MHB Proving Equation (1): Let r(t) be a Vector in $\mathbb{R^3}$

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To prove equation (1), the derivative of the expression \( r \times (v \times r) \) can be computed using the product rule for derivatives and the vector triple product identity. The initial steps involve applying the derivative to both components, resulting in terms that include \( v \) and \( a \). The discussion highlights the need to simplify these terms using vector identities, specifically the relationship between cross products and dot products. Participants are encouraged to continue the derivation from the provided expressions, utilizing the hint about the squared magnitude of vectors. The conversation emphasizes collaborative problem-solving in vector calculus.
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Let r(t) be the position vector for a particle moving in $\mathbb{R^3}.$ How to show that
$$\frac{d}{dt}(r \times (v\times r))=||r||^2 *a+ (r\cdot v)*v-(||v||^2+ r\cdot a)*r \tag{1}$$

Where r(t) is a position vector (x(t),y(t),z(t)), $v(t)=\frac{dr}{dt}=(x'(t),y'(t),z'(t)), a(t)=\frac{dv}{dt}=\frac{d^2r}{dt^2}=(x''(t),y''(t),z''(t))$ Note: v=velocity, a= acceleration

My attempt:-
I know $\frac{d}{dt}(f\times g)=\frac{df}{dt}\times g +f\times \frac{dg}{dt}$. I also know for any vectors u, v, w in $\mathbb{R^3}, u\times (v\times w)=(u\cdot w)*v-(u\cdot v)*w $

But I don't understand how to use these formulas to prove equation (1)?

If any member knows answer to this question may reply with correct answer.
 
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Hi Dhamnekar,

You're correct that those two formulas need to be used, nicely done. I will start the computation below. See if you can finish it. If not, I'm happy to answer any other questions you may have. Note: I will use the dot notation to denote a derivative with respect to time.

\begin{align*}

\frac{d}{dt}(r\times (v\times r)) &= \dot{r}\times (v\times r) + r\times \dot{(v\times r)}\\
& = v\times (v\times r) + r\times (\dot{v}\times r + v\times\dot{r})\\
& = v\times (v\times r) + r\times (a\times r + v\times v)

\end{align*}

Can you proceed from here? Hint: $x\cdot x = \| x\|^{2}.$
 
[math]\textbf{a} \times ( \textbf{b} \times \textbf{c} ) = ( \textbf{a} \cdot \textbf{c} ) \textbf{b} + ( \textbf{a} \cdot \textbf{b} ) \textbf{[/math]
Dhamnekar Winod said:
Let r(t) be the position vector for a particle moving in $\mathbb{R^3}.$ How to show that
$$\frac{d}{dt}(r \times (v\times r))=||r||^2 *a+ (r\cdot v)*v-(||v||^2+ r\cdot a)*r \tag{1}$$

Where r(t) is a position vector (x(t),y(t),z(t)), $v(t)=\frac{dr}{dt}=(x'(t),y'(t),z'(t)), a(t)=\frac{dv}{dt}=\frac{d^2r}{dt^2}=(x''(t),y''(t),z''(t))$ Note: v=velocity, a= acceleration

My attempt:-
I know $\frac{d}{dt}(f\times g)=\frac{df}{dt}\times g +f\times \frac{dg}{dt}$. I also know for any vectors u, v, w in $\mathbb{R^3}, u\times (v\times w)=(u\cdot w)*v-(u\cdot v)*w $

But I don't understand how to use these formulas to prove equation (1)?

If any member knows answer to this question may reply with correct answer.
Is this a part of a derivation? The [math]\textbf{r} \times ( \textbf{v} \times \textbf{r} )[/math] looks familiar but I can't seem to find it.

-Dan