Question

In Exercises 59-66, 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. $$ D_{t}[c \mathbf{r}(t)]=c \mathbf{r}^{\prime}(t) $$

Short answer

The property \(D_{t}[c \mathbf{r}(t)]=c \mathbf{r}^{\prime}(t)\) holds true, which is a reflection of the fact that the derivative of a scalar multiple of a vector-valued function equals the scalar multiple of the derivative of that function. The proof is based on the fundamental rule of differentiation regarding the derivative of a scalar multiple of a function.

Step-by-step solution

Understand the scalar multiplication

The given property is \(D_{t}[c \mathbf{r}(t)]=c \mathbf{r}^{\prime}(t)\). Here, \(c\) is a scalar and \(\mathbf{r}(t)\) is a differentiable vector-valued function. Note that \(c \mathbf{r}(t)\) represents the scalar multiplication of a vector. The action of scalar multiplying a vector is to scale (change length of) the vector, without changing the direction. Also, recall that the derivative of a scalar multiple of a function equals the scalar multiple of the derivative of the function.

Apply the rule for the derivative of a scalar multiple of a function

According to the rule for obtaining the derivative of a scalar multiple, we differentiate the scalar function, keeping the vector function constant and then multiply by the vector function. Here the scalar function is \(c\) (which is a constant), and its derivative is 0. So, \(D_{t}(c) = 0\). Now we differentiate the vector function \(\mathbf{r}(t)\), denoted by \(\mathbf{r}^{\prime}(t)\). According to the rule stated, \(D_{t}[c \mathbf{r}(t)] = c * \mathbf{r}^{\prime}(t)\).

Conclusion from the proven rule

From the steps described above, we can see that by applying the rule of scalar multiplication in derivative, that the derivative of the scalar multiple of a vector-valued function equals to the scalar multiple of the derivative of the vector-valued function. Therefore, the given property \(D_{t}[c \mathbf{r}(t)]=c \mathbf{r}^{\prime}(t)\) is proven.