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To solve the given problem, let's work through it step by step. Here, we have a parametric representation of a curve:

\[
\begin{cases}
x(t) = 3t + 1 \\
y(t) = \ln(9t^2 + 1), \quad t \in \mathbb{R}
\end{cases}
\]

### Part (a): Find the point where the tangent has a slope of \(-1\).

1. **Calculate \( \frac{dy}{dx} \) using the parametric derivatives:**

   The derivative of \( x(t) \) is:
   \[
   \frac{dx}{dt} = 3
   \]

   The derivative of \( y(t) \) is:
   \[
   \frac{dy}{dt} = \frac{d}{dt}\left(\ln(9t^2 + 1)\right) = \frac{18t}{9t^2 + 1}
   \]

   Thus, the slope of the tangent \( \frac{dy}{dx} \) is:
   \[
   \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{18t}{9t^2 + 1}}{3} = \frac{6t}{9t^2 + 1}
   \]

2. **Set the derivative equal to \(-1\) to find \( t \):**
   \[
   \frac{6t}{9t^2 + 1} = -1
   \]

   Solve for \( t \):
   \[
   6t = - (9t^2 + 1)
   \]
   \[
   6t + 9t^2 + 1 = 0 \Rightarrow 9t^2 + 6t + 1 = 0
   \]

   Solve this quadratic equation using the quadratic formula:
   \[
   t = \frac{-6 \pm \sqrt{36 - 36}}{18} = \frac{-6}{18} = -\frac{1}{3}
   \]

   Thus, the parameter value is \( t = -\frac{1}{3} \).

3. **Find the corresponding point \((x, y)\):**

   Substituting \( t = -\frac{1}{3} \) into \( x(t) \):
   \[
   x\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1 = 0
   \]

   Substituting \( t = -\frac{1}{3} \) into \( y(t) \):
   \[
   y\left(-\frac{1}{3}\right) = \ln\left(9\left(-\frac{1}{3}\right)^2 + 1\right) = \ln\left(9 \cdot \frac{1}{9} + 1\right) = \ln(2)
   \]

   So, the point where the tangent slope is \(-1\) is \((0, \ln(2))\).

### Part (b): Convert the curve to the form \( y = y(x) \).

1. **Express \( t \) in terms of \( x \):**

   From \( x(t) = 3t + 1 \):
   \[
   t = \frac{x - 1}{3}
   \]

2. **Substitute \( t \) into \( y(t) \):**

   \[
   y = \ln\left(9\left(\frac{x - 1}{3}\right)^2 + 1\right)
   \]
   Simplify this:
   \[
   y = \ln\left(9 \cdot \frac{(x - 1)^2}{9} + 1\right) = \ln\left((x - 1)^2 + 1\right)
   \]

   Thus, the curve in terms of \( y = y(x) \) is:
   \[
   y(x) = \ln\left((x - 1)^2 + 1\right)
   \]

### Final Answers:

- **a)** \( t = -\frac{1}{3} \), and the point is \( (0, \ln(2)) \).
- **b)** \( y(x) = \ln\left((x - 1)^2 + 1\right) \).

Would you like a more detailed explanation or have any other questions?

### Related Questions:
1. How do you find the derivative of a parametric equation?
2. What is the meaning of the tangent's slope in terms of curve behavior?
3. How do logarithmic functions behave when composed with quadratic terms?
4. What is the significance of solving a quadratic equation in terms of parameterization?
5. How does the derivative \(\frac{dy}{dx}\) change if \( y(t) \) or \( x(t) \) is modified?

### Tip:
When dealing with parametric equations, always differentiate \( x(t) \) and \( y(t) \) separately before combining them into \( \frac{dy}{dx} \) for the tangent's slope!

Olkoon xy-tason käyrän parametriesitys { x(t) = 3t + 1, y(t) = ln(9t^2 + 1), t ∈ R }. a) Määrää käyrän piste, jossa tangentin kulmakerroin on -1. b) Muunna käyrän yhtälö muotoon y = y(x).

Learn how to solve for a point on a parametric curve with tangent slope -1, and convert the parametric equation into the form y = y(x). This problem covers parametric differentiation, quadratic equations, and logarithmic functions, suitable for university-level calculus.

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