# Length of a Curve

How to find the length of a curve (of y = f(x), of a parametric function): 2 formulas, 2 examples, and their solutions.

## Formula: Curve of a Parametric Function

### Formula

For a parametric function x = f(t) and y = g(t),

the length of a curve from t = a to t = b is

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

## Example 1

### Example

### Solution

x = t - sin t

So dx/dt = 1 - cos t.

Derivative of a Polynomial

Derivative of sin x

y = 1 - cos t

So dy/dt = sin t.

Derivative of cos x

t is from 0 to 2π.

dx/dt = 1 - cos t

dy/dt = sin t

So l = ∫_{0}^{2π} √(1 - cos t)^{2} + sin^{2} t dt.

(1 - cos t)^{2} = 1 - 2 cos t + cos^{2} t

Square of a Difference: (a - b)^{2}

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

Pythagorean Identity

1 - 2 cos t + 1 = 2 - 2 cos t

2 - 2 cos t = 2(1 - cos t)

2(1 - cos t) = 4⋅[(1 - cos t)/2]

√2^{2}⋅[sin^{2} t/2] = |2 sin t/2|

Simplify a Radical

You should check

if 2 sin t/2 ≥ 0

from t = 0 to t = 2π.

So write absolue value signs.

If t = 0,

2 sin t/2 = 2 sin 0/2 = 2⋅0 = 0.

If t = 2π,

2 sin t/2 = 2 sin 2π/2 = 2 sin π = 2⋅0 = 0.

And any t between 0 and 2π

makes 2 sin t/2 plus.

So you can remove the absolue value signs.

Sine: Graph

Take the constant 2

out from the integral.

Solve the integral.

Definite Integral: How to Solve

The integral of sin t/2 is

2⋅(-cos t/2).

Linear Change of Variable Rule

Integral of sin x

2[2⋅(-cos t/2)]_{0}^{2π} = -4[cos t/2]_{0}^{2π}

Put 0 and 2π

into cos t/2.

cos 2π/2 = cos π

-cos 0/2 = -cos 0

cos π = (-1)

-cos 0 = -1

Cosine Values of Commonly Used Angles

(-1) - 1 = -2

-4⋅(-2) = 8

So 8 is the answer.

## Formula: Curve of y = f(x)

### Formula

For a function y = f(x),

the length of a curve from x = a to x = b is

l = ∫_{a}^{b} √1 + [y']^{2} dx.

## Example 2

### Example

### Solution

y = [1/2]x^{2} - [1/4][ln x]

Then y' = x - 1/4x.

Derivative of ln x

x is from 1 to e.

y' = x - 1/4x

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

(x - 1/4x)^{2} = x^{2} - 2⋅x⋅[1/4x] + [1/4x]^{2}

-2⋅x⋅[1/4x] = -1/2

1 - 1/2 = +1/2

You've changed -2⋅x⋅[1/4x] to -1/2.

So, change +1/2 to +2⋅x⋅[1/4x].

x^{2} + 2⋅x⋅[1/4x] + [1/4x]^{2} = (x + 1/4x)^{2}

Factor a Perfect Square Trinomial

√(x + 1/4x)^{2} = |x + 1/4x|

Again, you should check

if x + 1/4x ≥ 0

from x = 1 to x = e.

So write absolue value signs.

Any number from x = 1 to x = e

satisfies x + 1/4x ≥ 0.

So you can remove the absolute value signs.

Solve the integral.

The integral of x is

[1/2]x^{2}.

Integral of a Polynomial

The integral of +1/4x is

+[1/4] ln |x|.

Integral of 1/x

Put e and 1

into x + [1/4] ln |x|.

[1/2]⋅e^{2} = e^{2}/2

+[1/4] ln e = +[1/4]⋅1

[1/2]⋅1^{2} = 1/2

+[1/4] ln e = +[1/4]⋅0

+[1/4]⋅1 = +1/4

+[1/4]⋅0 = +0

-(1/2 + 0) = -1/2

To solve +1/4 - 1/2,

multiply 2/2 to -1/2.

Then -1/2 = -[1/2]⋅[2/2] = -2/4.

+1/4 - 2/4 = -1/4

So e^{2}/2 - 1/4 is the answer.