The derivative of the dot product "distributes" over each vector

[brotherbobby]

Homework Statement

Prove that : $\boxed{\dfrac{d}{dt}(\mathbf{A}\cdot \mathbf{B})=\dfrac{d\mathbf{A}}{dt}\cdot \mathbf{B}+\mathbf{A}\cdot \dfrac{d\mathbf{B}}{dt}}$

Relevant Equations

The dot product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is given by $\mathbf{A}\cdot \mathbf{B}=AB\cos\theta$, where $\theta$ is the (smaller) angle between them.

Drawing : I draw a diagram explaining the situation to the right. The vectors $\mathbf{A}$ and $\mathbf{B}$ are drawn at some time $t$ making an angle $\theta$ between them. At some later time $t+dt$, the same vectors have changed to $\mathbf{A}(t+dt)$ and $\mathbf{B}(t+dt)$ while their angle has changed to some $\theta(t+dt)$.


Attempt :

We have $\dfrac{d}{dt}(\mathbf{A}\cdot \mathbf{B})=\dfrac{d}{dt}(AB\cos\theta)=\dfrac{dA}{dt}B\cos\theta+A\dfrac{dB}{dt}\cos\theta+AB\dfrac{d}{dt}\cos\theta$
$=\dfrac{dA}{dt}B\cos\theta+A\dfrac{dB}{dt}\cos\theta-AB\sin\theta \dfrac{d\theta}{dt}$
$=\underbrace{\dfrac{d\mathbf{A}}{dt}\cdot \mathbf{B}}_{\text{Is this correct?}}+\underbrace{\mathbf{A}\cdot \dfrac{d\mathbf{B}}{dt}}_{\text{Is this correct?}}-AB\sin\theta \dfrac{d\theta}{dt}$

Even if I am correct in the first two terms, the third term is a violation of the required statement of the problem.

Request : A hint as to where am going wrong would be very welcome.

 

[Orodruin]

The step where you have asked if it is correct is not correct. I would go so far as to say that the expression $\vec A \cdot \vec B = AB \cos(\theta)$ will not be very useful for this. Instead, I suggest that you write your inner product in terms of the vector components, i.e.,
$$
\vec A \cdot \vec B = \sum_{i = 1}^3 A_i B_i
$$
and work from there.

 

[brotherbobby]

The step where you have asked if it is correct is not correct.

I wrote $\dfrac{dA}{dt}B\cos\theta = \dfrac{d\mathbf{A}}{dt}\cdot \mathbf{B}$. Why is it wrong?

I would go so far as to say that the expression A→⋅B→=ABcos⁡(θ) will not be very useful for this

Yes, so it seems, but surely there's got to be a reason why it doesn't work out. Perhaps it is connected to the previous point above.

I have found a solution to the problem in a different way, not far from what you suggested. I'd post it presently. For the moment, I am curious as to why doesn't my method above doesn't work.

 

[Ibix]

I wrote $\dfrac{dA}{dt}B\cos\theta = \dfrac{d\mathbf{A}}{dt}\cdot \mathbf{B}$. Why is it wrong?

Consider the case where $\mathbf{A}$ is changing direction but not magnitude.

 

[Orodruin]

Yes, so it seems, but surely there's got to be a reason why it doesn't work out.

A more relevant question is: Why you think that it would work out?
You yourself have indicated that you are uncertain and therefore trying to work out why it is or isn't the case is a good place to start. The hint by @Ibix above will provide you with a scenario where it is clearly false.

 

[PeroK]

I wrote $\dfrac{dA}{dt}B\cos\theta = \dfrac{d\mathbf{A}}{dt}\cdot \mathbf{B}$. Why is it wrong?

First $A = |\vec A| = \sqrt{\vec A \cdot \vec A}$ is the magnitide of the vector $A$. In general $\frac{dA}{dt} \ne \big | \frac{d \vec A}{dt}\big |$. The obvious example being, as above, when $\vec A$ is changing but not in magnitude.

In any case, assuming that $\frac{dA}{dt} = \big | \frac{d \vec A}{dt}\big |$ is a fundamental misconception.

Also, the angle between $\vec A$ and $\vec B$ is not the same as the angle between $\frac{d\vec A}{dt}$ and $\vec B$. That's another major misconception.