Parametrized curves and their derivatives
If is a vector-valued function in two dimensions (i.e., ), then parametrizes a curve in the plane. (This is similar to how we parametrized a line.) For example, the function parametrizes an ellipse, as illustrated in the following demo. I’ve drawn as a blue arrow. As you change from 0 to (by dragging the blue point on the slider), the head of the arrow traces out the ellipse.
Can you show mathematically that traces out an ellipse? To do so, let be the point defined by for some value of . Since , we conclude that and . Can you see that and satisfy
which is the equation for the above ellipse? (You remember that for any . You can calculate expresions and , and add them together.)
The derivative of a parametrized curve
The definition of the derivative of a parametrized curve is analogous to what you learned in one-variable calculus:
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 is the tangent vector of the curve traced by . In this way, the direction of the derivative specifies the slope of the curve traced by . The length (or magnitude) of the derivative specifies how fast traces out the curve as you change .
Let’s see why the derivative defined by equation (1) is tangent to the curve. Below, I’ve plotted in blue and in green. In red, I’ve plotted the estimate of the derivative for a given , which we’ll denote by
Note that is defined so that
For any value of , you can see that when , then (since the red vector joins the heads of the blue and green vectors). As you move toward zero (by moving the second slider), the red vector approaches the tangent of the ellipse. When is zero, the blue and green vectors are identical. In this limit, the red vector is the derivative and is tangent to the ellipse.
We can use the fact that is tangent to the curve to calculate the equation for the tangent line to a curve.