The slope of the tangent to the curves `x=3t^2+1,y=t^3-1` at t=1 is

Options

A. 0
B. `1/2`
C. 1
D. -2

Answer

To find the slope of the tangent to the curve defined by the parametric equations \( x = 3t^2 + 1 \) and \( y = t^3 - 1 \) at \( t = 1 \), we will follow these steps:

### Step-by-Step Solution:

1. **Find \( \frac{dx}{dt} \)**:
- Given \( x = 3t^2 + 1 \), we differentiate with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(3t^2 + 1) = 6t \]

2. **Find \( \frac{dy}{dt} \)**:
- Given \( y = t^3 - 1 \), we differentiate with respect to \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(t^3 - 1) = 3t^2 \]

3. **Find \( \frac{dy}{dx} \)**:
- The slope of the tangent \( m \) is given by \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \):
\[ \frac{dy}{dx} = \frac{3t^2}{6t} = \frac{3}{6} \cdot \frac{t^2}{t} = \frac{1}{2} t \]

4. **Evaluate \( \frac{dy}{dx} \) at \( t = 1 \)**:
- Substitute \( t = 1 \) into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \bigg|_{t=1} = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

5. **Conclusion**:
- The slope of the tangent to the curve at \( t = 1 \) is \( \frac{1}{2} \).

### Final Answer:
The slope of the tangent to the curves at \( t = 1 \) is \( \frac{1}{2} \).