Direction in which directional derivative is zero

  • #1
songoku
2,319
331
Homework Statement
Please see below
Relevant Equations
Partial Derivative

Direction derivative in the direction of unit vector u = <a, b, c>:
Du f(x,y,z) = fx (x,y,z) a + fy (x,y,z) b + fz (x, y, z)
1697812128978.png

I want to ask about the direction in which ##D_v## is zero at point (1, 2, 1)

My attempt:
$$w_x=yz+\frac{1}{x}$$
$$w_y=xz+\frac{1}{y}$$
$$w_z=xy+\frac{1}{z}$$

At point (1, 2, 1), the ##\nabla w=<3, \frac{3}{2}, 3>##

$$D_v w=0$$
$$\nabla w \cdot v=0$$
$$
\begin{pmatrix}
3 \\
\frac{3}{2} \\
3
\end{pmatrix}
\cdot
\begin{pmatrix}
p \\
q \\
r
\end{pmatrix}
=0
$$
$$3p+\frac{3}{2} q+3r=0$$
$$2p+q+2r=0$$

Another equation is ##p^2+q^2+r^2=1##

But I can't find ##p, q## and ##r## from these equations. Is my working wrong?

Thanks
 
Last edited:
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  • #2
songoku said:
Homework Statement: Please see below
Relevant Equations: Partial Derivative

Direction derivative in the direction of unit vector u = <a, b, c>:
Du f(x,y,z) = fx (x,y,z) a + fy (x,y,z) b + fz (x, y, z)

View attachment 333900
I want to ask about the direction in which ##D_v## is zero at point (1, 2, 1)

My attempt:
$$w_x=yz+\frac{1}{x}$$
$$w_y=xz+\frac{1}{y}$$
$$w_z=xy+\frac{1}{z}$$

At point (1, 2, 1), the ##\nabla w=<3, \frac{3}{2}, 3>##

$$D_v w=0$$
$$\nabla w \cdot v=$$
$$
\begin{pmatrix}
3 \\
\frac{3}{2} \\
3
\end{pmatrix}
\cdot
\begin{pmatrix}
p \\
q \\
r
\end{pmatrix}
=0
$$
$$3p+\frac{3}{2} q+3r=0$$
$$2p+q+2r=0$$

Another equation is ##p^2+q^2+r^2=1##

But I can't find ##p, q## and ##r## from these equations. Is my working wrong?

Thanks
No. I can't find a mistake, except that I have no idea what ##\vec{u}=\vec{i}+\vec{j}## is since your coordinates seem to be ##x,y,z## and not ##\vec{i},\vec{j},\vec{k},## and why you did not calculate ##\nabla_{(1,2,1)}(\omega )\cdot \vec{u}## as requested.

The points in
$$
\left\{(p,q,r)\,|\,\nabla_{(1,2,1)}(\omega )\cdot (p,q,r)=\vec{0}\right\} =\left\{(p,q,r)\in \mathbb{R}^3\,|\,2p+q+2r=0\right\}=(2,1,2)^\perp
$$
form a hyperplane, i.e. a plane in the ##3##-dimensional Euclidean space. This makes sense because it is the kernel of a linear function of rank one, the kernel of the multiplication with the gradient, the kernel of the nabla operator.

Finally, you want to consider only unit vectors (why?). This means to intersect the plane with the unit sphere. The result will be an ellipse somewhere in space. Unfortunately, the emphasis is on "somewhere". It doesn't have a nice equation, but you can write ##q=-2p-2r,## plug it into ##p^2+q^2+r^2=1## and get the equation of the ellipse, which you can normalize or not.
 
Last edited:
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  • #3
I am really sorry for late reply.

fresh_42 said:
No. I can't find a mistake, except that I have no idea what ##\vec{u}=\vec{i}+\vec{j}## is since your coordinates seem to be ##x,y,z## and not ##\vec{i},\vec{j},\vec{k},## and why you did not calculate ##\nabla_{(1,2,1)}(\omega )\cdot \vec{u}## as requested.
I have calculated ##\nabla_{(1,2,1)}(\omega )\cdot \vec{u}##

fresh_42 said:
Finally, you want to consider only unit vectors (why?).
Because the formula involves the dot product with unit vector. I am thinking the final answer would be something like ##k <p, q, r> ## (all vectors parallel to unit vector ##<p, q, r>## but I can't get that form.

fresh_42 said:
This means to intersect the plane with the unit sphere. The result will be an ellipse somewhere in space. Unfortunately, the emphasis is on "somewhere". It doesn't have a nice equation, but you can write ##q=-2p-2r,## plug it into ##p^2+q^2+r^2=1## and get the equation of the ellipse, which you can normalize or not.
$$p^2+q^2+r^2=1$$
$$p^2+(-2p-2r)^2+r^2=1$$
$$p^2+4p^2+8pr+4r^2=1$$
$$5p^2+8pr+4r^2=1$$

Then I don't know what to do.

Is it possible the answer in the form like:
$$v=
\begin{pmatrix}
p \\
q \\
r
\end{pmatrix}
$$
$$=
\begin{pmatrix}
p \\
-2p-2r \\
r
\end{pmatrix}
$$
$$=p
\begin{pmatrix}
1 \\
-2 \\
0
\end{pmatrix}
+ r
\begin{pmatrix}
0 \\
-2 \\
1
\end{pmatrix}
$$

Thanks
 
  • #4
songoku said:
I am really sorry for late reply.I have calculated ##\nabla_{(1,2,1)}(\omega )\cdot \vec{u}##Because the formula involves the dot product with unit vector. I am thinking the final answer would be something like ##k <p, q, r> ## (all vectors parallel to unit vector ##<p, q, r>## but I can't get that form.$$p^2+q^2+r^2=1$$
$$p^2+(-2p-2r)^2+r^2=1$$
$$p^2+4p^2+8pr+4r^2=1$$
$$5p^2+8pr+4r^2=1$$

Then I don't know what to do.

Is it possible the answer in the form like:
$$v=
\begin{pmatrix}
p \\
q \\
r
\end{pmatrix}
$$
$$=
\begin{pmatrix}
p \\
-2p-2r \\
r
\end{pmatrix}
$$
$$=p
\begin{pmatrix}
1 \\
-2 \\
0
\end{pmatrix}
+ r
\begin{pmatrix}
0 \\
-2 \\
1
\end{pmatrix}
$$

Thanks
I would say yes. But it might be a matter of convention. There are a few ways to describe a plane and I can't know whether you are expected to use any or a specific one. The same with an ellipse. It looks like
1698509228861.png


and you can calculate a rotation (coordinate transformation) so that it looks like a standrad ellipse. The new coordinates would be ##p'=\sqrt{5}p+4r/\sqrt{5}## and ##r'=2r/\sqrt{5}## if I made no mistake.
 
  • #5
fresh_42 said:
I would say yes. But it might be a matter of convention. There are a few ways to describe a plane and I can't know whether you are expected to use any or a specific one.
Now that you said it, I just realised that the answer is equation of plane passing through origin.

fresh_42 said:
and you can calculate a rotation (coordinate transformation) so that it looks like a standrad ellipse. The new coordinates would be ##p'=\sqrt{5}p+4r/\sqrt{5}## and ##r'=2r/\sqrt{5}## if I made no mistake.
Are there any resources to learn about rotation for ellipse and other geometrical shapes?

Thanks
 
  • #6
I would search for "conic sections" + pdf or look it up in Wikipedia, e.g. https://en.wikipedia.org/wiki/Parabola or more general https://en.wikipedia.org/wiki/Conic_section. Here is a good article https://arxiv.org/pdf/1205.5935.pdf but I'm not sure if it fits you. It is a bit "dry".

The equation of your plane through the origin was correct. It is a parametric representation, mine was a geometric representation by its normal vector. There is no better or worse, only different equations with the same solution space.
 
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  • #7
Thank you very much for the help and explanation fresh_42
 
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