Derivative of a Vector-valued Function¶

Let $\vec v$ be a vector-valued function. If we just try to apply the definition of derivative to it, we get $$ \vec v'(t) = \lim_{h \to 0} \frac{\vec v(t+h) - \vec v(t)}{h}. $$ This works, although the meanings of the operations involved change a bit. On the top, we are now subtracting vectors instead of numbers, and the result is a vector. When we divide by $h$, we divide a vector by a number, which results in a vector, so we end up taking a limit of a vector. Just like before, we call this the derivative of $v$, and we write $$ \frac{d}{dt} \vec v(t) = \vec v'(t), $$ although now we use $t$ instead of $x$ everywhere, so we have $\frac{d}{dt}$ instead of $\frac{d}{dx}$.

The physical meaning of this derivative is similar to the derivative of a real-valued function; if $\vec v(t)$ is the location of an object at time $t$, thought of as a point in space, then $\vec v'(t)$ is the velocity vector, indicating in which direction and how fast the object is moving. To see why, do the things that we did with real-valued functions, replacing the walked distance with a vector representing the location. It's very common to use the letter $\vec v$ for the velocity vector, and some other letter such as $\vec r$ for the location.

Taking derivatives of $\I$ and $\J$ parts separately¶

Let $\vec v(t) = (f(t), g(t))$. With the definition of derivative, we get $$ \begin{align} \vec v'(t) &= \lim_{h \to 0} \frac{\vec v(t+h) - \vec v(t)}{h} \\ &= \lim_{h \to 0} \frac{(f(t+h), g(t+h)) - (f(t), g(t))}{h}. \end{align} $$ Let's switch to using the $\I$ and $\J$ notation of vectors. Now we get $$ \begin{align} \vec v'(t) &= \lim_{h \to 0} \frac{( f(t+h)\I + g(t+h)\J ) - ( f(t)\I + g(t)\J )}{h} \\ &= \lim_{h \to 0} \frac{f(t+h)\I + g(t+h)\J - f(t)\I - g(t)\J}{h} \\ &= \lim_{h \to 0} \frac{(f(t+h) - f(t))\I + (g(t+h) - g(t))\J}{h} \\ &= \lim_{h \to 0} \left( \frac{f(t+h) - f(t)}{h}\I + \frac{g(t+h) - g(t)}{h}\J \right). \end{align} $$ By taking limits of $\I$ and $\J$ parts separately, we get $$ \vec v'(t) = \underbrace{\left( \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \right)}_{f'(t)}\I + \underbrace{\left( \lim_{h \to 0} \frac{g(t+h) - g(t)}{h} \right)}_{g'(t)}\J. $$ So, we can not only calculate any limit of a vector-valued function as two limits of real-valued functions, but also calculate any derivative of a vector-valued function as two derivatives of real-valued functions. Here the derivative $\vec v'(t)$ exists if and only if $f'(t)$ and $g'(t)$ exist; this follows from the "if and only if" in the result for taking limits of $\I$ and $\J$ parts separately.