- #1
WMDhamnekar
MHB
- 378
- 28
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.
$$\frac{d}{dt}(r \times (v\times r))=||r||^2 *a+ (r\cdot v)*v-(||v||^2+ r\cdot a)*r \tag{1}$$
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.