To find the tangent to each curve at the point where $t = 3$, we need to know the parameterized equations of the curves. Typically, these would be given as functions of $t$. For this example, let's consider we have two parameterized curves $C_1$ and $C_2$ given by:

$C_1: \begin{cases} x_1(t) = f(t) \\ y_1(t) = g(t) \end{cases}$ $C_2: \begin{cases} x_2(t) = h(t) \\ y_2(t) = k(t) \end{cases}$

To find the tangent to each curve at $t = 3$, we follow these steps:

1. Evaluate the point on each curve at $t = 3$: $P_1 = (x_1(3), y_1(3))$ $P_2 = (x_2(3), y_2(3))$

2. Compute the derivatives of the parameterized functions to find the slopes: $\frac{dx_1}{dt} \Bigg|_{t=3} = f'(3)$ $\frac{dy_1}{dt} \Bigg|_{t=3} = g'(3)$ $\frac{dx_2}{dt} \Bigg|_{t=3} = h'(3)$ $\frac{dy_2}{dt} \Bigg|_{t=3} = k'(3)$

3. The slope of the tangent line to each curve at $t = 3$ is given by: $m_1 = \frac{dy_1}{dx_1} = \frac{g'(3)}{f'(3)}$ $m_2 = \frac{dy_2}{dx_2} = \frac{k'(3)}{h'(3)}$

4. Write the equations of the tangents using the point-slope form of the line equation: $y - y_1(3) = m_1 (x - x_1(3))$ $y - y_2(3) = m_2 (x - x_2(3))$

Now let's illustrate this with specific parameterized functions. Assume:

$C_1: \begin{cases} x_1(t) = t^2 \\ y_1(t) = t^3 \end{cases}$ $C_2: \begin{cases} x_2(t) = \sin(t) \\ y_2(t) = \cos(t) \end{cases}$

1. Evaluate the points at $t = 3$: $P_1 = (x_1(3), y_1(3)) = (9, 27)$ $P_2 = (x_2(3), y_2(3)) = (\sin(3), \cos(3)) \approx (0.1411, -0.98999)$

2. Compute the derivatives at $t = 3$: $\frac{dx_1}{dt} = 2t \Bigg|_{t=3} = 6$ $\frac{dy_1}{dt} = 3t^2 \Bigg|_{t=3} = 27$

   $\frac{dx_2}{dt} = \cos(t) \Bigg|_{t=3} \approx -0.98999$ $\frac{dy_2}{dt} = -\sin(t) \Bigg|_{t=3} \approx -0.1411$

3. Find the slopes of the tangents: $m_1 = \frac{dy_1}{dx_1} = \frac{27}{6} = 4.5$ $m_2 = \frac{dy_2}{dx_2} = \frac{-0.1411}{-0.98999} \approx 0.1425$

4. Write the equations of the tangents: $y - 27 = 4.5(x - 9)$ $y - (-0.98999) = 0.1425(x - 0.1411)$

Simplify these to get: $y = 4.5x - 13.5$ $y = 0.1425x - 1.009$

These are the equations of the tangents to the curves $C_1$ and $C_2$ at $t = 3$.

Would you like further details or have any questions about this solution?

Here are some related questions you might want to ask:

1. How do you find the intersection points of two parametric curves?
2. What is the chain rule, and how is it used in parametric differentiation?
3. How do you convert a parametric equation to Cartesian form?
4. How do you find the arc length of a parametric curve?
5. What are some applications of parametric equations in real life?

Tip: When dealing with parametric equations, always check the derivatives to understand the behavior and slope of the curve at any given point.