Calculus with Parametric Curves¶
We now turn our focus to performing calculus along parametric curves.
Tangent Lines¶
While parametric curves might not be the graphs of functions, they can still have well-defined tangent lines at given points. The slope of the tangent line can be computed via the chain rule.
Suppose $x = x(t)$ and $y = y(t)$ are differentiable functions, and suppose $\tfrac{dx}{dt} \neq 0$ at the point we're interested in. Then
$$
\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt},
$$
by the chain rule, and solving for $\tfrac{dy}{dx}$ we obtain
$$
\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.
$$
This notation is a bit cumbersome. To simplify it, we introduce new notation for a derivative with respect to a parameter.
Notation. Suppose $f = f(t)$ is a function of a parameter $t$. Then we define
$$
\dot{f}(t) = \dfrac{d}{dt}\big[f(t)\big] = \frac{df}{dt}.
$$
This notation is known as "Newton's notation," or simply "dot-notation." It is used frequently in Physics to represent derivatives with respect to time.
For us, this dot-notation for derivative will always indicate the derivative with respect to a parameter. In Calculus III, we'll set even more restrictions for when we can use the dot-notation.
Derivative notations [optional]
In dot-notation, the slope of the tangent line to a parametric curve can be written as
$$
\frac{dy}{dx} = \frac{\dot y}{\dot x},
$$
provided $\dot x \neq 0$.
Concavity¶
As we know from Calculus I, the concavity of a curve is at a point is given by its second derivative,
$$
\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right].
$$
We derive a formula for $\tfrac{d^2y}{dx^2}$ for parametrized curves.
Video lecture
Formula for $\tfrac{d^2y}{dx^2}$
$$\frac{d^2y}{dx^2} = \dfrac{\dot{x}\ddot{y} - \ddot{x}\dot{y}}{\dot{x}^3}$$
1. Example¶
Find $\tfrac{dy}{dx}$ and $\tfrac{d^2y}{dx^2}$ for the cycloid,
$$
\begin{cases}
x = r(\theta - \sin\theta), \\[0.5 ex] y = r(1 - \cos\theta).
\end{cases}
$$
Check your answer
$$
\frac{dy}{dx} = \dfrac{\sin\theta}{1 - \cos\theta},
$$
$$
\frac{d^2y}{dx^2} = \dfrac{-1}{r(1-\cos\theta)^2}.
$$
Video solution
Error Alert! I lost a minus sign during my computation in this video. When you see it happen, yell at me and tell me I'm wrong! But mostly, just make sure you do it correctly.
Areas¶
Next, we discuss how to compute the area under a parametric curve; that is, the area ofthe region bounded between the curve and the $x$-axis.
Video lecture
2. Example¶
Compute the area enclosed under one cycle of the cycloid,
$$
\begin{cases}
x = r(\theta - \sin\theta), \\[0.5 ex] y = r(1 - \cos\theta).
\end{cases}
$$
Check your answer
$$3\pi r^2$$
Video solution
Arc Length¶
Recall that we derived the arc length element $ds$ in Chapter 8.
$$
ds = \sqrt{dx^2 + dy^2}
$$
Suppose $x = x(t)$ and $y = y(t)$ are parametric functions. Using our dot-notation and the chain rule, we have
$$
\begin{cases}
dx = \dot x\, dt, \\[0.5 ex]
dy = \dot y\, dt.
\end{cases}
$$
The arc length element for a parametrized curve is then given by
$$
ds = \sqrt{\dot{x}^2 + \dot{y}^2}\ dt.
$$
Implicit Exercise. Fill in the details here.
3. Example¶
Compute the arc length of one cycle of the cycloid,
$$
\begin{cases}
x = r(\theta - \sin\theta), \\[0.5 ex] y = r(1 - \cos\theta).
\end{cases}
$$
Check your answer
$$8r$$
Video solution
Surface Area¶
Suppose $\mathbf{r}(t) = \left\langle x(t),y(t) \right\rangle$ is a parametrized curve that passes the vertical line test on an interval $a \leq t \leq b$.
In this context, the vertical line test can be stated as follows: If $x(t_1) = x(t_2)$, then $t_1 = t_2$.
Now, consider the surface obtained by rotating the portion of the curve $\mathbf{r}(t) = \left\langle x(t),y(t) \right\rangle$, $a \leq t \leq b$, about the $x$-axis. The surface area of this surface is given by
$$
\begin{array}{}
S & = & \displaystyle \int_a^b 2\pi y\, ds \\[0.5 ex]
& = & \displaystyle \int_a^b 2\pi y(t)\, \sqrt{\dot{x}^2(t) + \dot{y}^2(t)}\, dt
\end{array}
$$
4. Exercise¶
Write out an integral that represents the surface area of the surface obtained by rotating the parametric curve $\mathbf{r}(t) = \left\langle x(t), y(t) \right\rangle$, $a \leq t \leq b$, about the $y$-axis.
Solution
$$S = \displaystyle \int_a^b 2\pi x(t)\, \sqrt{\dot{x}^2(t) + \dot{y}^2(t)}\, dt$$
5. Example¶
Use a parametrization of a circle of radius $r$ to compute the surface area of a sphere of radius $r$.
Check your answer
$$4\pi r^2$$
Video solution
Discussion¶
Questions? You can ask them below.