# Dot Product of a Vector and its Derivative- Reality

1. Mar 29, 2014

### mjxcrowley

Hey everyone,
This has been bugging me for a bit. I think I'm probably missing something pretty easy.

A dot B= ABcos(θ), where θ is the angle between A and B.

There is the little shortcut that says where B is the derivative of A, A dot B= AB. Clearly then cos(θ) = 1, and the angle between a vector and its derivative is 2nPi, where n=0, 1, 2... Intuitively this would not be true, and is clearly not true for an orbit (which is what I use it for). There the angle between R and V where V is the derivative of R is 90°+the flight path angle.

What am I missing? How are both these things true?

Note: I don't want to mathematical derivation of the shortcut. I have that. I want to understand it in reality.

I hope this is in the correct forum

Last edited: Mar 29, 2014
2. Mar 29, 2014

### D H

Staff Emeritus
Both of these things are false.

The inner product of a vector ${\bf A}(t)$ and it's derivative ${\bf B}(t) = \frac{d {\bf A}(t)}{dt}$ is equal to $||{\bf A}|| \, ||{\bf B}||$ only in the special case that ${\bf A}(t) = A(t)\hat {\bf a}$, where $A(t)$ is a scalar and $\hat {\bf a}$ is a constant vector.

The inner product of a vector ${\bf A}(t)$ and it's derivative ${\bf B}(t) = \frac{d {\bf A}(t)}{dt}$ is zero only in the special case that ${\bf A}(t)$ is a constant length vector. Only circular orbits satisfy this condition; not all orbits are circular (in fact, almost all orbits are not circular).

3. Mar 29, 2014

### mjxcrowley

My confusion I guess is that the relationship A dot B= AB where B is the derivative wrt time of A is often used generally in Astrodynamics texts. It doesn't seem generally correct, but it appears everywhere as a mathematical shortcut.

4. Mar 29, 2014

### D H

Staff Emeritus
Where have you seen that? Are you sure you aren't confusing that with A cross B, which is equal in magnitude to AB in the case of a circular orbit?

5. Mar 29, 2014

### mjxcrowley

I'm having a very hard time with the symbols. I'll try and explain then.

While deriving the trajectory equation for the general two body problem, it says:

r(vector) dot r(vector, dot) = r r(dot)

where the dot within the parenthesis represents the dot written above the variable to indicate derivative. The dot without the parenthesis is the dot product.

I apologize that that was so clunky but I tried to use the sigma button, and could not get it to work.

Last edited: Mar 29, 2014
6. Mar 29, 2014

### D H

Staff Emeritus
That's different. You mean $\vec r \cdot \dot{\vec r} = r\dot r$.

Note very well: The $\dot r$ in that expression is not the magnitude of the velocity vector. It is instead the time derivative of the radial distance between the two orbiting bodies.

Here's the derivation. I'll work in cylindrical coordinates, where $\hat r$ is the unit vector along the radial vector, $\hat z$ is the unit vector along the angular velocity vector, and $\hat{\theta}$ completes the right hand coordinate system (with $\hat r \times \hat {\theta} = \hat z$). With this, the radial vector is $\vec r = r \hat r$. Taking the time derivative yields $\dot{\vec r} = \dot r \hat r + r\dot{\hat r}$. The second term is necessarily orthogonal to radial vector. Thus $\vec r \cdot \dot{\vec r} = r \dot r$.

Last edited: Mar 29, 2014
7. Mar 29, 2014

### TheOldHag

From the calculus of vector valued functions a vector valued function and its derivative are orthogonal. In euclidean n-space this would mean cos Θ = 1 and hence the dot product of A and B would be the norm of A times the norm of B.

So my understanding of your question is you want to know why.

To see why this is true consider R(t) = f(t) i + g(t) j where i and j are standard basis vector and f and g are parametric equations describing a curve in the plane. The derivative would then be

R'(t) = lim Δt → 0 $R(t + Δt) - R(t)/Δt$

So now draw a vector from the origin to the point on the curve at t + Δt and then draw another vector a t. These two vector would be R(t + Δt) and R(t) respectively. Now draw a vector pointing from t to t + Δt on the curve at those values. This arrow would be the numerator in the limit above. Now decrease Δt and you will see that this vector will soon be tangent to the curve at t and it will be visually clear that it is orthogonal.

8. Mar 29, 2014

### D H

Staff Emeritus
That's just, on wrong both accounts. A vector valued function and its derivative are not necessarily orthogonal, and when they are, the inner product would be zero, not the norm of A times the norm of B.

9. Mar 29, 2014

### TheOldHag

You're right. I must have had brain leakage. I believe I'm a bit off on orthogonality too. I believe they are orthogonal only when the norm of R(t) is constant on an interval.

In general, I'm not seeing how the statement A dot B = norm(A) * norm(B) can be true since a vector valued function and it's derivative are not generally scalar multiple of one another.

Sorry for the confusion.

10. Mar 29, 2014

### D H

Staff Emeritus
As I wrote in post #2, it's not true.

What is true is that $\displaystyle{\vec A \cdot \frac{d\vec A}{dt} = ||\vec A||\,\frac{d\,||\vec A||}{dt}}$

11. Apr 1, 2014

### mjxcrowley

Thank you so much DH. I was confused and you were extremely helpful.