# 7.2 Calculus of parametric curves  (Page 3/6)

 Page 3 / 6

Then a Riemann sum for the area is

${A}_{n}=\sum _{i=1}^{n}y\left(x\left({\stackrel{–}{t}}_{i}\right)\right)\phantom{\rule{0.2em}{0ex}}\left(x\left({t}_{i}\right)-x\left({t}_{i-1}\right)\right).$

Multiplying and dividing each area by ${t}_{i}-{t}_{i-1}$ gives

${A}_{n}=\sum _{i=1}^{n}y\left(x\left({\stackrel{–}{t}}_{i}\right)\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{x\left({t}_{i}\right)-x\left({t}_{i-1}\right)}{{t}_{i}-{t}_{i-1}}\right)\left({t}_{i}-{t}_{i-1}\right)=\sum _{i=1}^{n}y\left(x\left({\stackrel{–}{t}}_{i}\right)\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{x\left({t}_{i}\right)-x\left({t}_{i-1}\right)}{\text{Δ}t}\right)\text{Δ}t.$

Taking the limit as $n$ approaches infinity gives

$A=\underset{n\to \infty }{\text{lim}}{A}_{n}={\int }_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$

This leads to the following theorem.

## Area under a parametric curve

Consider the non-self-intersecting plane curve defined by the parametric equations

$x=x\left(t\right),\phantom{\rule{1em}{0ex}}y=y\left(t\right),\phantom{\rule{1em}{0ex}}a\le t\le b$

and assume that $x\left(t\right)$ is differentiable. The area under this curve is given by

$A={\int }_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$

## Finding the area under a parametric curve

Find the area under the curve of the cycloid defined by the equations

$x\left(t\right)=t-\text{sin}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=1-\text{cos}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}0\le t\le 2\pi .$

$\begin{array}{cc}\hfill A& ={\int }_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }\left(1-\text{cos}\phantom{\rule{0.2em}{0ex}}t\right)\left(1-\text{cos}\phantom{\rule{0.2em}{0ex}}t\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }\left(1-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t+{\text{cos}}^{2}t\right)dt\hfill \\ & ={\int }_{0}^{2\pi }\left(1-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t+\frac{1+\text{cos}\phantom{\rule{0.2em}{0ex}}2t}{2}\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }\left(\frac{3}{2}-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t+\frac{\text{cos}\phantom{\rule{0.2em}{0ex}}2t}{2}\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\frac{3t}{2}-2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t+\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}2t}{4}|}_{0}^{2\pi }\hfill \\ & =3\pi .\hfill \end{array}$

Find the area under the curve of the hypocycloid defined by the equations

$x\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t+\text{cos}\phantom{\rule{0.2em}{0ex}}3t,\phantom{\rule{1em}{0ex}}y\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t-\text{sin}\phantom{\rule{0.2em}{0ex}}3t,\phantom{\rule{1em}{0ex}}0\le t\le \pi .$

$A=3\pi$ (Note that the integral formula actually yields a negative answer. This is due to the fact that $x\left(t\right)$ is a decreasing function over the interval $\left[0,2\pi \right];$ that is, the curve is traced from right to left.)

## Arc length of a parametric curve

In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

Given a plane curve defined by the functions $x=x\left(t\right),y=y\left(t\right),a\le t\le b,$ we start by partitioning the interval $\left[a,b\right]$ into n equal subintervals: ${t}_{0}=a<{t}_{1}<{t}_{2}<\text{⋯}<{t}_{n}=b.$ The width of each subinterval is given by $\text{Δ}t=\left(b-a\right)\text{/}n.$ We can calculate the length of each line segment:

$\begin{array}{}\\ {d}_{1}=\sqrt{{\left(x\left({t}_{1}\right)-x\left({t}_{0}\right)\right)}^{2}+{\left(y\left({t}_{1}\right)-y\left({t}_{0}\right)\right)}^{2}}\hfill \\ {d}_{2}=\sqrt{{\left(x\left({t}_{2}\right)-x\left({t}_{1}\right)\right)}^{2}+{\left(y\left({t}_{2}\right)-y\left({t}_{1}\right)\right)}^{2}}\phantom{\rule{0.2em}{0ex}}\text{etc}.\hfill \end{array}$

Then add these up. We let s denote the exact arc length and ${s}_{n}$ denote the approximation by n line segments:

$s\approx \sum _{k=1}^{n}{s}_{k}=\sum _{k=1}^{n}\sqrt{{\left(x\left({t}_{k}\right)-x\left({t}_{k-1}\right)\right)}^{2}+{\left(y\left({t}_{k}\right)-y\left({t}_{k-1}\right)\right)}^{2}}.$

If we assume that $x\left(t\right)$ and $y\left(t\right)$ are differentiable functions of t, then the Mean Value Theorem ( Introduction to the Applications of Derivatives ) applies, so in each subinterval $\left[{t}_{k-1},{t}_{k}\right]$ there exist ${\stackrel{^}{t}}_{k}$ and ${\stackrel{˜}{t}}_{k}$ such that

$\begin{array}{}\\ x\left({t}_{k}\right)-x\left({t}_{k-1}\right)={x}^{\prime }\left({\stackrel{^}{t}}_{k}\right)\left({t}_{k}-{t}_{k-1}\right)={x}^{\prime }\left({\stackrel{^}{t}}_{k}\right)\text{Δ}t\hfill \\ y\left({t}_{k}\right)-y\left({t}_{k-1}\right)={y}^{\prime }\left({\stackrel{˜}{t}}_{k}\right)\left({t}_{k}-{t}_{k-1}\right)={y}^{\prime }\left({\stackrel{˜}{t}}_{k}\right)\text{Δ}t.\hfill \end{array}$

$\begin{array}{cc}\hfill s& \approx \sum _{k=1}^{n}{s}_{k}\hfill \\ & =\sum _{k=1}^{n}\sqrt{{\left({x}^{\prime }\left({\stackrel{^}{t}}_{k}\right)\text{Δ}t\right)}^{2}+{\left({y}^{\prime }\left({\stackrel{˜}{t}}_{k}\right)\text{Δ}t\right)}^{2}}\hfill \\ & =\sum _{k=1}^{n}\sqrt{{\left({x}^{\prime }\left({\stackrel{^}{t}}_{k}\right)\right)}^{2}{\left(\text{Δ}t\right)}^{2}+{\left({y}^{\prime }\left({\stackrel{˜}{t}}_{k}\right)\right)}^{2}{\left(\text{Δ}t\right)}^{2}}\hfill \\ & =\left(\sum _{k=1}^{n}\sqrt{{\left({x}^{\prime }\left({\stackrel{^}{t}}_{k}\right)\right)}^{2}+{\left({y}^{\prime }\left({\stackrel{˜}{t}}_{k}\right)\right)}^{2}}\right)\text{Δ}t.\hfill \end{array}$

This is a Riemann sum that approximates the arc length over a partition of the interval $\left[a,b\right].$ If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives

$\begin{array}{cc}\hfill s& =\underset{n\to \infty }{\text{lim}}\sum _{k=1}^{n}{s}_{k}\hfill \\ & =\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^{n}\sqrt{{\left({x}^{\prime }\left({\stackrel{^}{t}}_{k}\right)\right)}^{2}+{\left({y}^{\prime }\left({\stackrel{˜}{t}}_{k}\right)\right)}^{2}}\right)\text{Δ}t\hfill \\ & ={\int }_{a}^{b}\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt.\hfill \end{array}$

When taking the limit, the values of ${\stackrel{^}{t}}_{k}$ and ${\stackrel{˜}{t}}_{k}$ are both contained within the same ever-shrinking interval of width $\text{Δ}t,$ so they must converge to the same value.

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