Question

Let $f\left(t\right)$be a differentiable scalar function and $r\left(t\right)$be a differentiable vector function. Prove that $\frac{d}{dt}\left(f\right(t\left)r\right(t\left)\right)=f\text{'}\left(t\right)r\left(t\right)+f\left(t\right)r\text{'}\left(t\right)$. (This is Theorem 11.11 (b).)

Short answer

Ans:

$$\begin{gathered} \frac{d}{dt}\left(f\right(t\left)r\right(t\left)\right)=\frac{d}{dt}\left(f\right(t)⟨x(t),y(t),z(t)⟩) \\ =f^{\text{'}}\left(t\right)r\left(t\right)+f\left(t\right)r^{\text{'}}\left(t\right) \end{gathered}$$

Step-by-step solution

Step 1. Given information: 

$f\left(t\right)$is differentiable scalar function and

$r\left(t\right)$is differentiable vector function.

Step 2. Proving :

Consider

$\frac{d}{dt}\left(f\right(t\left)r\right(t\left)\right)=\frac{d}{dt}\left(f\right(t)⟨x(t),y(t),z(t)⟩)$

$=\frac{d}{dt}(⟨f(t\left)x\right(t),f(t\left)y\right(t),f(t\left)z\right(t)⟩)$

$=\frac{d}{dt}\left(f\right(t\left)x\right(t\left)\right)i+\frac{d}{dt}\left(f\right(t\left)y\right(t\left)\right)j+\frac{d}{dt}\left(f\right(t\left)z\right(t\left)\right)k$

$=\left(f^{\text{'}}\left(t\right)x\left(t\right)+f\left(t\right)x^{\text{'}}\left(t\right)\right)i+\left(f^{\text{'}}\left(t\right)y\left(t\right)+f\left(t\right)y^{\text{'}}\left(t\right)\right)j+$$\left(f^{\text{'}}\left(t\right)z\left(t\right)+f\left(t\right)z^{\text{'}}\left(t\right)\right)k$

$=\left(f^{\text{'}}\left(t\right)x\left(t\right)i+f^{\text{'}}\left(t\right)y\left(t\right)j+f^{\text{'}}\left(t\right)z\left(t\right)k\right)+\left(\begin{array}{l}f\left(t\right)x^{\text{'}}\left(t\right)i+f\left(t\right)y^{\text{'}}\left(t\right)j+\\ f\left(t\right)z^{\text{'}}\left(t\right)\end{array}\right)k$

$=f^{\text{'}}\left(t\right)\left(x\right(t)i+y(t)j+z(t\left)k\right)+f\left(t\right)\left(x^{\text{'}}\left(t\right)o+y^{\text{'}}\left(t\right)j+z^{\text{'}}\left(t\right)k\right)$

$=f^{\text{'}}\left(t\right)r\left(t\right)+f\left(t\right)r^{\text{'}}\left(t\right)$

Thus$\frac{d}{dt}\left(f\right(t\left)r\right(t\left)\right)=f^{\text{'}}\left(t\right)r\left(t\right)+f\left(t\right)r^{\text{'}}\left(t\right)$