The derivative of the function represented parametrically as `x=2t=|t|,y=t^3+t^2|t|a tt=0` is a. -1 b. 1 c. 0 d. does not exist

Options

A. `-1`
B. 0
C. 1
D. does not exist

Explanation

To find the derivative of the function represented parametrically as \( x = 2t - |t| \) and \( y = t^3 + t^2 |t| \) at \( t = 0 \), we will follow these steps:

### Step 1: Define the functions
We have the parametric equations:
\[ x = 2t - |t| \]
\[ y = t^3 + t^2 |t| \]

### Step 2: Analyze the function for \( t < 0 \)
For \( t < 0 \):
- The absolute value \( |t| \) becomes \( -t \).
- Therefore, we can rewrite \( x \) and \( y \):
\[ x = 2t - (-t) = 2t + t = 3t \]
\[ y = t^3 + t^2(-t) = t^3 - t^3 = 0 \]

### Step 3: Analyze the function for \( t \geq 0 \)
For \( t \geq 0 \):
- The absolute value \( |t| \) is simply \( t \).
- Thus, we can rewrite \( x \) and \( y \):
\[ x = 2t - t = t \]
\[ y = t^3 + t^2(t) = t^3 + t^3 = 2t^3 \]

### Step 4: Find \( \frac{dy}{dx} \)
To find the derivative \( \frac{dy}{dx} \), we will use the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

#### For \( t < 0 \):
- Calculate \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = 3 \]
- Calculate \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = 0 \]
- Therefore:
\[ \frac{dy}{dx} = \frac{0}{3} = 0 \]

#### For \( t \geq 0 \):
- Calculate \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = 1 \]
- Calculate \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = 6t^2 \]
- Therefore:
\[ \frac{dy}{dx} = \frac{6t^2}{1} = 6t^2 \]

### Step 5: Evaluate at \( t = 0 \)
Now, we need to evaluate \( \frac{dy}{dx} \) at \( t = 0 \):
- From the left (as \( t \) approaches 0 from negative side), \( \frac{dy}{dx} = 0 \).
- From the right (as \( t \) approaches 0 from positive side), \( \frac{dy}{dx} = 6(0)^2 = 0 \).

Since both limits are equal, we conclude that:
\[ \frac{dy}{dx} \text{ at } t = 0 = 0 \]

### Final Answer
The derivative of the function at \( t = 0 \) is:
\[ \boxed{0} \]