A curve parametrically given by `x=t+t^(3)" and "y=t^(2)," where "t in R." For what vlaue(s) of t is "(dy)/(dx)=(1)/(2)`?

Options

A. `(1)/(3)`
B. 2
C. 3
D. 1

Explanation

To solve the problem, we need to find the values of \( t \) for which \( \frac{dy}{dx} = \frac{1}{2} \) given the parametric equations \( x = t + t^3 \) and \( y = t^2 \).

### Step-by-Step Solution:

1. **Identify the Parametric Equations**:
\[ x = t + t^3 \]
\[ y = t^2 \]

2. **Differentiate \( x \) and \( y \) with respect to \( t \)**:
- Differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(t + t^3) = 1 + 3t^2 \]
- Differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(t^2) = 2t \]

3. **Find \( \frac{dy}{dx} \)**:
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{1 + 3t^2} \]

4. **Set \( \frac{dy}{dx} \) equal to \( \frac{1}{2} \)**:
\[ \frac{2t}{1 + 3t^2} = \frac{1}{2} \]

5. **Cross-multiply to eliminate the fraction**:
\[ 2(2t) = 1 + 3t^2 \]
This simplifies to:
\[ 4t = 1 + 3t^2 \]

6. **Rearrange the equation**:
\[ 3t^2 - 4t + 1 = 0 \]

7. **Solve the quadratic equation using the quadratic formula**:
The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3 \), \( b = -4 \), and \( c = 1 \):
\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]
\[ t = \frac{4 \pm \sqrt{16 - 12}}{6} \]
\[ t = \frac{4 \pm \sqrt{4}}{6} \]
\[ t = \frac{4 \pm 2}{6} \]

8. **Calculate the two possible values for \( t \)**:
- First value:
\[ t = \frac{4 + 2}{6} = \frac{6}{6} = 1 \]
- Second value:
\[ t = \frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3} \]

9. **Final Values**:
The values of \( t \) for which \( \frac{dy}{dx} = \frac{1}{2} \) are:
\[ t = 1 \quad \text{and} \quad t = \frac{1}{3} \]