Question

Find the derivative of the vector function

r ( t ) = atcos3ti + bsin³tj +ccos³ tk

Short answer

The derivative of the vector function is

\[r'(t) = ( - 3at\sin (3t) + t\cos 3t,3b{\sin ^2}t\cos t + {\sin ^3}t - 3t{\cos ^2}t\sin t + \cos 2t)\]

Step-by-step solution

Step 1: Applied the derivative function

Use the formula for derivative

\[\frac{d}{{dx}} = [f(x)g(x)] = f(x)g'(x) + f'(x)g(x)\]

Step 2: Solution of the vector function

Let’s applied the formula

\[\begin{array}{l}\frac{{d(at\cos 3t)}}{{dt}} = at(\cos 3t)' + (at)'(\cos 3t)\\ = at( - 3\sin (3t)) + t(\cos 3t)\\ = - 3at\sin (3t) + t\cos 3t\\\frac{{d(b{{\sin }^3}t)}}{{dt}} = b({\sin ^3}t)' + (b)'({\sin ^3}t)\\ = 3b{\sin ^2}t\cos t + {\sin ^3}t\\\frac{{d(c{{\cos }^3}t)}}{{dt}} = t(\cos 2t)' + (t)'\cos 2t\\ = - 3t{\cos ^2}t\sin t + \cos 2t\\r'(t) = ( - 3at\sin (3t) + t\cos 3t,3b{\sin ^2}t\cos t + {\sin ^3}t - 3t{\cos ^2}t\sin t + \cos 2t)\end{array}\]