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 = ∫02π √(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]
√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)]02π = -4[cos t/2]02π
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 = ∫ab √1 + [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 = ∫1e √1 + (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.