To find the area enclosed by the curve given by the parametric equations x = t² – 2t and y = t, as well as the y-axis, we start by determining the points of intersection between the curve and the y-axis. The y-axis corresponds to x = 0.
Setting the equation for x equal to zero:
t² – 2t = 0
This can be factored as:
t(t – 2) = 0
From this equation, we find that t = 0 and t = 2 are the points where the curve intersects the y-axis.
Next, we can set up the integral to find the area between the curve and the y-axis from t = 0 to t = 2. The area A can be represented as:
A = ∫[y(t) * dx/dt] dt
Calculating dx/dt:
From x = t² – 2t, we differentiate:
dx/dt = 2t – 2
Given y = t, we can now write:
A = ∫[t * (2t – 2)] dt from 0 to 2
Now, simplifying the integrand:
A = ∫[2t² – 2t] dt
This integrates to:
A = [(2/3)t³ – t²] evaluated from 0 to 2.
Calculating at t = 2:
A = (2/3)(2)³ – (2²) = (2/3)(8) – 4 = 16/3 – 4 = 16/3 – 12/3 = 4/3.
Now, evaluating at t = 0 gives us 0. Therefore, the total area A is:
A = 4/3 square units.
Thus, the area enclosed by the curve and the y-axis is 4/3 square units.