recall that the arc length is
s = ∫[0,2π] √(dx/dt)^2 + (dy/dt)^2) dt
= ∫[0,2π] √(cos^2t + sin^2t) dt
= ∫[0,2π] 1 dt
= 2π
surprised? Note that this is just the circle of radius 1 with center at (2,1)
x=2+sint and y=1-cost
x-2 = sint, y-1 = -cost
(x-2)^2 + (y-1)^2 = sin^2t + cos^2t = 1
Consider the curve represented by the parametric equations x(t)= 2+sin(t) and y(t)=1-cos(t) when answering the following questions.
A) Find Dy/Dx in terms of t
I got tan(t)
B) Find all values of t where the curve has a horizontal tangent.
I got (+-)(pi)(n)
C) Find all values of t where the curve has a vertical tangent.
I got (pi/2)(+-)(pi)(n)
D) Write an integral that represents the arc length of the curve on the interval 0 ≤ t ≤ 2π. Evaluate the integral
I'm kind of confused on this?
1 answer