Subject: Calculus


In a Euclidean plane, the length of a curve can be measured by approximation using multiple linear segments that interconnects the through a curve. The total approximated length of the curve can then be determined by adding all these linear segments. The reason why we use linear segments in the approximation is that lines are easy to measure using Pythagorean Theorem. Better approximations to the curve can be obtained by following the shape of the curve increasingly more closely. The approach is to use an increasingly larger number of segments of smaller lengths.
Consider the figure at the right. As you see, the curve (in black) is interconnected by straight line segments (in violet). Now, to measure the length of the curve, we will add all the lengths of each line segment. Before that, let us first determine the length of the line segments by considering one line segment first. As seen in the figure. the length of the first segment is determined using Pythagorean Theorem.
Since we mentioned before that better approximations are obtained by making the line segments very small, then we let the infinitesimal length ds be the length of the first line segment, then according to the Pythagorean Theorem, with the assumption that {$f(x)$) represents the curve in the plane,

(ds)^2 = (dx)^2 + (dy)^2

we then divide by (dx)^2,

\frac{(ds)^2}{(dx)^2 } = 1 + \frac{(dy)^2}{(dx)^2 }

taking the square root, we obtain

\frac{ds}{dx } = \sqrt{1 + \frac{(dy)^2}{(dx)^2 }}
ds = \left (\sqrt{1 + \frac{(dy)^2}{(dx)^2 }}\right )dx

So, we now have the length of the first line segment. If we add up all these line segments that interconnects the curve from point x_1 to x_2, we end up doing the Reimann sum which is just the integral shown below.

s = \int_{x_1}^{x_2} \sqrt{1 + \frac{(dy)^2}{(dx)^2 }}dx

And since \frac{dy}{dx} = f'(x) then,

s = \int_{x_1}^{x_2} \sqrt{1 + \left [ f'(x) \right ]^2 }dx

So, this is it. The length of the curve is just the quantity s given that the curve is defined by the function f on the interval x_1 to x_2. Now, consider a curve defined by y = x^{\frac{2}{3}}. Let us find the length of this curve from point x=1 to x=8 using the above formula. First, let is find f'(x).

f(x)=y = x^{\frac{2}{3}}
f'(x)=dy/dx = D_{x}x^{\frac{2}{3}}
(f'(x))^2=\frac{4}{9} x^{-\frac{2}{3}}

Putting this into the formula, we get

s = \int_{1}^{8} \sqrt{1 + \left [ \frac{4}{9}x^{-\frac{2}{3}} \right ]^2 }dx

With the aid of Table of integrals, we can directly evaluate the integral and find its answer to be

s \approx 7.634

Thus, the length of the curve y = x^{\frac{2}{3}} from x=1 to x=8 is approximately 7.634 units.

Example #1

Find the length of the curve f(x) = x^{3} from point x=0 to x=2.

s \approx 8.630

Example #2

Find the length of the curve y = 6\cosh{\frac{x}{6}} from the origin to x = 10.75.

s = 17.5

Example #3

Find the length of the curve y = x^{2/3} from the point (1, 1) to (8, 4).

First, we get the first derivative of y.

y' = \frac{2}{3}x^{1/3}

Then we calculate the length L as;

L = \int_{1}^{8} \sqrt{1+\frac{4}{9x^2}}

Letting u = 9x^{2/3}

L = \frac{1}{18}\int_{13}^{40} u^{1/2}du

= 7.634 units

Therefore, the length of the arc is 7.634 units

Example #4

Find the length of the curve x^2+y^2= 16 from x= -4 to x = 4.

Obviously, the curve is a circle with radius=4.

Also, the circumference of a circle is given by C=\theta r and clearly, the problem calls for the length of the half-circle which has an angle of \pi.


C=\pi r = \pi \cdot 4

C= 4\pi

Example #5

Find the length of the arc represented by f(x= x^{3/2} from x=0 to x=1.

Answer: L = 1.44 units

BACK TO: Integral Calculus