Find `(d^(2)y)/(dx^(2))` in the following
`x=at^(2),y=2at`

Answer

To find \(\frac{d^2y}{dx^2}\) for the given parametric equations \(x = at^2\) and \(y = 2at\), we will follow these steps:

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

Given:
- \(x = at^2\)
- \(y = 2at\)

We differentiate both equations with respect to \(t\):

\[ \frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at \]

\[ \frac{dy}{dt} = \frac{d}{dt}(2at) = 2a \]

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

Using the chain rule, we can find \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} \]

### Step 3: Find \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(\frac{dy}{dx}\) with respect to \(x\):

Using the formula for the second derivative in parametric form:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]

First, we differentiate \(\frac{dy}{dx}\) with respect to \(t\):

\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2} \]

Next, we need \(\frac{dt}{dx}\):

From \(\frac{dx}{dt} = 2at\), we find:

\[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2at} \]

Now substituting back into the formula for \(\frac{d^2y}{dx^2}\):

\[ \frac{d^2y}{dx^2} = -\frac{1}{t^2} \cdot \frac{1}{2at} = -\frac{1}{2at^3} \]

### Final Answer

Thus, the second derivative \(\frac{d^2y}{dx^2}\) is:

\[ \frac{d^2y}{dx^2} = -\frac{1}{2at^3} \]