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. $$ D_{t}[f(t) \mathbf{r}(t)]=f(t) \mathbf{r}^{\prime}(t)+f^{\prime}(t) \mathbf{r}(t) $$

Short answer

The derivative of the product of a scalar function \(f(t)\) and a vector-valued function \(\mathbf{r}(t)\) is verified as the sum of the original function multiplied by the derivative of the vector function and then added to the derivative of the original function multiplied by the vector function itself.

Step-by-step solution

Understanding the Derivative of a Function

In order to solve this exercise, use the definition of derivative for a vector function: \( D_t f(t) = f'(t) \), for a scalar function \( f(t) \), and \( D_t \mathbf{r}(t) = \mathbf{r}'(t) \), for a vector function \( \mathbf{r}(t) \).

Applying Chain Rule

The chain rule for differentiation of the product of functions is given by \( (u(v(t))' = u'(v(t)) * v'(t) \) where \( u \) and \( v \) are the functions and \( v(t) \) is a differentiable real-valued function. This is analogous to a scalar function. In this step, apply the chain rule to the problem \( f(t) \mathbf{r}(t) \), treating \( f(t) \) as the scalar function, and \( \mathbf{r}(t) \) as the vector function.

Differentiation of Vector-Valued Functions

The derivative of the product of a scalar and a vector function is computed as: \( D_t[f(t) \mathbf{r}(t)]=f(t) \mathbf{r}'(t)+f'(t) \mathbf{r}(t) \). Now, replacing \( f(t) \) with \( f \), and \( \mathbf{r}(t) \) with \( \mathbf{r} \), the derivative yields \( D_t[f(t) \mathbf{r}(t)]=f \mathbf{r}'+f' \mathbf{r} \). Hence, this verifies the given property.