What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (15 points)
A) 8
B) 4 times the square root of 2
C) 4
D) 8 times the square root of 2
I got 4√2
s = ∫[0,π] √((dx/dt)^2 + (dy/dt)^2) dt
= ∫[0,π] √((1+sint)^2 + (cost)^2) dt
= ∫[0,π] √(1+2sint+sin^2t + cos^2t) dt
= ∫[0,π] √(2+2sint) dt
Now you can use your sum-to product formulas to get the job done.
I got 4, can you check this? Please!
To find the length of a curve with parametric equations, you can use the arc length formula:
L = ∫ √(dx/dt)^2 + (dy/dt)^2 dt
In this case, the parametric equations are:
x = t - cos(t)
y = 1 - sin(t)
To find dx/dt, take the derivative of x with respect to t:
dx/dt = 1 + sin(t)
To find dy/dt, take the derivative of y with respect to t:
dy/dt = -cos(t)
Now, square both derivatives:
(dx/dt)^2 = (1 + sin(t))^2
(dy/dt)^2 = (-cos(t))^2
Add them together:
(dx/dt)^2 + (dy/dt)^2 = (1 + sin(t))^2 + (-cos(t))^2
Simplify the expression:
(dx/dt)^2 + (dy/dt)^2 = 1 + 2sin(t) + sin(t)^2 + cos(t)^2
Since sin(t)^2 + cos(t)^2 = 1, the expression becomes:
(dx/dt)^2 + (dy/dt)^2 = 2 + 2sin(t)
Now take the square root of the expression:
√(dx/dt)^2 + (dy/dt)^2 = √(2 + 2sin(t))
To find the length of the curve, we need to integrate this expression from t = 0 to t = π:
L = ∫[0 to π] √(2 + 2sin(t)) dt
This integral does not have a simple closed-form solution. However, you can approximate its value using numerical techniques or calculators.
The correct answer choice based on the given options is not known without evaluating the integral or using approximation methods.