If a plane curve is given by the equation
and the derivative
is continuous, then the length of an arc of this curve is expressed by the
integral

where
and
are the abscissas of the end-points of the given arc.

Compute the length of the arc of the semicubical parabola between the points and .

**Solution.**

The function
is defined for
Since the given points lie in the first quadrant,
Hence,

and

Consequently,

Compute the arc length of the curve between the points with the abscissas

**Solution.**

Since
,
then

Hence,

Find the arc length of the curve between the points with the ordinates and

** Solution.**

Here it is convenient to adopt
as the independent variable; then

and

Hence,

Find the length of the astroid

**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 and the -axis and multiply the result by

In the first quadrant and at at

Further

and

Consequently,

If a curve is given by the equations in the parametric form
and the derivatives
are continuous on the interval
,
then the arc length of the curve is expressed by the integral

Compute the arc length of the involute of a circle from to

**Solution.**

Differentiating with respect to
we obtain

whence
Hence,

Compute the length of the astroid:

**Solution.**

Differentiating with respect to
we obtain

Hence

Since the function
has a period

**Note.** If we forget that we have to take the arithmetic value
of the root and put
,
we shall obtain the wrong result, since

Compute the length of the loop of the curve

**Solution.**

Let us find the limits of integration. Both functions and are defined for all values of Since the function the curve lies in the right half-plane. Since with a change in sign of the parameter remains unchanged, while changes sign, the curve is symmetrical about the -axis. Furthermore, the function 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 -axis, i.e., at

But at Since the point is the only point of self-intersection of the curve. Consequently, we must integrate within the limits and

Differentiating the parametric equations of the curve with respect to
we get

whence

Consequently,