 Find the tangent to each curve at the point where t = 3:

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.

Learn how to find the tangent lines to parametric curves at t = 3 with detailed steps and examples. This problem covers parametric equations, derivatives, and tangent line equations. Suitable for advanced high school students and anyone interested in calculus.

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