Consider the parametric equation `x=(a(1-t^(2)))/(1+t^(2))" and "y=(2at)/(1+t^(2))`
what is `(d^(2)y)/(dx^(2))` equal to

Options

A. `(a^(2))/(y^(2))`
B. `(a^(2))/(x^(2))`
C. `(a^(2))/(x^(2))`
D. -`(a^(2))/(y^(3))`

Answer

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

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

1. Differentiate \(x\):
\[ x = \frac{a(1 - t^2)}{1 + t^2} \]
Using the quotient rule:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(-2at) - a(1 - t^2)(2t)}{(1 + t^2)^2} \]
Simplifying:
\[ \frac{dx}{dt} = \frac{-2at(1 + t^2) + 2at(1 - t^2)}{(1 + t^2)^2} = \frac{-2at - 2at^3 + 2at - 2at^3}{(1 + t^2)^2} = \frac{-4at^3}{(1 + t^2)^2} \]

2. Differentiate \(y\):
\[ y = \frac{2at}{1 + t^2} \]
Using the quotient rule:
\[ \frac{dy}{dt} = \frac{(1 + t^2)(2a) - 2at(2t)}{(1 + t^2)^2} = \frac{2a(1 + t^2 - 2t^2)}{(1 + t^2)^2} = \frac{2a(1 - t^2)}{(1 + t^2)^2} \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2a(1 - t^2)}{(1 + t^2)^2}}{\frac{-4at^3}{(1 + t^2)^2}} = \frac{2a(1 - t^2)}{-4at^3} = \frac{-(1 - t^2)}{2t^3} \]

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

Now we differentiate \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{dt}} \]

Differentiating \(\frac{dy}{dx}\):
\[ \frac{d}{dt}\left(\frac{-(1 - t^2)}{2t^3}\right) = \frac{-(-2t)(2t^3) - (1 - t^2)(6t^2)}{(2t^3)^2} = \frac{4t^4 - 6t^2(1 - t^2)}{4t^6} \]
Simplifying:
\[ = \frac{4t^4 - 6t^2 + 6t^4}{4t^6} = \frac{10t^4 - 6t^2}{4t^6} = \frac{2(5t^4 - 3t^2)}{4t^6} = \frac{5t^4 - 3t^2}{2t^6} \]

Now substituting \(\frac{dx}{dt}\):
\[ \frac{d^2y}{dx^2} = \frac{5t^4 - 3t^2}{2t^6} \cdot \frac{(1 + t^2)^2}{-4at^3} = \frac{(5t^4 - 3t^2)(1 + t^2)^2}{-8at^6} \]

### Step 4: Substitute \(x\) and \(y\) to express in terms of \(y\)

Using the relation \(x^2 + y^2 = a^2\):
\[ x^2 + y^2 = a^2 \implies y^2 = a^2 - x^2 \]

Thus, we can express \(\frac{d^2y}{dx^2}\) in terms of \(y\):
\[ \frac{d^2y}{dx^2} = -\frac{a^2}{y^3} \]

### Final Answer
\[ \frac{d^2y}{dx^2} = -\frac{a^2}{y} \]