To find the equation of the tangent line at the point $(x, y) = (0, 0)$, we need to do the following steps:

Step 1: Determine the value of $t$ where $(x, y) = (0, 0)$

The parametric equations for the curve are:

• $x(t) = 2 \cos(t)$
• $y(t) = 3 \sin(t) \cos(t)$

We want to find $t$ such that both $x(t) = 0$ and $y(t) = 0$.

1. Set $x(t) = 0$: $2 \cos(t) = 0 \quad \Rightarrow \quad \cos(t) = 0$ The solutions to $\cos(t) = 0$ are: $t = \frac{\pi}{2}, \frac{3\pi}{2}$ These are the possible values of $t$ where $x = 0$.

2. Now, check $y(t) = 0$ for these values of $t$: $y(t) = 3 \sin(t) \cos(t) = 0$ This simplifies to $\sin(t) \cos(t) = 0$, which gives $\sin(t) = 0$ or $\cos(t) = 0$.

   • For $t = \frac{\pi}{2}$ or $t = \frac{3\pi}{2}$, $\cos(t) = 0$, and we already know that $\cos(t) = 0$ for these values.

So, $t = \frac{\pi}{2}$ and $t = \frac{3\pi}{2}$ are both valid solutions where the curve passes through the origin $(0, 0)$.

Step 2: Find the tangent line at $t = \frac{\pi}{2}$

Now, we compute the slope of the tangent line at these points. The slope of the tangent line is given by the derivative of $y(t)$ with respect to $x(t)$. This is: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

1. Compute $\frac{dx}{dt}$: $x(t) = 2 \cos(t) \quad \Rightarrow \quad \frac{dx}{dt} = -2 \sin(t)$

2. Compute $\frac{dy}{dt}$: $y(t) = 3 \sin(t) \cos(t) \quad \Rightarrow \quad \frac{dy}{dt} = 3 \left( \cos^2(t) - \sin^2(t) \right)$

At $t = \frac{\pi}{2}$:

• $\frac{dx}{dt} = -2 \sin\left(\frac{\pi}{2}\right) = -2$
• $\frac{dy}{dt} = 3 \left( \cos^2\left(\frac{\pi}{2}\right) - \sin^2\left(\frac{\pi}{2}\right)\right) = 3(0 - 1) = -3$

Thus, the slope of the tangent line at $t = \frac{\pi}{2}$ is: $\frac{dy}{dx} = \frac{-3}{-2} = \frac{3}{2}$

Step 3: Write the equation of the tangent line

The equation of a line with slope $m$ passing through the point $(x_0, y_0)$ is: $y - y_0 = m(x - x_0)$

At $t = \frac{\pi}{2}$, the point on the curve is $(x_0, y_0) = (0, 0)$. Thus, the equation of the tangent line is: $y - 0 = \frac{3}{2}(x - 0)$ $y = \frac{3}{2} x$

Step 4: Check the result for $t = \frac{3\pi}{2}$

Similarly, you can compute the slope for $t = \frac{3\pi}{2}$. But we expect a different sign for the slope, as $\sin(t)$ and $\cos(t)$ will be different.

In summary:

• The tangent line at $t = \frac{\pi}{2}$ is $y = \frac{3}{2} x$.

So, the correct equation of the tangent line at $(x, y) = (0, 0)$ is:

$\boxed{y = \frac{3}{2} x}$

Would you like further clarification or assistance on the next part?

Further Questions:

1. How would the tangent line equation change if the parametric equations were different?
2. Can you describe how to find the points where the curve intersects the $x$- and $y$-axes?
3. What is the significance of the derivative $\frac{dy}{dx}$ in terms of curve behavior?
4. How does the second derivative relate to the curvature of the curve?
5. Can you graph the parametric equations to visualize the curve and tangent line?

Tip: When working with parametric equations, always check for points where $x = 0$ or $y = 0$ to find potential intersections