Finding the directional derivative

In summary: The thing to notice here is that the magnitude of gradient (the maximum directional derivative) is 540.
  • #1
Saptarshi Sarkar
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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.

IMG_20200131_233638.jpg
 
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  • #2
Is this the exact problem formulation?
 
  • #3
Orodruin said:
Is this the exact problem formulation?

Yes

Screenshot_20200201-005036~2.png
 
  • #4
Saptarshi Sarkar said:
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.
 
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  • #5
Saptarshi Sarkar said:

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.
 
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  • #6
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##?
 
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  • #7
vanhees71 said:
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.
 
  • #8
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).
 
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  • #9
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.
 
Last edited:
  • #10
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!
 
  • #11
vanhees71 said:
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".
 
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  • #12
Ok, with the mentioned corrections the problem is well posed, and the solution is (at least conceptually) correct.
 

Related to Finding the directional derivative

1. What is the directional derivative?

The directional derivative is a measure of the rate of change of a function in a particular direction. It tells us how much the function changes as we move in a specific direction from a given point.

2. How is the directional derivative calculated?

The directional derivative is calculated by taking the dot product of the gradient vector of the function and the unit vector in the desired direction. This can be represented mathematically as:
Duf(x,y) = ∇f(x,y) · u
where ∇f is the gradient vector and u is the unit vector in the desired direction.

3. What is the significance of the directional derivative?

The directional derivative is important in many areas of mathematics and science, particularly in vector calculus and optimization problems. It allows us to determine the direction in which a function is changing the fastest at a given point, and is therefore useful in solving problems involving maximum and minimum values.

4. Can the directional derivative be negative?

Yes, the directional derivative can be negative. This indicates that the function is decreasing in the direction of the unit vector u. A positive directional derivative indicates an increase in the function in the direction of u.

5. How is the directional derivative used in real-world applications?

The directional derivative has many applications in fields such as physics, engineering, and economics. It is used to analyze the rate of change of physical quantities, such as velocity and acceleration, and can also be used to optimize processes and solve optimization problems in various industries.

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