If `x=(3at)/ (1+t^3), y=(3at^2)/(1+t^3), then dy/dx` is

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = \frac{3at}{1+t^3}\) and \(y = \frac{3at^2}{1+t^3}\), we will use the chain rule for differentiation.

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

Given:
\[ x = \frac{3at}{1+t^3} \]

Using the quotient rule:
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
where \(u = 3at\) and \(v = 1 + t^3\).

First, we find \(u'\) and \(v'\):
- \(u' = 3a\)
- \(v' = 3t^2\)

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

Simplifying:
\[ \frac{dx}{dt} = \frac{3a(1+t^3) - 9at^3}{(1+t^3)^2} = \frac{3a - 6at^3}{(1+t^3)^2} \]

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

Given:
\[ y = \frac{3at^2}{1+t^3} \]

Using the quotient rule again:
\[ \frac{dy}{dt} = \frac{(6at)(1+t^3) - (3at^2)(3t^2)}{(1+t^3)^2} \]

Simplifying:
\[ \frac{dy}{dt} = \frac{6at + 6at^3 - 9at^4}{(1+t^3)^2} = \frac{6at - 3at^4}{(1+t^3)^2} = \frac{3at(2 - t^3)}{(1+t^3)^2} \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the values we found:
\[ \frac{dy}{dx} = \frac{\frac{3at(2 - t^3)}{(1+t^3)^2}}{\frac{3a - 6at^3}{(1+t^3)^2}} \]

The \((1+t^3)^2\) cancels out:
\[ \frac{dy}{dx} = \frac{3at(2 - t^3)}{3a - 6at^3} \]

### Step 4: Simplify the expression

Factoring out \(3a\) from the denominator:
\[ \frac{dy}{dx} = \frac{3at(2 - t^3)}{3a(1 - 2t^3)} = \frac{t(2 - t^3)}{1 - 2t^3} \]

Thus, the final result is:
\[ \frac{dy}{dx} = \frac{t(2 - t^3)}{1 - 2t^3} \]