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
B) Find all values of t where the curve has a horizontal tangent.
C) Find all values of t where the curve has a vertical tangent.
D) Write an integral that represents the arc length of the curve on the interval 0 ≤ t ≤ 2π. Evaluate the integral.
Just plug and chug.
A) dy/dx = (dy/dt) / (dx/dt) = sin(t)/cos(t) = tan(t)
B and C should now be clear
Now you can easily write the arc length integral in terms of t - it might surprise you.
So the answer to B is t=+- (pi)(n)
and for C it is (pi/2)(+-)(pi)(n)
A) To find Dy/Dx in terms of t, we need to calculate the derivatives of x(t) and y(t) with respect to t and then divide dy/dt by dx/dt.
First, let's find dx/dt:
dx/dt = d/dt (2 + sin(t))
= 0 + cos(t) = cos(t)
Next, let's find dy/dt:
dy/dt = d/dt (1 - cos(t))
= 0 + sin(t) = sin(t)
Now, let's divide dy/dt by dx/dt:
Dy/Dx = (dy/dt) / (dx/dt)
= (sin(t))/(cos(t))
B) To find all values of t where the curve has a horizontal tangent, we need to find the values of t for which dy/dt = 0.
Since dy/dt = sin(t), it equals zero when sin(t) = 0. This happens at t = 0, π, 2π, etc.
C) To find all values of t where the curve has a vertical tangent, we need to find the values of t for which dx/dt = 0.
Since dx/dt = cos(t), it equals zero when cos(t) = 0. This happens at t = π/2, 3π/2, 5π/2, etc.
D) To write an integral representing the arc length of the curve on the interval 0 ≤ t ≤ 2π, we use the following formula for arc length:
L = ∫[a,b] √[dx/dt)^2 + (dy/dt)^2] dt
In this case, a = 0 and b = 2π. Substituting the dx/dt and dy/dt values we found earlier, the integral becomes:
L = ∫[0,2π] √[(cos(t))^2 + (sin(t))^2] dt
= ∫[0,2π] √(1) dt
= ∫[0,2π] 1 dt
= t |[0,2π]
= 2π - 0
= 2π
Evaluating the integral, we find that the arc length of the curve on the interval 0 ≤ t ≤ 2π is equal to 2π.