1.2 Calculus of parametric curves (Page 3/6)

Calculus volume 3 Page 3 / 6

Then a Riemann sum for the area is

A_{n} = \sum_{i = 1}^{n} y (x ({\bar{t}}_{i})) (x (t_{i}) - x (t_{i - 1})) .

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

A_{n} = \sum_{i = 1}^{n} y (x ({\bar{t}}_{i})) (\frac{x (t_{i}) - x (t_{i - 1})}{t_{i} - t_{i - 1}}) (t_{i} - t_{i - 1}) = \sum_{i = 1}^{n} y (x ({\bar{t}}_{i})) (\frac{x (t_{i}) - x (t_{i - 1})}{Δ t}) Δ t .

Taking the limit as $n$ approaches infinity gives

A = lim_{n \to \infty} A_{n} = \int_{a}^{b} y (t) x^{'} (t) d t .

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 (t), y = y (t), a \leq t \leq b

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

A = \int_{a}^{b} y (t) x^{'} (t) d t .

Finding the area under a parametric curve

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

x (t) = t - sin t, y (t) = 1 - cos t, 0 \leq t \leq 2 π .

Using [link] , we have

\begin{matrix} A & = \int_{a}^{b} y (t) x^{'} (t) d t \\ = \int_{0}^{2 π} (1 - cos t) (1 - cos t) d t \\ = \int_{0}^{2 π} (1 - 2 cos t + {cos}^{2} t) d t \\ = \int_{0}^{2 π} (1 - 2 cos t + \frac{1 + cos 2 t}{2}) d t \\ = \int_{0}^{2 π} (\frac{3}{2} - 2 cos t + \frac{cos 2 t}{2}) d t \\ = {\frac{3 t}{2} - 2 sin t + \frac{sin 2 t}{4} |}_{0}^{2 π} \\ = 3 π . \end{matrix}

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

x (t) = 3 cos t + cos 3 t, y (t) = 3 sin t - sin 3 t, 0 \leq t \leq π .

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

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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.

A curved line in the first quadrant with points marked for x = 1, 2, 3, 4, and 5. These points have values roughly 2.1, 2.7, 3, 2.7, and 2.1, respectively. The points for x = 1 and 5 are marked A and B, respectively. — Approximation of a curve by line segments.

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

\begin{matrix} d_{1} = \sqrt{{(x (t_{1}) - x (t_{0}))}^{2} + {(y (t_{1}) - y (t_{0}))}^{2}} \\ d_{2} = \sqrt{{(x (t_{2}) - x (t_{1}))}^{2} + {(y (t_{2}) - y (t_{1}))}^{2}} etc . \end{matrix}

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{{(x (t_{k}) - x (t_{k - 1}))}^{2} + {(y (t_{k}) - y (t_{k - 1}))}^{2}} .

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

\begin{matrix} x (t_{k}) - x (t_{k - 1}) = x^{'} ({\hat{t}}_{k}) (t_{k} - t_{k - 1}) = x^{'} ({\hat{t}}_{k}) Δ t \\ y (t_{k}) - y (t_{k - 1}) = y^{'} ({\tilde{t}}_{k}) (t_{k} - t_{k - 1}) = y^{'} ({\tilde{t}}_{k}) Δ t . \end{matrix}

Therefore [link] becomes

\begin{matrix} s & \approx \sum_{k = 1}^{n} s_{k} \\ = \sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}) Δ t)}^{2} + {(y^{'} ({\tilde{t}}_{k}) Δ t)}^{2}} \\ = \sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}))}^{2} {(Δ t)}^{2} + {(y^{'} ({\tilde{t}}_{k}))}^{2} {(Δ t)}^{2}} \\ = (\sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}))}^{2} + {(y^{'} ({\tilde{t}}_{k}))}^{2}}) Δ t . \end{matrix}

This is a Riemann sum that approximates the arc length over a partition of the interval $[a, b] .$ 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{matrix} s & = lim_{n \to \infty} \sum_{k = 1}^{n} s_{k} \\ = lim_{n \to \infty} (\sum_{k = 1}^{n} \sqrt{{(x^{'} ({\hat{t}}_{k}))}^{2} + {(y^{'} ({\tilde{t}}_{k}))}^{2}}) Δ t \\ = \int_{a}^{b} \sqrt{{(x^{'} (t))}^{2} + {(y^{'} (t))}^{2}} d t . \end{matrix}

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

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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