7.2 Calculus of parametric curves  (Page 10/6)

Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve?

Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve $\left(x\left(t\right),y\left(t\right)\right),$ then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time.

Derivatives of parametric equations

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations

The graph of this curve appears in [link] . It is a line segment starting at $\left(\mathrm{-1},\mathrm{-10}\right)$ and ending at $(9,5).$

We can eliminate the parameter by first solving the equation $x(t)=2t+3$ for t :

Substituting this into $y(t),$ we obtain

The slope of this line is given by $\frac{dy}{dx}=\frac{3}{2}.$ Next we calculate $x^{\prime }\left(t\right)$ and $y^{\prime }\left(t\right).$ This gives $x^{\prime }\left(t\right)=2$ and $y^{\prime }\left(t\right)=3.$ Notice that $\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}=\frac{3}{2}.$ This is no coincidence, as outlined in the following theorem.

Proof

This theorem can be proven using the Chain Rule. In particular, assume that the parameter t can be eliminated, yielding a differentiable function $y=F(x).$ Then $y\left(t\right)=F\left(x\left(t\right)\right).$ Differentiating both sides of this equation using the Chain Rule yields

so

But $F^{\prime }\left(x\left(t\right)\right)=\frac{dy}{dx},$ which proves the theorem.

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[link] can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function $y=f\left(x\right)$ is any point $x=x_0$ such that either $f^{\prime }\left(x_0\right)=0$ or $f^{\prime }\left(x_0\right)$ does not exist. [link] gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function $y=f\left(x\right)$ or not.