FIND d2y/dx2 FROM THE FOLLOWING PARAMETRIC EQUATIONS: x=6 - 2t-3 y= 1- 6t-2

Solution:
1. Given parametric equations:
- x = 6 - 2t - 3 = 3 - 2t
- y = 1 - 6t - 2 = -1 - 6t

2. Find the first derivatives with respect to t:
- \frac{dx}{dt} = \frac{d}{dt}(3 - 2t) = -2
- \frac{dy}{dt} = \frac{d}{dt}(-1 - 6t) = -6

3. Find the first derivative \frac{dy}{dx} using the chain rule:
- \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-6}{-2} = 3

4. Differentiate \frac{dy}{dx} with respect to t to find \frac{d}{dt}\left(\frac{dy}{dx}\right):
- Since \frac{dy}{dx} = 3 is constant, \frac{d}{dt}\left(\frac{dy}{dx}\right) = 0

5. Use the formula for the second derivative of a parametric equation:
- \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}
- Substitute the known values:
- \frac{d^2y}{dx^2} = \frac{0}{-2} = 0

6. The second derivative \frac{d^2y}{dx^2} is:
- \frac{d^2y}{dx^2} = 0