# How Does the Dot Product of Vector Derivatives Relate to Their Original Vectors?

• I
• Tegdif
So in other words, a, a', b, and b' are all functions of the same variable, and c and c' are also functions of that variable.
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.

ab = a⋅c = b⋅c = 0

b' = αa + βc

a' ⋅ c = - a ⋅ c' (1)

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

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 ?

Tegdif
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.

HallsofIvy and 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

fresh_42 said:
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?

WWGD said:
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.

Tegdif said:
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'
Tegdif said:
a, b, c are Vectors.
And a', b', c' are the derivative of them.
c and c' don't enter the picture here
Tegdif said:
ab = 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.

Mark44 said:
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?

WWGD said:
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.
Tegdif said:
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.

lekh2003 and fresh_42

## 1. What is the dot product with derivative?

The dot product with derivative is a mathematical operation that combines two vectors to produce a scalar value. It is calculated by multiplying the corresponding components of two vectors and then summing the products.

## 2. What is the significance of the dot product with derivative in science?

The dot product with derivative is used in a variety of scientific fields, including physics, engineering, and computer science. It is particularly useful in calculating work, determining the direction of forces, and finding the rate of change of a vector.

## 3. How is the dot product with derivative different from the cross product?

The dot product with derivative is a scalar value, while the cross product is a vector. This means that while the dot product with derivative gives information about the magnitude of two vectors, the cross product gives information about their direction.

## 4. Can the dot product with derivative be negative?

Yes, the dot product with derivative can be negative. This occurs when the angle between the two vectors is greater than 90 degrees, resulting in a negative value for the dot product with derivative.

## 5. How is the dot product with derivative used in machine learning?

In machine learning, the dot product with derivative is used to calculate the gradient of a function. This gradient is then used to update the parameters of a model in order to minimize a cost function and improve the model's performance.

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