If `x=2t-3t^(2)` and `y=6t^(3)` then `(dy)/(dx)` at point `(-1,6)` is

Options

A. `(1)/(3)`
B. `-(9)/(2)`
C. `(8)/(3)`
D. 0

Answer

To find \(\frac{dy}{dx}\) at the point \((-1, 6)\) given the parametric equations \(x = 2t - 3t^2\) and \(y = 6t^3\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

1. **Differentiate \(y\) with respect to \(t\)**:
\[ y = 6t^3 \implies \frac{dy}{dt} = 18t^2 \]

2. **Differentiate \(x\) with respect to \(t\)**:
\[ x = 2t - 3t^2 \implies \frac{dx}{dt} = 2 - 6t \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{18t^2}{2 - 6t} \]

### Step 3: Find the value of \(t\) at the point \((-1, 6)\)

To find \(t\) that corresponds to the point \((-1, 6)\), we will use the equation for \(y\):
\[ y = 6t^3 \implies 6 = 6t^3 \implies t^3 = 1 \implies t = 1 \]

### Step 4: Verify \(t\) using \(x\)

Now, we will verify if \(t = 1\) gives us the corresponding \(x\) value:
\[ x = 2(1) - 3(1^2) = 2 - 3 = -1 \]
This confirms that at \(t = 1\), \(x = -1\) and \(y = 6\).

### Step 5: Substitute \(t = 1\) into \(\frac{dy}{dx}\)

Now we can substitute \(t = 1\) into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{18(1^2)}{2 - 6(1)} = \frac{18}{2 - 6} = \frac{18}{-4} = -\frac{9}{2} \]

### Conclusion

Thus, the value of \(\frac{dy}{dx}\) at the point \((-1, 6)\) is:
\[ \boxed{-\frac{9}{2}} \]