Question

if r(t) and r'(t) are differentiable functions of (t), then find the value of d/dt[2r(t) x r'(t)]

Answer 1

r(t)
r'(t) = v(t)
r''(t) = v'(t) = a'(t)

so :

2r(t) X r'(t) =

(2ry(t)vz(t) - 2rz(t)vy(t))i + (2rz(t)vx(t) - 2rx(t)vz(t))j + (2rx(t)vy(t) - 2ry(t)vx(t))k

take the derivative of it :

(2vy(t)vz(t) + 2ry(t)az(t) - 2vz(t)vy(t) - 2rz(t)ay(t))i + (2vz(t)vx(t) + 2rz(t)ax(t) - 2vx(t)vz(t) - 2rx(t)az(t))j + (2vx(t)vy(t) + 2rx(t)ay(t) -2vy(t)vx(t) -2ry(t)ax(t))k

=

(2ry(t)az(t) - 2rz(t)ay(t))i + (2rz(t)ax(t)- 2rx(t)az(t))j + (2rx(t)ay(t) - 2ry(t)ax(t))k

Answer 2

or in your notations :

(2ry(t)r''z(t) - 2rz(t)r''y(t))i + (2rz(t)r''x(t)- 2rx(t)r''z(t))j + (2rx(t)r''y(t) - 2ry(t)r''x(t))k

Answer 3

please see the attachment and tell me the solution.

Answer 4

The cross-product follows the chain rule; that is, let a(t) and b(t) be vector fields. Then

\[\frac{d \ }{dt} [ a(t) \times b(t) ] = \frac{da}{dt} \times b + a \times \frac{db}{dt}\]

Using this result, your problem should be straight forward.