Length of a Curve
See how to find the length of a curve
(y = f(x), parametric function).
2 examples and their solutions.
Length of a Curve: y = f(x)
Formula
l = ∫ab√1 + [f'(x)]2 dx
dl2 = (dx)2 + (dy)2 - [1]
dl = √(dx)2 + (dy)2
dldx = √(dx)2 + (dy)2dx
= √(dx)2 + (dy)2(dx)2 - [2]
= √(dx)2(dx)2 + (dy)2(dx)2
= √1 + (dydx)2
= √1 + [f'(x)]2 - [3]
∴ l = ∫ab√1 + [f'(x)]2 dx - [4]
[1]
[3]
dy/dx = f'(x)
Derivative Rules
Derivative Rules
[4]
Close
Definite Integral
Example
Find the length of the given curve from x = 1 to x = e.
y = 12x2 - 14 ln x
Solution y = 12x2 - 14 ln x
f(x) = 12x2 - 14 ln x
f'(x) = 12⋅2x1 - 14⋅1x - [1]
= x - 14x
l = ∫1e√1 + (x - 14x)2 dx
= ∫1e√1 + x2 - 2⋅x⋅14x + (14x)2 dx - [2]
= ∫1e√1 + x2 - 12 + (14x)2 dx
= ∫1e√x2 + 12 + (14x)2 dx
= ∫1e√x2 + 2⋅x⋅14x + (14x)2 dx - [3]
= ∫1e√(x + 14x)2 dx - [4]
= ∫1e|x + 14x| dx - [5]
= ∫1e(x + 14x) dx - [6]
= [12x2 + 14 ln |4x|]1e - [7]
= 12e2 + 14 ln |4e| - [12⋅12 + 14 ln |4⋅1|]
= e22 + ln 4e4 - [12⋅1 + ln |4|4]
= e22 + ln 4 + ln e4 - [12 + ln 44]
= e22 + ln 4 + 14 - 12 - ln 44 - [8]
= e22 + ln 44 + 14 - 12 - ln 44
= e22 - 14
f'(x) = 12⋅2x1 - 14⋅1x - [1]
= x - 14x
l = ∫1e√1 + (x - 14x)2 dx
= ∫1e√1 + x2 - 2⋅x⋅14x + (14x)2 dx - [2]
= ∫1e√1 + x2 - 12 + (14x)2 dx
= ∫1e√x2 + 12 + (14x)2 dx
= ∫1e√x2 + 2⋅x⋅14x + (14x)2 dx - [3]
= ∫1e√(x + 14x)2 dx - [4]
= ∫1e|x + 14x| dx - [5]
= ∫1e(x + 14x) dx - [6]
= [12x2 + 14 ln |4x|]1e - [7]
= 12e2 + 14 ln |4e| - [12⋅12 + 14 ln |4⋅1|]
= e22 + ln 4e4 - [12⋅1 + ln |4|4]
= e22 + ln 4 + ln e4 - [12 + ln 44]
= e22 + ln 4 + 14 - 12 - ln 44 - [8]
= e22 + ln 44 + 14 - 12 - ln 44
= e22 - 14
[2]
[3]
-2⋅x⋅[1/4x] = -1/2
→ +1/2 = +2⋅x⋅[1/4x]
→ +1/2 = +2⋅x⋅[1/4x]
[6]
[8]
Close
Length of a Curve: Parametric Function
Formula
l = ∫ab√(dxdt)2 + (dydt)2 dt
dl2 = (dx)2 + (dy)2 - [1]
dl = √(dx)2 + (dy)2
dldt = √(dx)2 + (dy)2dt
= √(dx)2 + (dy)2(dt)2 - [2]
= √(dx)2(dt)2 + (dt)2(dt)2
= √(dxdt)2 + (dydt)2
∴ l = ∫ab√(dxdt)2 + (dydt)2 dt - [3]
[1]
[3]
Close
Definite Integral
Example
Find the length of the given curve from t = 0 to t = 2π.
x = t - sin t
y = 1 - cos t
Solution x = t - sin t
y = 1 - cos t
x = t - sin t
y = 1 - cos t
dxdt = 1 - cos t
dydt = -(-sin t)
= sin t - [1]
l = ∫02π√(1 - cos t)2 + (sin t)2 dt
= ∫02π√12 - 2⋅1⋅cos t + cos2 t + sin2 t dt - [2]
= ∫02π√1 - 2 cos t + 1 dt - [3]
= ∫02π√2 - 2 cos t dt
= ∫02π√2(1 - cos t) dt
= ∫02π√4⋅1 - cos t2 dt
= ∫02π√4⋅sin2 t2 dt
= ∫02π√22 sin2 t2 dt - [4]
= ∫02π|2 sin t2| dt - [5]
y = sin t2 (0 ≤ t ≤ 2π)
(period) = 2π|12|
= 2π12
= 4π
→ y = sin t2 ≥ 0
→ 2 sin t2 ≥ 0
= ∫02π(2 sin t2) dt - [6]
= [2⋅2⋅(-cos t2)]02π - [7]
= -4[cos t2]02π
= -4[cos 2π2 - (cos 02)]
= -4[cos π - cos 0]
= -4[(-1) - 1] - [8]
= -4⋅(-2)
= 8
y = 1 - cos t
dxdt = 1 - cos t
dydt = -(-sin t)
= sin t - [1]
l = ∫02π√(1 - cos t)2 + (sin t)2 dt
= ∫02π√12 - 2⋅1⋅cos t + cos2 t + sin2 t dt - [2]
= ∫02π√1 - 2 cos t + 1 dt - [3]
= ∫02π√2 - 2 cos t dt
= ∫02π√2(1 - cos t) dt
= ∫02π√4⋅1 - cos t2 dt
= ∫02π√4⋅sin2 t2 dt
= ∫02π√22 sin2 t2 dt - [4]
= ∫02π|2 sin t2| dt - [5]
y = sin t2 (0 ≤ t ≤ 2π)
(period) = 2π|12|
= 2π12
= 4π
→ y = sin t2 ≥ 0
→ 2 sin t2 ≥ 0
= ∫02π(2 sin t2) dt - [6]
= [2⋅2⋅(-cos t2)]02π - [7]
= -4[cos t2]02π
= -4[cos 2π2 - (cos 02)]
= -4[cos π - cos 0]
= -4[(-1) - 1] - [8]
= -4⋅(-2)
= 8
[2]
[3]
sin2 t + cos2 t = 1
Derivative Rules
Derivative Rules
[6]
t: 0 ~ 2π
y = sin t/2 ≥ 0
2 sin t/2 ≥ 0
→ |2 sin t/2| = 2 sin t/2
Trigonometric Function Graph
Absolute Value Equation
y = sin t/2 ≥ 0
2 sin t/2 ≥ 0
→ |2 sin t/2| = 2 sin t/2
Trigonometric Function Graph
Absolute Value Equation
Close