, c$ such that the maximal directional derivative of $$f(x,y,z) = a x y^2 + b y z + c z^2 x^3$$ at point $(1,2-1)$ is $64$ in a direction parallel to the $z$ axis.

My work so far:

Claim 1: I know the directional derivative can only be in the z direction because to must be parallel to the z axis. so out of the vector $<i,j,k>$ only k will be used

I also know that the maximum directional derivative will concern the unit vector, but thiis means that the length of the vector in the direction of j will be 1 ( I think this is right? )

so $\frac{\partial f}{\partial z} = by +2cx^3$ and given that the unit vector should be <0,0,1>

I get the equation:

$64$ = $\frac{\partial f}{\partial z} * 1$

$64 = (by +2cx^3) * 1$ and this is at the point (1,2,-1)

so my equation becomes $64 = 2b-2c$ so I know that the equality $32=b-c$ exists for the values a,b,c

but I cannot get further than that.

I am an adult learner, not in school. just doing this for fun, and I have noone to ask. Please help! would be apreciated!

$\endgroup$
1
  • $\begingroup$ You’re going astray when you say that “only $k$ will be used.” What the condition really means is that $i$ and $j$ must be zero, which is rather different. $\endgroup$
    – amd
    Sep 4 '17 at 23:22
1
$\begingroup$

The formula for directional derivative of $f$ in the direction of $\vec u$ is $$D_{\ \vec u} \ f = \vec \nabla f \ \bullet \frac{\vec u}{|\vec u|} $$

where $\vec \nabla f := <f_x,f_y,f_z> \ $ (the vector of partial derivatives) and "$\bullet$" is the dot product.

Observe that you can choose $\vec u$ in whathever length you want, because it will be divided by its length. In other words, if $\vec u$ is unit, $D_{\ \vec u} \ f = \vec \nabla f \ \bullet \ \vec u $.

To maximize the directional derivative, $\vec \nabla f$ must be parallel to $\vec u$.

So, $f_x$ and $f_y$ must be equal to zero at the point $(1,2,-1)$. And $f_z$ can be equal to anything.

In addition to the equality you have already found, solving $\ f_x \big|_{(1,2,-1)} =0$ and $\ f_y \big|_{(1,2,-1)}=0$ will give you all equations you need.

$\endgroup$

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.