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 = ∫ab(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(1 - cos t)2 + sin2 t dt.

(1 - cos t)2 = 1 - 2 cos t + cos2 t

Square of a Difference: (a - b)2

cos2 t + sin2 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]

4 = 22

(1 - cos t)/2 = sin2 t/2

sin A/2

22⋅[sin2 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 = -4[cos t/2]0

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 = ∫ab1 + [y']2 dx.

Example 2

Example

Solution

y = [1/2]x2 - [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 = ∫1e1 + (x - 1/4x)2 dx.

(x - 1/4x)2 = x2 - 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].

x2 + 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]x2.

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]⋅e2 = e2/2
+[1/4] ln e = +[1/4]⋅1

[1/2]⋅12 = 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 e2/2 - 1/4 is the answer.