Question

Prove the property. In each case, assume that \(\mathbf{r}, \mathbf{u},\) and \(\mathbf{v}\) are differentiable vector-valued functions of \(t,\) \(f\) is a differentiable real-valued function of \(t,\) and \(c\) is a scalar. $$ D_{t}\left[\mathbf{r}(t) \times \mathbf{r}^{\prime}(t)\right]=\mathbf{r}(t) \times \mathbf{r}^{\prime \prime}(t) $$

Short answer

The derivative of the cross product of a vector-valued function \(\mathbf{r}(t)\) with its first derivative \(\mathbf{r}^{\prime}(t)\) equals the cross product of \(\mathbf{r}(t)\) with its second derivative \(\mathbf{r}^{\prime \prime}(t)\). The property is proved by applying the product rule, replacing the derivatives, then using the property that the cross product of a vector with itself is a zero vector.

Step-by-step solution

Write given equation and apply product rule

We have to prove the equation \(D_{t}\left[\mathbf{r}(t) \times \mathbf{r}^{\prime}(t)\right]=\mathbf{r}(t) \times \mathbf{r}^{\prime \prime}(t)\). We start by applying the product rule to the left hand side of the given equation: \(D_{t}\left[\mathbf{r}(t) \times \mathbf{r}^{\prime}(t)\right] = D_{t} \mathbf{r}(t) \times \mathbf{r}^{\prime}(t) + \mathbf{r}(t) \times D_{t} \mathbf{r}^{\prime}(t)\).

Substitute the derivative of a function

We can substitute \(D_{t} \mathbf{r}(t)\) with \(\mathbf{r}'(t)\) and \(D_{t} \mathbf{r}^{\prime}(t)\) with \(\mathbf{r}''(t)\). This will change the equation to \(\mathbf{r}'(t) \times \mathbf{r}^{\prime}(t) + \mathbf{r}(t) \times \mathbf{r}''(t)\).

Use properties of cross product

The cross product of a vector with itself is a zero vector. Therefore, \(\mathbf{r}'(t) \times \mathbf{r}^{\prime}(t)\) equals to the zero vector and we are left with \(\mathbf{r}(t) \times \mathbf{r}''(t)\) which is the right hand side of the original equation.