I have some cumbersome with derivative problems

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In summary, the key difference between the two formulae is that in one, the nabla operator acts on vector a while in the other, it acts on vector b. This results in the two formulae having different vectors being operated on. The first formula is known as the bidirectional form, while the second formula is the nonstandard notation with a minus sign. These two forms arise in the gradient of dot product identity. Additionally, there is no specific name for the first identity mentioned, but it is important to understand the difference between the two formulae. Other vector calculus identities can be found online for further reference.
  • #1
fdbjruitoirew
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I am confronting with how to realize the difference between two formulae belows
[itex]\left( {\vec a \times \nabla } \right) \times \vec b[/itex], and [itex]\left( {\nabla \times \vec a} \right) \times \vec b[/itex], in here
[itex]\nabla [/itex] is Del (also known as nabla)
Thanks
 
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  • #2
Why these results are so different in formulae
 
  • #3
Ah yes recall that
$$(\vec{a}\cdot \nabla)\vec{b}+\vec{a} \times (\nabla \times \vec{b})=\vec{a}(\nabla \cdot \vec{b})+(\vec{a} \times \nabla) \times \vec{b}$$

By convention the nabla operator acts to the right. For this reason we must be careful. In your formula
$$(\vec{a} \times \nabla )\times \vec{b}$$
Since nabla acts to the right it acts on b and not a
The left acting nabla would act on a and not b
$$(\vec{a} \times \overset{\leftarrow}{\nabla} )\times \vec{b}$$
This is nonstandard notation, but we can write it as
$$-(\nabla \times \vec{a} ) \times \vec{b}=\vec{b} \times (\nabla \times \vec{a} )$$
which is your second formula (with a minus sign)
The sum of the left and right forms is the bidirectional form
$$(\vec{a} \times \overset{\leftrightarrow}{\nabla} )\times \vec{b}=(\vec{a} \times \nabla )\times \vec{b}+\vec{b} \times (\nabla \times \vec{a} )$$
This is useful when you get to integrals because you will have
$$\int_{\partial V} (\vec{a} \times n )\times \vec{b} \, \mathrm{d}S=\int_V ((\vec{a} \times \nabla )\times \vec{b}+\vec{b} \times (\nabla \times \vec{a} )) \, \mathrm{d}V$$
Where we change a surface integral into a volume integral.

The short version if any of that was confusing is that since nabla acts to the right it acts on a and not b in one formula and b and not a in the other, they are very different.
 
  • #4
I am new and unfamiliar with these operation, so what is the name of the first identity you wrote, then I could look up online for reference
thank you
 
  • #5
I don't know if that identity has a name. I brought it up to illustrate that
$$( {\vec a \times \nabla } ) \times \vec b \text{ and } ( {\nabla \times \vec a} ) \times \vec b$$
have different vectors being operated on while
$$( {\vec a \times \nabla } ) \times \vec b \text{ and } ( {\nabla \times \vec b} ) \times \vec a$$
have the same vector being operated on
These two forms arise in the gradient of dot product identity
$$\nabla_\vec{b} (\vec{a} \cdot \vec{b})=(\vec{a}\cdot \nabla)\vec{b}+\vec{a} \times (\nabla \times \vec{b})=\vec{a}(\nabla \cdot \vec{b})+(\vec{a} \times \nabla) \times \vec{b}$$
The nabla subscript b means that a is not operated on.

In your original question the key point was that one expression had a being operated on while the other had b being operated on. This is similar to in single variable calculus if we have
(D indicates the derivative)
D(uv)=uDv+vDu
uDv and vDu are not equal because
in uDv
v is being operated on while u is not
and
in vDu
u is being operated on while v is not

http://en.wikipedia.org/wiki/Vector_calculus_identities
Does not include the identity I mention, but it includes others that may be of interest.
 

FAQ: I have some cumbersome with derivative problems

What is a derivative?

A derivative is a mathematical concept that represents the instantaneous rate of change of a function with respect to one of its variables. It essentially measures how much a function is changing at a specific point.

How do I solve derivative problems?

To solve a derivative problem, you need to use the rules of differentiation. These include the power rule, product rule, quotient rule, and chain rule. You also need to understand the concept of limits and how to take derivatives of common functions such as polynomials, trigonometric functions, and exponential functions.

Why are derivative problems important?

Derivatives are important because they have numerous real-world applications in fields such as physics, engineering, economics, and finance. They are used to model rates of change, find optimum solutions, and make predictions about future behavior.

What are some common mistakes when solving derivative problems?

Some common mistakes when solving derivative problems include forgetting to apply the correct rule, making algebraic errors, and not simplifying the final answer. It's also important to pay attention to the domain of the function and any given constraints.

How can I improve my understanding of derivatives?

To improve your understanding of derivatives, practice solving problems and make sure you understand the underlying concepts. It may also be helpful to seek out additional resources such as textbooks, online tutorials, or working with a tutor. Don't be afraid to ask questions and seek clarification when needed.

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