Finding the directional derivative

[Saptarshi Sarkar]

Homework Statement

Find the directional derivative of $V^2$ where $\vec V = \hat ixy^2 + \hat jzy^2 + \hat kxz^2$ at the point (2,0,3), in the direction of the outward normal to the sphere $x^2+y^2+z^2=14$ at the point (3,4,0).

Relevant Equations

Directional Derivative = $\nabla f.\hat n$

I tried to calculate the directional derivative but the answer that I found was 194.4 but the answer marked in the book was 540. I tried a lot but couldn't understand what my mistake was.
Please let me know what mistake I did.

 

[Orodruin]

Is this the exact problem formulation?

 

[Saptarshi Sarkar]

Is this the exact problem formulation?

Yes

 

[SammyS]

Homework Statement:: Find the directional derivative of $V^2$ where $\vec V = \hat ixy^2 + \hat jzy^2 + \hat kxz^2$ at the point (2,0,3), in the direction of the outward normal to the sphere $x^2+y^2+z^2=14$ at the point (3,4,0).
Relevant Equations:: Directional Derivative = $\nabla f.\hat n$

I tried to calculate the directional derivative but the answer that I found was 194.4 but the answer marked in the book was 540. I tried a lot but couldn't understand what my mistake was.
Please let me know what mistake I did.

View attachment 256393

Your solution looks good.

Notice what you obtain if you use (3, 0, 4) instead of the point (3, 4, 0) .

People working out answer keys have been known to make mistakes.

 

[PeroK]

Yes

View attachment 256396

There's a peculiar randomness to this question. Note that the outward normal to a sphere (centred at the origin) is in the direction $\hat r$. So, you could just write down the normal in this case: $(\frac 3 5, \frac 4 5, 0)$, or otherwise.

 

[vanhees71]

I'm a bit puzzled by the problem statement too, because $(3,4,0)$ isn't on the sphere $x^2+y^2+z^2=14$... Could it be that the author wanted to consider the sphere $x^2+y^2+z^2=25$?

 

[Saptarshi Sarkar]

I'm a bit puzzled by the problem statement too, because $(3,4,0)$ isn't on the sphere $x^2+y^2+z^2=14$... Could it be that the author wanted to consider the sphere $x^2+y^2+z^2=25$?

It's very much possible. But even if (3,4,0) is not on the sphere, I can draw a normal to that point from the sphere's centre. But, most sphere equations that I see have a squared term on the RHS, so I think it was a typing mistake there.

 

[vanhees71]

The question is a normal to what? You need a surface to define a unit normal vector (in 3D) (up to a sign of course).

 

[SammyS]

Another thing to consider:

As O.P. determined, the gradient of V² at the point (2, 0, 3) is:
$\displaystyle \left. \vec{\nabla} V^2 \right|_{(2,\ 0,\ 3)} = 324 \hat{i} + 0 \hat{j} + 432 \hat{k}$

Now notice that the magnitude of that is 540 .

Added in Edit:
After reading the next post (#10, from @vanhees71 ) in this thread: I suppose that I should have been more complete in this post.

The thing to notice here is that the magnitude of gradient (the maximum directional derivative) is 540.

I was not trying to find the directional derivative along any normal to any surface.

 

[vanhees71]

Yes, but that's not a normal vector on the surface of the given sphere, because the point $(2,0,3)$ is not located at that sphere!

 

[PeroK]

Yes, but that's not a normal vector on the surface of the given sphere, because the point $(2,0,3)$ is not located at that sphere!

These questions that the OP finds are often like this. First, there are two mistakes (perhaps typos) in the question: it should be $25$ instead of $14$; and, the point on the sphere should be $(3, 0, 4)$.

There's also the peculiar aspect that the normal to a sphere at a given point has nothing to do with the gradient of the given function at another point. It looks like two different questions "cobbled together".

 

[vanhees71]

Ok, with the mentioned corrections the problem is well posed, and the solution is (at least conceptually) correct.