# 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* = ∫_{a}^{b} √1 + [*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* = √*dx*^{2} + *dy*^{2}.

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* = ∫_{a}^{b} *dl*/*dx* *dx*

Put √1 + [*f*(*x*)]^{2} into *dl*/*dx*.

Then *l* = ∫_{a}^{b} √1 + [*f*(*x*)]^{2} *dx*.

## Example 1

Find *y*'.

6*x* is the derivative of the inner part: 3*x*^{2} - 2.

Chain rule in differentiation

Power rule in differentiation (Part 3)

Put *y*' = √3*x*(3*x*^{2} - 2)^{1/2} into 1 + (*y*')^{2}.

Try to make a perfect square.

(3*x*^{2})^{2} + 2⋅3*x*^{2}⋅1 + 1 = (3*x*^{2} - 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} = (3*x*^{2} - 1)^{2}

And the integral interval is [1, 5].

So *l* = ∫_{1}^{5} √(3*x*^{2} - 1)^{2} *dx*.

Cancel the square root and the square.

Then the integrand becomes |3*x*^{2} - 1|.

The absolute value signs are added

because of the case when (3*x*^{2} - 1) is (-).

Simplifying a radical (Part 2)

*x* moves from 1 to 5.

So (3*x*^{2} - 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 *y*'.

Derivative of *e*^{x}

Chain rule in differentiation

Put *y*' = (1/2)(*e*^{x} - *e*^{-x}) into 1 + (*y*')^{2}.

To make a perferct square,

change the middle term +2 into +2⋅*e*^{x}⋅*e*^{-x}.

Then this trinomial can be factored.

Factoring a perfect square trinomial

1 + (*y*')^{2} = (1/4)(*e*^{x} + *e*^{-x})^{2}

And the integral interval is [-1, 1].

So *l* = ∫_{-1}^{1} √(1/4)(*e*^{x} + *e*^{-x})^{2} *dx*.

Cancel the square root and the square.

Then the integrand becomes |(1/2)(*e*^{x} + *e*^{-x})|.

Simplifying a radical (Part 2)

Both *e*^{x} and *e*^{-x} are (+).

So (1/2)(*e*^{x} + *e*^{-x}) is always (+).

So you can remove the absolute value signs.

Solve the integral.

Indefinite integration of *e*^{x}

## Example 2 (Another Solution)

Let's solve this example

by using the hyperbolic functions.

(*e*^{x} - *e*^{-x})/2 is cosh *x*.

So *y* = cosh *x*.

Put *y*' = sinh *x* into 1 + (*y*')^{2}.

cosh^{2} *x* - sinh^{2} *x* = 1

So 1 + sinh^{2} *x* = cosh^{2} *x*.

Hyperbolic functions - property

1 + (*y*')^{2} = cosh^{2} *x*

And the integral interval is [-1, 1].

So *l* = ∫_{-1}^{1} √cosh^{2} *x* *dx*.

Cancel the square root and the square.

Then the integrand becomes |cosh *x*|.

Simplifying a radical (Part 2)

cosh *x* = (*e*^{x} + *e*^{-x})/2

Both *e*^{x} 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* = (*e*^{x} - *e*^{-x})/2

Hyperbolic functions

So sinh 1 = (*e*^{1} - *e*^{-1})/2.

And sinh 0 = (*e*^{0} - *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* = ∫_{a}^{b} √(*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* = √*dx*^{2} + *dy*^{2}.

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* = ∫_{a}^{b} *dl*/*dt* *dt*

Put √(*dx*/*dt*)^{2} + (*dy*/*dt*)^{2} into *dl*/*dt*.

Then *l* = ∫_{a}^{b} √(*dx*/*dt*)^{2} + (*dy*/*dt*)^{2} *dt*.

## Example 3

Find *x*'.*x*' is equal to,

the derivative of *e*^{t}, *e*^{t}

times cos 2*t*

plus *e*^{t}

times, the derivative of cos 2*t*, (-sin 2*t*)⋅2.

Derivative of *e*^{x}

Derivative of cos *x*

Product rule in differentiation

So *x*' = *e*^{t}(cos 2*t* - 2 sin 2*t*)

Find (*x*')^{2}.

Square of a difference

Find *y*'.*y*' is equal to,

the derivative of *e*^{t}, *e*^{t}

times sin 2*t*

plus *e*^{t}

times, the derivative of sin 2*t*, (cos 2*t*)⋅2.

Derivative of sin *x*

So *y*' = *e*^{t}(sin 2*t* + 2 cos 2*t*)

Find (*y*')^{2}.

Square of a sum

Find (*x*')^{2} + (*y*')^{2}.

Add the right sides.

cos^{2} 2*t* + sin^{2} 2*t* = 1

The middle terms are cancelled.

And 4 sin^{2} 2*t* + 4 cos^{2} 2*t* = 4

Pythagorean identities

So (*x*')^{2} + (*y*')^{2} = 5*e*^{2t}.

(*x*')^{2} + (*y*')^{2} = 5*e*^{2t}

And the integral interval is [0, *π*].

So *l* = ∫_{0}^{π} √5*e*^{2t} *dt*.

Solve the integral.

Indefinite integration of *e*^{x}