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

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SUMMARY

The discussion centers on proving the equation $$\frac{d}{dt}(r \times (v\times r))=||r||^2 *a+ (r\cdot v)*v-(||v||^2+ r\cdot a)*r$$ where r(t) is the position vector of a particle in $\mathbb{R^3}$. Participants reference the product rule for derivatives of cross products and the vector triple product identity. The computation begins with the derivative of the cross product and applies these principles to derive the equation, indicating the necessity of understanding vector calculus and derivatives in three-dimensional space.

PREREQUISITES
  • Understanding of vector calculus, specifically cross products and derivatives.
  • Familiarity with the vector triple product identity in $\mathbb{R^3}$.
  • Knowledge of derivatives with respect to time in the context of motion.
  • Basic understanding of position, velocity, and acceleration vectors.
NEXT STEPS
  • Study the vector triple product identity in detail.
  • Learn about the product rule for derivatives of vector functions.
  • Explore applications of derivatives in physics, particularly in kinematics.
  • Practice proving similar equations involving derivatives of vector functions.
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Students and professionals in physics and mathematics, particularly those focusing on mechanics and vector calculus, will benefit from this discussion.

WMDhamnekar
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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
 

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