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

Options

A. `y//x`
B. `-y//x`
C. `x//y`
D. `-x//y`

Answer

To find \(\frac{dy}{dx}\) 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 \(x\) with respect to \(t\)

We start with the equation for \(x\):
\[ x = \frac{a(1 - t^2)}{1 + t^2} \]

Using the quotient rule, where \(u = a(1 - t^2)\) and \(v = 1 + t^2\):
\[ \frac{dx}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
- \(\frac{du}{dt} = a(-2t)\)
- \(\frac{dv}{dt} = 2t\)

Now substituting into the quotient rule:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(-2at) - a(1 - t^2)(2t)}{(1 + t^2)^2} \]
\[ = \frac{-2at - 2at^3 - 2at + 2at^3}{(1 + t^2)^2} \]
\[ = \frac{-4at}{(1 + t^2)^2} \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Now we differentiate \(y\):
\[ y = \frac{2at}{1 + t^2} \]

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

### Step 3: Find \(\frac{dy}{dx}\)

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\frac{2a(1 - t^2)}{(1 + t^2)^2}}{\frac{-4at}{(1 + t^2)^2}} \]
The \((1 + t^2)^2\) cancels out:
\[ \frac{dy}{dx} = \frac{2a(1 - t^2)}{-4at} \]
\[ = \frac{1 - t^2}{-2t} \]
\[ = -\frac{1 - t^2}{2t} \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\frac{1 - t^2}{2t} \]