If `x=a(t-sint), y=a(1-cost)` then find `(d^2y)/(dx^2)`.

Answer

To find \(\frac{d^2y}{dx^2}\) given the parametric equations \(x = a(t - \sin t)\) and \(y = a(1 - \cos t)\), we will follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)

1. Differentiate \(x\):
\[ \frac{dx}{dt} = a\left(1 - \cos t\right) \]

2. Differentiate \(y\):
\[ \frac{dy}{dt} = a\sin t \]

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

Using the chain rule, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a\sin t}{a(1 - \cos t)} = \frac{\sin t}{1 - \cos t} \]

### Step 3: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)

Now we need to differentiate \(\frac{dy}{dx}\) with respect to \(t\):
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{\sin t}{1 - \cos t}\right) \]

Using the quotient rule:
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
where \(u = \sin t\) and \(v = 1 - \cos t\).

Calculating \(du/dt\) and \(dv/dt\):
- \(du/dt = \cos t\)
- \(dv/dt = \sin t\)

Now substituting into the quotient rule:
\[ \frac{d}{dt}\left(\frac{\sin t}{1 - \cos t}\right) = \frac{(1 - \cos t)(\cos t) - (\sin t)(\sin t)}{(1 - \cos t)^2} \]
Simplifying the numerator:
\[ = \frac{\cos t - \cos^2 t - \sin^2 t}{(1 - \cos t)^2} = \frac{\cos t - 1}{(1 - \cos t)^2} \]

### Step 4: Find \(\frac{d^2y}{dx^2}\)

Using the chain rule again, we have:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]

We already calculated \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\). Now we need \(\frac{dt}{dx}\):
\[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{a(1 - \cos t)} \]

Thus,
\[ \frac{d^2y}{dx^2} = \left(\frac{\cos t - 1}{(1 - \cos t)^2}\right) \cdot \left(\frac{1}{a(1 - \cos t)}\right) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{\cos t - 1}{a(1 - \cos t)^3} \]

### Final Answer:
\[ \frac{d^2y}{dx^2} = \frac{\cos t - 1}{a(1 - \cos t)^3} \]