Dot Product with Derivative

[Tegdif]

TL;DR Summary

The Dot Product of a Vector 'a' and a Derivative 'b' is the same like the negative of the Derivative 'a' and the Vector 'b'.

Summary: The Dot Product of a Vector 'a' and a Derivative 'b' is the same like the negative of the Derivative 'a' and the Vector 'b'.

Hello, I have the following Problem. The Dot Product of a Vector 'a' and a Derivative 'b' is the same like the negative of the Derivative 'a' and the Vector 'b'.
a, b, c are Vectors.
And a', b', c' are the derivative of them.

a ⋅ b = a⋅c = b⋅c = 0

b' = αa + βc

⇒ a' ⋅ c = - a ⋅ c' (1)

I don't understand how you get the last Formula (1).

 

[BvU]

Hello teg, !

If one can do differentiation, there must be some independent variable wrt which to differentiate. What is that in your case ?

What do you get when you diferentiate both sides of a⋅c = 0 ?

 

[fresh_42]

It's the product rule. $\left(\mathbf{a\cdot c}\right)'= \mathbf{a}'\cdot \mathbf{c}+\mathbf{a}\cdot \mathbf{c}'$ and if the product is zero, the formula follows.

 

[Tegdif]

Diferentiate both sides require the product rule, like Fresh_42 mentioned it.
so it should be (a⋅b)' = a'⋅c + a⋅c'. Diferentiate 0 gives me 0.
⇒ 0 = a'⋅c + a⋅c' ⇔ a'⋅c = -a⋅c'

Thank you, BvU and Fresh_42

 

[WWGD]

It's the product rule. $\left(\mathbf{a\cdot c}\right)'= \mathbf{a}'\cdot \mathbf{c}+\mathbf{a}\cdot \mathbf{c}'$ and if the product is zero, the formula follows.

But arent these vectors constants or do we assume they are both a function of the same variable t?

 

[fresh_42]

But arent these vectors constants or do we assume they are both a function of the same variable t?

There isn't sufficient information. I concluded by Occam's razor: it explains what is given. The product rule is present on so many levels, that it is more than likely to be the answer.

 

[Mark44]

Hello, I have the following Problem. The Dot Product of a Vector 'a' and a Derivative 'b' is the same like the negative of the Derivative 'a' and the Vector 'b'.

Or in symbols, a' ⋅ b = - a ⋅ b'

a, b, c are Vectors.
And a', b', c' are the derivative of them.

c and c' don't enter the picture here

a ⋅ b = a⋅c = b⋅c = 0

b' = αa + βc

⇒ a' ⋅ c = - a ⋅ c' (1)

I don't understand how you get the last Formula (1).

From your problem statement, the above isn't what you needed to show, which involves only a and b and their derivatives, but doesn't involve c or its derivative.
There seems to be a lot more information here than is needed. If we get rid of all the extra stuff, we're given that a ⋅ b = 0.
Differentiate both sides of this equation and you get a' ⋅ b + a ⋅ b' = 0, by using the product rule, as mentioned by other posters. From there, isolate a ⋅ b'.

I don't see the purpose of including this: b' = αa + βc. It doesn't have any bearing on what you're trying to prove, as far as I can see.

 

[WWGD]

Or in symbols, a' ⋅ b = - a ⋅ b'
c and c' don't enter the picture here
From your problem statement, the above isn't what you needed to show, which involves only a and b and their derivatives, but doesn't involve c or its derivative.
There seems to be a lot more information here than is needed. If we get rid of all the extra stuff, we're given that a ⋅ b = 0.
Differentiate both sides of this equation and you get a' ⋅ b + a ⋅ b' = 0, by using the product rule, as mentioned by other posters. From there, isolate a ⋅ b'.

I don't see the purpose of including this: b' = αa + βc. It doesn't have any bearing on what you're trying to prove, as far as I can see.

Derivative with respect to what? A derivative is a concept that applies to functions. How is this the case here?

 

[Mark44]

Derivative with respect to what? A derivative is a concept that applies to functions. How is this the case here?

Based on the initial post, we are given the existence of derivatives.

a, b, c are Vectors.
And a', b', c' are the derivative of them.

We can infer that these vectors are all functions of some unstated independent variable, and since the derivatives exist, these must be differentiable functions of that variable.