To find the area enclosed by the curve given by the parametric equations:

$x(t) = 2 \cos(t), \quad y(t) = 7 \cos(t) \sin(t)$

over the interval $0 \leq t \leq 2\pi$, we can use the formula for the area enclosed by a parametric curve:

$A = \int_{a}^{b} y(t) \frac{dx}{dt} \, dt$

Here, $x(t) = 2 \cos(t)$ and $y(t) = 7 \cos(t) \sin(t)$, and the interval is from $0$ to $2\pi$.

Step 1: Calculate $\frac{dx}{dt}$

$\frac{dx}{dt} = \frac{d}{dt} (2 \cos(t)) = -2 \sin(t)$

Step 2: Set up the integral

Substitute $y(t)$ and $\frac{dx}{dt}$ into the formula for the area:

$A = \int_{0}^{2\pi} 7 \cos(t) \sin(t) (-2 \sin(t)) \, dt$

This simplifies to:

$A = -14 \int_{0}^{2\pi} \cos(t) \sin^2(t) \, dt$

Step 3: Use symmetry to simplify the calculation

Notice that $\cos(t) \sin^2(t)$ is an odd function over the interval $0 \leq t \leq 2\pi$. Since the function is symmetric about $\pi$, the integral over the full period will be zero:

$\int_{0}^{2\pi} \cos(t) \sin^2(t) \, dt = 0$

Thus, the area enclosed by the curve is:

$A = 0$

The curve’s symmetry results in the net area calculation canceling out to zero over one complete period from $0$ to $2\pi$.

Would you like further details, or do you have any questions? Here are some related questions to consider:

1. How do we determine if a parametric curve encloses any area?
2. What role does symmetry play in evaluating integrals of trigonometric functions?
3. Can the area calculation for a parametric curve be done using polar coordinates?
4. How does the parametrization of a curve affect its geometric properties?
5. What are the conditions for using Green’s Theorem to find the area of a parametric curve?

Tip: When working with trigonometric integrals, always consider symmetry to simplify the calculations.