Arc length

ARC LENGTH

We know what is meant by the length of a straight line segment, but without calculus, we have no precise definition of the length of a general winding curve. If the curve is the graph of a continuous function defined over an interval, then we can find the length of the curve using a procedure similar to that we used for defining the area between the curve and the x‒axis. This procedure results in a division of the curve from point A to point B into many pieces and joining successive points of division by straight line segments. We then sum the lengths of all these line segments and define the length of the curve to be the limiting value of this sum as the number of segments goes to infinity.

Length of a Curve y = f(x)

Suppose the curve whose length we want to find is the graph of the function y = f(x) from x = a to x = b. In order to derive an integral formula for the length of the curve, we assume that f has a continuous derivative at every point of [a, b]. Such a function is called smooth, and its graph is a smooth curve because it does not have any breaks, corners, or cusps.

The length of the polygonal path P₀P₁P₂ ... P_n approximates the length of the curve y = f(x) from point A to point B.

We partition the interval [a, b] into n subintervals with a = x₀ < x₁ < x₂ <…. < x_a = b. If y_k = f(x_k) then the [a, b] corresponding point P_k(x_k, y_k) lies on the curve. Next we connect successive points P_k‒1 and P_k with straight line segments that, taken together, form a polygonal path whose length approximates the length of the curve. If ∆x_k = x_k ‒ x_k‒1 and ∆y_k = y_k ‒ y_k‒1, then a representative line segment in the path has length.

 L_k = √[(∆x_k)² + (∆y_k)²]

The arc P_k‒1P_k of the curve y = f(x) is approximated by the straight‒line segment shown here, which has length L_k = √[(∆x_k)² + (∆y_k)²].

So the length of the curve is approximated by the sum

          ……….(1)

We expect the approximation to improve as the partition of [a, b] becomes finer. Now by the Mean Value Theorem, there is a point c_k with x_k‒1 < c_k < x_k such that

 ∆y_k = f '(c_k)∆x_k

With this substitution for ∆y_k, the sums in Equation (1) take the form

         ……….(2)

Because √[1+f '(x)] is continuous on [a, b] the limit of the Riemann sum on the right‒hand side of Equation (2) exists as the norm the partition goes to zero, giving

We define the value of this limiting integral to be the length of the curve.

DEFINITION:

If f ' is continuous on [a, b], then the length (arc length) of the curve y = f(x) from the point A =  (a, f(a)) to the point B = (b, f(b)) is the value of the integral

       ……….(3)

Dealing with Discontinuities in dy/dx

At a point on a curve where dy/dx fails to exist, dx/dy may exist. In this case, we may be able to find the curve's length by expressing x as a function of y and applying the following analogue of Equation (3).

Formula for the Length of x = g(y), c ≤ y ≤ d

If g' is continuous on [c, d], the length of the curve x = g(y) from the point A = (g(c), c) to B = (g(d), d) is

       ……….(4)

Example 108. Find the length of the curve.

The length of the curve is slightly larger than the length of the line segment joining points A and B.

We use Equation (3) with a = 0, b = 1 and

 = 13/6 ~ 2.17

Example 109. Set up an integral that represents the length of the curve. y = sinx: 0 < x < π.

Solution:

Given that We know is y = sinx ⇒ dy/dx = cos x

We know that the arc length is

Example 110. Find the length of the arc of the curve y = log(cos x) on [0,π/4]

Solution:

Given that y = log(cos x)

dy / dx = 1/cosx(‒sinx) = ‒ sinx / cosx

The arc length formula is given by

 = log(√2+1) ‒ log 1

L = log(√2+1)

Example 111. Find the length of the arc of the parabola x² = 4ay mensured from the vertex to one extremity of the latus rectum.

Solution:

Let A be the vertex and L an extremity of the latus‒rectum so that at A, x = 0 and at L, x = 2a

= a[√2 + sinh^‒11]

= a[√2 + log(1 + √2)]

 [ sinh^‒1x= log[x + √(1+x²)]

Example 112. Find the perimeter of the loop of the curve

 3ay² = x(x ‒ a)².

 Solution:

The curve is symmetrical about the x‒axis and the loop lies between the limits x=0 and x = a

Example 113. Find the length of the arc of the curve y = 1 + 6x^3/2 0 ≤ x ≤ 1.

Solution:

Given that y = 1 + 6x^3/2

L= 6.103

Example 114. Compute the arc length of the given function f(x) = 2(x − 1)^3/2 on [1, 5].

Solution:

Given that y = 2(x − 1)^3/2

L = 16.597

Example 115. Find the arc length of the given function y = 2/3 (x² + 1)^3/2 on [1,4].

Solution:

Given that y = 2/3 (x² + 1)^3/2

L= 45

Example 116. Find the arc length of the curve y = x³/6 + 1/2x on [1,3].

Solution:

Given that y = x²/6 + 1/2x

Example 117. Find the area of the surface generating by rotating the curve y=e^x, 0 ≤x≤ 1, about the x‒axis.

Solution: Given y = e^x,

 dy/dx = e^x

The surface area is given by

Example 118. Find the length of the curve y=(x/2)^2/3 from x = 0 to x = 2.

Solution:

The derivative

is not defined at x = 0, so we cannot find the curve's length.

We therefore rewrite the equation to express x in terms of y.

y = (x/2)^2/3

x = 2y^3/2

From this we see that the curve whose length we want is also the graph of x = 2y^3/2 from y = 0 to y = 1. The graph of y = (x/2)^2/3 from x=0 to x=2 is also the graph of x = 2y^2/3 from y = 0 to y = 1. The derivative

L=2.27

Example 119. Find the length of the arc of the parabola y² = x from (0,0) to (1,1)

Solution:

Given that x = y²

⇒ dx/dy = 2y

We know that

Since we have considered tanα = 2

We have sec²α = 1 + tan²α

⇒ sec²α = 1 + (2)² = 5

secα = √5

Hence L = ¼ [ (√5. 2) + log|√5+2| ]

 = ¼2√5 + ¼log(√5+2)

L = ½ √5 + ¼log(√5+2)

Example 120. Find the length of one arch of the cycloid

 x = a(t − sint), y = a(1 ‒ cost)

Solution:

As a point moves from one end 0 to the other end of its first arch, the parameter t increases from 0 to 2π..

Also dx/dt = a(1 ‒ cost), dy/dt = a sint

L = 8a

Example 121. Find the entire length of the cardioid r = a(1 + cosθ). Also show that the upper half is bisected by θ = π/3.

Solution: The cardioid is symmetrical about the initial line and for its upper half, 6 increases from 0 to π.

Length of upper half of the curve is 4a. Also length of the arc AP from 0 to π/3.

= 2a = half the length of upper half of the cardioid.

EXERCISE

14. Find the length of the arc of the parabola y² = 4ax (i) from the vertex to one end of the latus‒rectum. (ii) cut off by the line 3y = 8x.

15. Find the length of the curve y² = (2x‒1)³ cut off by the line x = 4.

(iii) x = cost + tsint, y = sint‒t cost, (‒π≤t≤π)

(iv) x = e^t(sint + cost), y = e^t(cost‒ sint) (3≤t≤4)

(v) x = e^tcost, y = e^tsint, (0 ≤ t ≤ π)