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?
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
To find the arc length of a curve represented by parametric equations, you can use the integral of the square root of the sum of the squares of the derivatives of x(t) and y(t) with respect to t.
For the given curve with parametric equations x(t) = 2 + sin(t) and y(t) = 1 - cos(t), we can find the derivatives dx/dt and dy/dt and then substitute them into the arc length formula.
Step 1: Find dx/dt and dy/dt
dx/dt = cos(t)
dy/dt = sin(t)
Step 2: Find (dx/dt)^2 and (dy/dt)^2
(dx/dt)^2 = cos^2(t)
(dy/dt)^2 = sin^2(t)
Step 3: Find the integral
The integral representing the arc length is given by:
∫ [sqrt((dx/dt)^2 + (dy/dt)^2)] dt
Substituting the values we obtained:
∫ [sqrt(cos^2(t) + sin^2(t))] dt
Since cos^2(t) + sin^2(t) = 1, the integral simplifies to:
∫ [sqrt(1)] dt
Simplifying further, the integral becomes:
∫ dt
Evaluating the integral from t = 0 to 2π:
∫ dt = t | from 0 to 2π
= 2π - 0
= 2π
Therefore, the arc length of the given curve on the interval 0 ≤ t ≤ 2π is 2π.