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Can someone please check my work below to confirm whether or not I got the correct answer? This is question 13.2.16 in the 7th edition of Stewart Calculus.
Find the derivative of the vector function: $\vec{r}(t)=t\vec{a}\times(\vec{b}+t\vec{c})$

Use: $\frac{d}{dt}[\vec{u}(t)\times\vec{v}(t)]=\vec{u'}(t)\times\vec{v}(t)+\vec{u}(t)\times\vec{v'}(t)$

$\vec{u}(t)=t\vec{a}$
$\vec{v}(t)=(\vec{b}+t\vec{c})$
$\vec{u'}(t)=\vec{a}$
$\vec{v'}(t)=\vec{c}$

Thus, $\vec{r'}(t)=\vec{a}\times(\vec{b}+t\vec{c})+t\vec{a}\times\vec{c}$

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up vote 0 down vote accepted

Yes, your result is correct, provided $\bf a$, $\bf b$, and $\bf c$ do not depend on time. An equivalent (simpler?) way to do this is: $$ {\bf r} = t \ {\bf a} \times {\bf b} + t^2 \ {\bf a} \times {\bf c} \ \ \ \Rightarrow \ \ \ \frac{d {\bf r}}{dt} = {\bf a} \times {\bf b} + 2t \ {\bf a} \times {\bf c} $$

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