# Math Insight

### Derivatives of parameterized curves

The derivative of a vector-valued function is the matrix of partial derivatives. For vector-valued functions of a single variable, $\dllp: \R \to \R^n$ (confused?), this matrix is an $n \times 1$ column matrix, which each component is the derivative of a component of $\dllp$ with respect to that single variable. Since the matrix has only one column, we could view it as a vector. For example, if $\dllp(t)$ had three components, we could write the derivative as \begin{align*} D\dllp(t) = \left[ \begin{array}{c} \diff{\dllpc_1}{t}(t)\\ \diff{\dllpc_2}{t}(t)\\ \diff{\dllpc_3}{t}(t) \end{array} \right] = (\dllpc_1'(t),\dllpc_2'(t),\dllpc_3'(t)) =\dllp'(t). \end{align*}

We could also have equivalently defined the derivative using the limit definition that we use in one-variable. This definition will be useful for obtaining a geometric interpretation of the derivative as a derivative of the curve parametrized by the vector-valued function.

The limit definition of the derivative is \begin{align*} \dllp'(t) = \lim_{h \rightarrow 0} \frac{\dllp(t+h) - \dllp(t)}{h}. \end{align*} In one-variable calculus, the derivative was the slope of the graph. Is this true for parametrized curves? In this case, the derivative is a vector, so it can't just be the slope (which is a scalar). Instead, the derivative $\dllp'(t)$ is the tangent vector of the curve traced by $\dllp(t)$. In this way, the direction of the derivative $\dllp'(t)$ specifies the slope of the curve traced by $\dllp(t)$. The length (or magnitude) of the derivative $\|\dllp'(t)\|$ specifies how fast $\dllp(t)$ traces out the curve as you change $t$.

The below applet illustrates why the derivative defined by the limit defintion is tangent to the curve. We could denote the estimate of the derivative shown by the red arrow as \begin{align*} \vc{g}_h(t) = \frac{\dllp(t+h) - \dllp(t)}{h}. \end{align*} Note that $\vc{g}_h(t)$ is defined so that \begin{align*} \lim_{h \to 0} \ \vc{g}_h(t) = \dllp'(t). \end{align*}

The derivative of a parametrized curve. The function $\dllp(t)=(3\cos t)\vc{i} + (2 \sin t) \vc{j}$ parametrizes an ellipse. For given values of $t$ and $h$ (changeable by the sliders), the blue vector is $\dllp(t)$ and the green vector is $\dllp(t+h)$. The red vector is an estimate for the derivative: $\dllp'(t) \approx (\dllp(t+h) - \dllp(t))/h$. As $h$ decreases to zero, this estimate approaches the value of the derivative $\dllp'(t)$, which is tangent to the ellipse.

For any value of $t$, you can see that when $h=1$, then $\vc{g}_h(t) = \dllp(t+h) - \dllp(t)$ (since the red vector joins the heads of the blue and green vectors). As you move $h$ toward zero, the red vector approaches the tangent of the ellipse. When $h$ is zero, the blue and green vectors are identical. In this limit, the red vector is the derivative $\dllp'(t)$ and is tangent to the ellipse.

We can use the fact that $\dllp'(t)$ is tangent to the curve $\dllp(t)$ to calculate the equation for the tangent line to a curve.