Product rule for vector derivative

In summary, the conversation discusses the use of a position vector and base frame in calculating velocity. The question is raised about how to apply the product rule for functions to a product of a base frame and functions. The concern is whether the base frame has the nature of a function or how the product rule for functions extends to non-function products.
  • #1
Bullwinckle
10
0
Say I have a position vector

p = e(t) p(t)

Where, in 2D, e(t) = (e1(t), e2(t)) and p(t) = (p1(t), p2(t))T

And if I conveniently point the FIRST base vector of the frame at the particle, I can use: p(t) = (r1(t), 0)T

I want the velocity, so I take

v = d(e(t))/dt p(t) + e(t) d(p(t))/dt

And from there... blah blah.. I can take the rate of change of the frame, etc... but that is not my concern.

My concern is that I KNOW the product rule for functions: I can prove it and use the rule and it is for functions.
But here, I am using it here NOT for a product of two functions but for a product of a base frame and functions.

So, I can ask my question two ways and I hope someone can answer it both ways.

First, what is it about the frame e(t) that can enable me to treat it like function and blithely apply the product rule.
OR
Second, what is it about the product rule for functions that can enable me to apply it to base vectors so expeditiously?

In other words: does the base frame have the nature of a function, OR how does product rule for functions extend to the product of things that are not functions?
 
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  • #2
I'm not sure that I see your concern. e is a unit vector? Is it dynamic or constant? If it's dynamic, I would refrain from expressing it as (r1(t),0), as the second compenent will not always be zero. Also, why is the product e*p not a function?
 

Related to Product rule for vector derivative

What is the product rule for vector derivative?

The product rule for vector derivative is a mathematical rule used to find the derivative of a product of two vector functions. It states that the derivative of the product of two vector functions is equal to the first function multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function.

Why is the product rule important in vector calculus?

The product rule is important in vector calculus because it allows us to find the derivative of more complex vector functions. It is a fundamental concept in calculus and is often used in applications such as physics and engineering.

What is the difference between the product rule for scalar functions and vector functions?

The product rule for scalar functions is a simplified version of the product rule for vector functions. In scalar calculus, the derivative of a product of two functions is equal to the first function multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function. However, in vector calculus, the derivative of a product of two vector functions also takes into account the direction and magnitude of the vectors.

How do you apply the product rule for vector derivative in practice?

To apply the product rule for vector derivative, you first identify the two vector functions that are being multiplied together. Then, you take the derivative of each function separately, and use the product rule to combine the results. This can be done using vector notation or by breaking the vectors into their components.

Are there any exceptions to the product rule for vector derivative?

Yes, there are some exceptions to the product rule for vector derivative. One exception is when the two vector functions being multiplied are parallel to each other. In this case, the product rule simplifies to the derivative of one function multiplied by the other function. Another exception is when one of the vector functions is a constant. In this case, the derivative of the constant is zero, so the product rule simplifies to just the derivative of the other function.

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