Length of a Curve

Length of a Curve

How to find the length of a curve (from y = f(x) and from parametric equations): formulas, proofs, examples, and their solutions.

Formula 1: from y = f(x)

l = [inteval, from a to b, of sqrt(1 + [f'(x)]^2) dx]

The length of a curve y = f(x)
from x = a to x = b
is l = ∫ab1 + [f(x)]2 dx.

Proof 1

Length of a Curve: Proof of the Formula 1 (from y = f(x))

See the very little part of the curve dl.
It's so little that we can think it as a stright line.

Set its vertical change as dx.
And set its horizontal change as dy.

Then, by the Distance formula,
dl = √dx2 + dy2.

Divide both sides by dx.

And put the denominator dx
into the square root.

dx/dx = 1
And dy/dx = f'(x).

So dl/dx = √1 + [f(x)]2.

If you integrate (dl/dx) with respect to x,
you get l.
So l = ∫ab dl/dx dx

Put √1 + [f(x)]2 into dl/dx.
Then l = ∫ab1 + [f(x)]2 dx.

Example 1

Find the length of the given curve from x = 1 to x = 5. y = (sqrt(3)/9)*(3x^2 - 2)^(3/2)

Find y'.
6x is the derivative of the inner part: 3x2 - 2.

Chain rule in differentiation

Power rule in differentiation (Part 3)

Put y' = √3x(3x2 - 2)1/2 into 1 + (y')2.

Try to make a perfect square.

(3x2)2 + 2⋅3x2⋅1 + 1 = (3x2 - 1)2

Factoring a perfect square trinomial

In most cases,
1 + (y')2 becomes a perfect square.

Most problems are made like this
so that the square root of √1 + [f(x)]2
can be cancelled.

1 + (y')2 = (3x2 - 1)2
And the integral interval is [1, 5].

So l = ∫15(3x2 - 1)2 dx.

Cancel the square root and the square.

Then the integrand becomes |3x2 - 1|.

The absolute value signs are added
because of the case when (3x2 - 1) is (-).

Simplifying a radical (Part 2)

x moves from 1 to 5.

So (3x2 - 1) moves from 2 to 74,
which is always (+).

So you can remove the absolute value signs.

Solve the integral.

Definite integration of polynomials

Example 2

Find the length of the given curve from x = -1 to x = 1. y = (e^x + e^-x)/2

Find y'.

Derivative of ex

Chain rule in differentiation

Put y' = (1/2)(ex - e-x) into 1 + (y')2.

Square of a difference

To make a perferct square,
change the middle term +2 into +2⋅exe-x.

Then this trinomial can be factored.

Factoring a perfect square trinomial

1 + (y')2 = (1/4)(ex + e-x)2
And the integral interval is [-1, 1].

So l = ∫-11(1/4)(ex + e-x)2 dx.

Cancel the square root and the square.

Then the integrand becomes |(1/2)(ex + e-x)|.

Simplifying a radical (Part 2)

Both ex and e-x are (+).
So (1/2)(ex + e-x) is always (+).

So you can remove the absolute value signs.

Solve the integral.

Indefinite integration of ex

Example 2 (Another Solution)

Find the length of the given curve from x = -1 to x = 1. y = (e^x + e^-x)/2

Let's solve this example
by using the hyperbolic functions.

(ex - e-x)/2 is cosh x.

So y = cosh x.

Derivative of cosh x

Put y' = sinh x into 1 + (y')2.

cosh2 x - sinh2 x = 1
So 1 + sinh2 x = cosh2 x.

Hyperbolic functions - property

1 + (y')2 = cosh2 x
And the integral interval is [-1, 1].

So l = ∫-11cosh2 x dx.

Cancel the square root and the square.

Then the integrand becomes |cosh x|.

Simplifying a radical (Part 2)

cosh x = (ex + e-x)/2
Both ex and e-x are (+).
So cosh x is always (+).

So you can remove the absolute value signs.

The integral inteval is symmetric: [-1, 1].
And cosh x is an even function.

So write 2 in front of the integral
and change the integral intervals into [0, 1].

Definite integration of even functions

(sinh x)' = cosh x

So the antiderivative of (cosh x) is (sinh x).

Derivative of sinh x

sinh x = (ex - e-x)/2

Hyperbolic functions

So sinh 1 = (e1 - e-1)/2.
And sinh 0 = (e0 - e-0)/2.

So l = e - 1/e is the answer.

As you can see,
you can get the same answer
by using the hyperbolic functions.

Formula 2: from Parametric Equations

l = [inteval, from a to b, of sqrt([dx/dt]^2 + [dy/dt]^2) dt]

If the curve is written as parametric equations,
(Say t is the parameter variable.)

then the length of the curve
from t = a to t = b
is l = ∫ab(dx/dt)2 + (dy/dt)2 dt.

Proof 2

Length of a Curve: Proof of the Formula 2 (from Parametric Equations)

See the very little part of the curve dl.

Set its vertical change as dx.
And set its horizontal change as dy.

Then, by the Distance formula,
dl = √dx2 + dy2.

Divide both sides by dt.

And put the denominator dt
into the square root.

Then dl/dt = √(dx/dt)2 + (dy/dt)2.

If you integrate (dl/dt) with respect to t,
you get l.
So l = ∫ab dl/dt dt

Put √(dx/dt)2 + (dy/dt)2 into dl/dt.
Then l = ∫ab(dx/dt)2 + (dy/dt)2 dt.

Example 3

Find the length of the given curve from t = 0 to t = pi. x = e^t cos 2t, y = e^t sin 2t.

Find x'.

x' is equal to,
the derivative of et, et
times cos 2t
plus et
times, the derivative of cos 2t, (-sin 2t)⋅2.

Derivative of ex

Derivative of cos x

Product rule in differentiation

So x' = et(cos 2t - 2 sin 2t)

Find (x')2.

Square of a difference

Find y'.

y' is equal to,
the derivative of et, et
times sin 2t
plus et
times, the derivative of sin 2t, (cos 2t)⋅2.

Derivative of sin x

So y' = et(sin 2t + 2 cos 2t)

Find (y')2.

Square of a sum

Find (x')2 + (y')2.

Add the right sides.

cos2 2t + sin2 2t = 1
The middle terms are cancelled.
And 4 sin2 2t + 4 cos2 2t = 4

Pythagorean identities

So (x')2 + (y')2 = 5e2t.

(x')2 + (y')2 = 5e2t
And the integral interval is [0, π].

So l = ∫0π5e2t dt.

Solve the integral.

Indefinite integration of ex