# Section9.3Calculus and Parametric Equations¶ permalink

The previous section defined curves based on parametric equations. In this section we'll employ the techniques of calculus to study these curves.

We are still interested in lines tangent to points on a curve. They describe how the $y$-values are changing with respect to the $x$-values, they are useful in making approximations, and they indicate instantaneous direction of travel.

The slope of the tangent line is still $\frac{dy}{dx}\text{,}$ and the Chain Rule allows us to calculate this in the context of parametric equations. If $x=f(t)$ and $y=g(t)\text{,}$ the Chain Rule states that \begin{equation*} \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}. \end{equation*}

Solving for $\frac{dy}{dx}\text{,}$ we get \begin{equation*} \frac{dy}{dx} = \frac{dy}{dt}\Bigg/\frac{dx}{dt} = \frac{g'(t)}{\fp(t)}, \end{equation*} provided that $\fp(t)\neq 0\text{.}$ This is important so we label it a Key Idea.

##### Key Idea9.3.1Finding $\frac{dy}{dx}$ with Parametric Equations.

Let $x=f(t)$ and $y=g(t)\text{,}$ where $f$ and $g$ are differentiable on some open interval $I$ and $\fp(t)\neq 0$ on $I\text{.}$ Then \begin{equation*} \frac{dy}{dx} = \frac{g'(t)}{\fp(t)}. \end{equation*}

We use this to define the tangent line.

##### Definition9.3.2Tangent and Normal Lines

Let a curve $C$ be parametrized by $x=f(t)$ and $y=g(t)\text{,}$ where $f$ and $g$ are differentiable functions on some interval $I$ containing $t=t_0\text{.}$ The tangent line to $C$ at $t=t_0$ is the line through $\big(f(t_0),g(t_0)\big)$ with slope $m=g'(t_0)/\fp(t_0)\text{,}$ provided $\fp(t_0)\neq 0\text{.}$

The normal line to $C$ at $t=t_0$ is the line through $\big(f(t_0),g(t_0)\big)$ with slope $m=-\fp(t_0)/g'(t_0)\text{,}$ provided $g'(t_0)\neq 0\text{.}$

The definition leaves two special cases to consider. When the tangent line is horizontal, the normal line is undefined by the above definition as $g'(t_0)=0\text{.}$ Likewise, when the normal line is horizontal, the tangent line is undefined. It seems reasonable that these lines be defined (one can draw a line tangent to the “right side” of a circle, for instance), so we add the following to the above definition.

1. If the tangent line at $t=t_0$ has a slope of 0, the normal line to $C$ at $t=t_0$ is the line $x=f(t_0)\text{.}$

2. If the normal line at $t=t_0$ has a slope of 0, the tangent line to $C$ at $t=t_0$ is the line $x=f(t_0)\text{.}$

##### Example9.3.3Tangent and Normal Lines to Curves

Let $x=5t^2-6t+4$ and $y=t^2+6t-1\text{,}$ and let $C$ be the curve defined by these equations.

1. Find the equations of the tangent and normal lines to $C$ at $t=3\text{.}$

2. Find where $C$ has vertical and horizontal tangent lines.

Solution
##### Example9.3.5Tangent and Normal Lines to a Circle
1. Find where the unit circle, defined by $x=\cos(t)$ and $y=\sin(t)$ on $[0,2\pi]\text{,}$ has vertical and horizontal tangent lines.

2. Find the equation of the normal line at $t=t_0\text{.}$

Solution
##### Example9.3.7Tangent lines when $\frac{dy}{dx}$ is not defined

Find the equation of the tangent line to the astroid $x=\cos^3(t)\text{,}$ $y=\sin^3(t)$ at $t=0\text{,}$ shown in Figure 9.3.8.

Solution

# Subsection9.3.1Concavity

We continue to analyze curves in the plane by considering their concavity; that is, we are interested in $\frac{d^2y}{dx^2}\text{,}$ “the second derivative of $y$ with respect to $x\text{.}$” To find this, we need to find the derivative of $\frac{dy}{dx}$ with respect to $x\text{;}$ that is, \begin{equation*} \frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{dy}{dx}\right], \end{equation*} but recall that $\frac{dy}{dx}$ is a function of $t\text{,}$ not $x\text{,}$ making this computation not straightforward.

To make the upcoming notation a bit simpler, let $h(t) = \frac{dy}{dx}\text{.}$ We want $\frac{d}{dx}[h(t)]\text{;}$ that is, we want $\frac{dh}{dx}\text{.}$ We again appeal to the Chain Rule. Note: \begin{equation*} \frac{dh}{dt} = \frac{dh}{dx}\cdot\frac{dx}{dt} \Rightarrow \frac{dh}{dx} = \frac{dh}{dt}\Bigg/\frac{dx}{dt}. \end{equation*}

In words, to find $\frac{d^2y}{dx^2}\text{,}$ we first take the derivative of $\frac{dy}{dx}$ with respect to $t$, then divide by $x'(t)\text{.}$ We restate this as a Key Idea.

##### Key Idea9.3.9Finding $\frac{d^2y}{dx^2}$ with Parametric Equations

Let $x=f(t)$ and $y=g(t)$ be twice differentiable functions on an open interval $I\text{,}$ where $\fp(t)\neq 0$ on $I\text{.}$ Then \begin{equation*} \frac{d^2y}{dx^2} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\frac{dx}{dt} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\fp(t). \end{equation*}

Examples will help us understand this Key Idea.

##### Example9.3.10Concavity of Plane Curves

Let $x=5t^2-6t+4$ and $y=t^2+6t-1$ as in Example 9.3.3. Determine the $t$-intervals on which the graph is concave up/down.

Solution
##### Example9.3.12Concavity of Plane Curves

Find the points of inflection of the graph of the parametric equations $x=\sqrt{t}\text{,}$ $y=\sin(t)\text{,}$ for $0\leq t\leq 16\text{.}$

Solution

# Subsection9.3.2Arc Length

We continue our study of the features of the graphs of parametric equations by computing their arc length.

Recall in Section 7.4 we found the arc length of the graph of a function, from $x=a$ to $x=b\text{,}$ to be \begin{equation*} L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx. \end{equation*}

We can use this equation and convert it to the parametric equation context. Letting $x=f(t)$ and $y=g(t)\text{,}$ we know that $\frac{dy}{dx} = g'(t)/\fp(t)\text{.}$ It will also be useful to calculate the differential of $x\text{:}$ \begin{equation*} dx = \fp(t)dt \qquad \Rightarrow \qquad dt = \frac{1}{\fp(t)}\cdot dx. \end{equation*}

Starting with the arc length formula above, consider: \begin{align*} L \amp = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx\\ \amp = \int_a^b \sqrt{1+\frac{g'(t)^2}{\fp(t)^2}}\ dx.\\ \end{align*} Factor out the $\fp(t)^2\text{:}$ \begin{align*} \amp = \int_a^b \sqrt{\fp(t)^2+g'(t)^2}\cdot\underbrace{\frac1{\fp(t)}\ dx}_{=dt}\\ \amp = \int_{t_1}^{t_2} \sqrt{\fp(t)^2+g'(t)^2}\ dt. \end{align*}

Note the new bounds (no longer “$x$” bounds, but “$t$” bounds). They are found by finding $t_1$ and $t_2$ such that $a= f(t_1)$ and $b=f(t_2)\text{.}$ This formula is important, so we restate it as a theorem.

As before, these integrals are often not easy to compute. We start with a simple example, then give another where we approximate the solution.

##### Example9.3.17Arc Length of a Circle

Find the arc length of the circle parametrized by $x=3\cos(t)\text{,}$ $y=3\sin(t)$ on $[0,3\pi/2]\text{.}$

Solution
##### Example9.3.18Arc Length of a Parametric Curve

The graph of the parametric equations $x=t(t^2-1)\text{,}$ $y=t^2-1$ crosses itself as shown in Figure 9.3.19, forming a “teardrop.” Find the arc length of the teardrop.

Solution

# Subsection9.3.3Surface Area of a Solid of Revolution

Related to the formula for finding arc length is the formula for finding surface area. We can adapt the formula found in Key Idea 7.4.13 from Section 7.4 in a similar way as done to produce the formula for arc length done before.

##### Key Idea9.3.20Surface Area of a Solid of Revolution

Consider the graph of the parametric equations $x=f(t)$ and $y=g(t)\text{,}$ where $\fp$ and $g'$ are continuous on an open interval $I$ containing $t_1$ and $t_2$ on which the graph does not cross itself.

1. The surface area of the solid formed by revolving the graph about the $x$-axis is (where $g(t)\geq~0$ on $[t_1,t_2]$): \begin{equation*} \text{ Surface Area } = 2\pi\int_{t_1}^{t_2} g(t)\sqrt{\fp(t)^2+g'(t)^2}\ dt. \end{equation*}

2. The surface area of the solid formed by revolving the graph about the $y$-axis is (where $f(t)\geq~0$ on $[t_1,t_2]$): \begin{equation*} \text{ Surface Area } = 2\pi\int_{t_1}^{t_2} f(t)\sqrt{\fp(t)^2+g'(t)^2}\ dt. \end{equation*}

##### Example9.3.21Surface Area of a Solid of Revolution

Consider the teardrop shape formed by the parametric equations $x=t(t^2-1)\text{,}$ $y=t^2-1$ as seen in Example 9.3.18. Find the surface area if this shape is rotated about the $x$-axis, as shown in Figure 9.3.22.

Solution

After defining a new way of creating curves in the plane, in this section we have applied calculus techniques to the parametric equation defining these curves to study their properties. In the next section, we define another way of forming curves in the plane. To do so, we create a new coordinate system, called polar coordinates, that identifies points in the plane in a manner different than from measuring distances from the $y$- and $x$- axes.

# Subsection9.3.4Exercises

Terms and Concepts

In the following exercises, parametric equations for a curve are given.

1. Find $\ds\frac{dy}{dx}\text{.}$

2. Find the equations of the tangent and normal line(s) at the point(s) given.

3. Sketch the graph of the parametric functions along with the found tangent and normal lines.

In the following exercises, find $t$-values where the curve defined by the given parametric equations has a horizontal tangent line. Note: these are the same equations as in Exercises 9.3.4.59.3.4.12.

In the following exercises, find $t=t_0$ where the graph of the given parametric equations is not smooth, then find $\lim\limits_{t\to t_0}\frac{dy}{dx}\text{.}$

In the following exercises, parametric equations for a curve are given. Find $\frac{d^2y}{dx^2}\text{,}$ then determine the intervals on which the graph of the curve is concave up/down. Note: these are the same equations as in Exercises 9.3.4.59.3.4.12.

In the following exercises, find the arc length of the graph of the parametric equations on the given interval(s).

In the following exercises, numerically approximate the given arc length.

In the following exercises, a solid of revolution is described. Find or approximate its surface area as specified.