Question

If \(u(t) = r(t) \times \left( {r'(t) \times r''(t)} \right)\), show that \(u'(t) = r(t) \times \left( {r'(t) \times r'''(t)} \right)\).

Short answer

Hence the required result \(u'(t) = r(t) \times \left( {r'(t) \times r'''(t)} \right)\)is proved.

Step-by-step solution

Step 1: Given and differentiate both sides

We have given that \(u(t) = r(t) \times \left( {r'(t) \times r''(t)} \right)\)and we have to show that,

\(u'(t) = r(t) \times \left( {r'(t) \times r'''(t)} \right)\)

Since \(u(t) = r(t) \times \left( {r'(t) \times r''(t)} \right)\)

Now differentiate u(t) by applying product rule on terms r(t) and \(r'(t) \times r''(t)\)

This implies that,

\(u'(t) = r'(t) \cdot \left( {r'(t) \times r''(t)} \right) + r(t) \cdot \left( {r'(t) \times r''(t)} \right)'\)

Again we have to find \(\left( {r'(t) \times r''(t)} \right)'\) using the product rule on terms r’(t) and r’’(t), we get

\(u'(t) = r'(t) \cdot \left( {r'(t) \times r''(t)} \right) + r(t) \cdot \left( {r''(t) \times r''(t)} \right) + r(t) \cdot (r'(t) \times r'''(t))\)

Step 2: Further Simplification

Now since there is a repeated unit r’(t), therefore, the triple product should be zero. Also, since the cross product of any vector with itself is zero, therefore, we have,

\(r'(t) \cdot \left( {r'(t) \times r''(t)} \right) = 0\)\(\)

And

\(r''(t) \times r''(t) = 0\)

By applying the above two equations in equation (*), we get,

\(u'(t) = 0 + r(t) \cdot 0 + r(t) \cdot (r'(t) \times r''(t))\)

This implies that, \(u'(t) = r(t) \times \left( {r'(t) \times r'''(t)} \right)\)

Hence the required result is being proved.