Arc Length of a Plane Curve in Rectangular Coordinates

If a plane curve is given by the equation $y=y(x)$ and the derivative $y\prime (x)$ is continuous, then the length of an arc of this curve is expressed by the integral
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where $a$ and $b$ are the abscissas of the end-points of the given arc.

Example 1

Compute the length of the arc of the semicubical parabola $y^{2}=x^{3}$ between the points $(0,0)$ and $(4,8)$.

Solution.

The function $y(x)$ is defined for $x\geq 0.$ Since the given points lie in the first quadrant, $y=x^{\frac{3}{2}}.$ Hence,
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and
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Consequently,
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Example 2

Compute the arc length of the curve $y=\ln \cos x$ between the points with the abscissas $x=0,x=\pi /4.$

Solution.

Since MATH, then
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Hence,
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Example 3

Find the arc length of the curve MATH between the points with the ordinates $y=1$ and $y=2.$

Solution.

Here it is convenient to adopt $y$ as the independent variable; then
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and
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Hence,
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Example 4

Find the length of the astroid MATH

Solution

The astroid is symmetrical about the axes of coordinates and the bisectors of the coordinate angles. Therefore, it is sufficient to compute the arc length of the astroid between the bisector $y=x$ and the $x$-axis and multiply the result by $8.$

In the first quadrant MATH and $y=0$ at $x=a,y=x$ at MATH

Further
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and
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Consequently,
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The Arc Length of a Curve Represented Parametrically

If a curve is given by the equations in the parametric form $x=x(t),y=y(t)$ and the derivatives MATH are continuous on the interval $[a,b]$, then the arc length of the curve is expressed by the integral
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Example 1

Compute the arc length of the involute of a circle MATH from $t=0$ to $t=2\pi .$

Solution.

Differentiating with respect to $t,$ we obtain
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whence MATH Hence,
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Example 2

Compute the length of the astroid:
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Solution.

Differentiating with respect to $t,$ we obtain
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Hence
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Since the function MATH has a period $\frac{\pi }{2},$
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Note. If we forget that we have to take the arithmetic value of the root and put MATH, we shall obtain the wrong result, since
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Example 3

Compute the length of the loop of the curve MATH

Solution.

Let us find the limits of integration. Both functions $x(t)$ and $y(t)$ are defined for all values of $t.$ Since the function MATH the curve lies in the right half-plane. Since with a change in sign of the parameter $t,x(t)$ remains unchanged, while $y(t)$ changes sign, the curve is symmetrical about the $x$-axis. Furthermore, the function $x(t)$ takes on one and the same value not more than twice. Hence, it follows that the points of self-intersection of the curve lie on the $x$-axis, i.e., at $y=0.$

But $y=0$ at $t=0,1,-1.$ Since MATH the point $(\sqrt{3},0) $ is the only point of self-intersection of the curve. Consequently, we must integrate within the limits $a=-1$ and $b=1.$

Differentiating the parametric equations of the curve with respect to $t,$ we get
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whence

Consequently,
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