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[SOLVED] Vector-valued function representing velocity, acceleration,force, momentum etc.

Dhamnekar Winod

Active member
Nov 17, 2018
103
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.
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
271
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: $x\cdot x = \| x\|^{2}.$
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,121
\(\displaystyle \textbf{a} \times ( \textbf{b} \times \textbf{c} ) = ( \textbf{a} \cdot \textbf{c} ) \textbf{b} + ( \textbf{a} \cdot \textbf{b} ) \textbf{\)
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 \(\displaystyle \textbf{r} \times ( \textbf{v} \times \textbf{r} )\) looks familiar but I can't seem to find it.

-Dan