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

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

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

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

So y = 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 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 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 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