If `x=3cos t and y=5sint`, where t is a parameter, then `9(d^(2)y)/(dx^(2))` at `t=-(pi)/(6)` is equal to

Answer

To solve the problem, we need to find the value of \( 9 \frac{d^2y}{dx^2} \) at \( t = -\frac{\pi}{6} \) given the parametric equations \( x = 3 \cos t \) and \( y = 5 \sin t \).

### Step-by-Step Solution:

1. **Find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)**:
\[ y = 5 \sin t \implies \frac{dy}{dt} = 5 \cos t \]
\[ x = 3 \cos t \implies \frac{dx}{dt} = -3 \sin t \]

2. **Find \( \frac{dy}{dx} \)**:
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{5 \cos t}{-3 \sin t} = -\frac{5}{3} \cot t \]

3. **Find \( \frac{d^2y}{dx^2} \)**:
To find \( \frac{d^2y}{dx^2} \), we differentiate \( \frac{dy}{dx} \) with respect to \( t \) and divide by \( \frac{dx}{dt} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left(-\frac{5}{3} \cot t\right) \cdot \frac{1}{\frac{dx}{dt}} \]
First, we differentiate \( -\frac{5}{3} \cot t \):
\[ \frac{d}{dt} \left(-\frac{5}{3} \cot t\right) = -\frac{5}{3} (-\csc^2 t) = \frac{5}{3} \csc^2 t \]
Now, substituting \( \frac{dx}{dt} = -3 \sin t \):
\[ \frac{d^2y}{dx^2} = \frac{\frac{5}{3} \csc^2 t}{-3 \sin t} = -\frac{5}{9} \frac{\csc^2 t}{\sin t} = -\frac{5}{9} \frac{1}{\sin^3 t} \]

4. **Evaluate \( \frac{d^2y}{dx^2} \) at \( t = -\frac{\pi}{6} \)**:
We know that:
\[ \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \quad \text{and} \quad \csc\left(-\frac{\pi}{6}\right) = -2 \]
Therefore:
\[ \frac{d^2y}{dx^2} = -\frac{5}{9} \cdot \frac{1}{\left(-\frac{1}{2}\right)^3} = -\frac{5}{9} \cdot \frac{1}{-\frac{1}{8}} = -\frac{5}{9} \cdot (-8) = \frac{40}{9} \]

5. **Calculate \( 9 \frac{d^2y}{dx^2} \)**:
\[ 9 \frac{d^2y}{dx^2} = 9 \cdot \frac{40}{9} = 40 \]

### Final Answer:
Thus, the value of \( 9 \frac{d^2y}{dx^2} \) at \( t = -\frac{\pi}{6} \) is \( \boxed{40} \).