if ` x = a cos^3 t, y = b sin^3 t, ` then ` (dy)/(dx)` =

Options

A. `3 ( b//a) tan^2 t sec ^2 t `
B. ` -(a//b) cot t`
C. ` -(b//a) tan t `
D. None

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = a \cos^3 t\) and \(y = b \sin^3 t\), we will use the chain rule for differentiation. Here are the steps:

### Step 1: Differentiate \(y\) with respect to \(t\)
We start by differentiating \(y\) with respect to \(t\):

\[ y = b \sin^3 t \]

Using the chain rule, we have:

\[ \frac{dy}{dt} = b \cdot 3 \sin^2 t \cdot \frac{d}{dt}(\sin t) = b \cdot 3 \sin^2 t \cdot \cos t \]

Thus,

\[ \frac{dy}{dt} = 3b \sin^2 t \cos t \]

### Step 2: Differentiate \(x\) with respect to \(t\)
Next, we differentiate \(x\) with respect to \(t\):

\[ x = a \cos^3 t \]

Using the chain rule again, we have:

\[ \frac{dx}{dt} = a \cdot 3 \cos^2 t \cdot \frac{d}{dt}(\cos t) = a \cdot 3 \cos^2 t \cdot (-\sin t) \]

Thus,

\[ \frac{dx}{dt} = -3a \cos^2 t \sin t \]

### Step 3: Find \(\frac{dy}{dx}\)
Now, we can find \(\frac{dy}{dx}\) using the formula:

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

Substituting the expressions we found:

\[ \frac{dy}{dx} = \frac{3b \sin^2 t \cos t}{-3a \cos^2 t \sin t} \]

### Step 4: Simplify the expression
Now, we simplify the expression:

\[ \frac{dy}{dx} = \frac{3b \sin^2 t \cos t}{-3a \cos^2 t \sin t} = \frac{b \sin t}{-a \cos t} \]

This simplifies to:

\[ \frac{dy}{dx} = -\frac{b}{a} \tan t \]

### Final Answer
Thus, the final result is:

\[ \frac{dy}{dx} = -\frac{b}{a} \tan t \]

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