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

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 formula:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

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

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

Using the quotient rule \(\frac{u}{v}\) where \(u = 3at^2\) and \(v = 1 + t^3\):

\[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
- \(\frac{du}{dt} = 6at\)
- \(\frac{dv}{dt} = 3t^2\)

Now substituting into the quotient rule:

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

Expanding the numerator:

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

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

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

Using the quotient rule again where \(u = 3at\) and \(v = 1 + t^3\):

\[ \frac{dx}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
- \(\frac{du}{dt} = 3a\)
- \(\frac{dv}{dt} = 3t^2\)

Now substituting into the quotient rule:

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

Expanding the numerator:

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

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

Now we can find \(\frac{dy}{dx}\):

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

The \((1 + t^3)^2\) terms cancel out:

\[ = \frac{3at(2 - t^3)}{3a(1 - 2t^3)} \]

The \(3a\) terms also cancel out:

\[ = \frac{t(2 - t^3)}{1 - 2t^3} \]

### Final Result

Thus, the final result is:

\[ \frac{dy}{dx} = \frac{t(2 - t^3)}{1 - 2t^3} \]

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