If `x=alpha"sin"^3 t` and `y=alpha cos^3t`, then what is the value of `(dy)/(dx)`?

Options

A. tan t
B. `-cot t`
C. `-tan t`
D. `cot t`

Answer

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

### Step 1: Differentiate \(y\) with respect to \(t\)
We start with the equation for \(y\):
\[ y = \alpha \cos^3 t \]
Using the chain rule, we differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = \alpha \cdot 3 \cos^2 t \cdot \frac{d}{dt}(\cos t) \]
Since \(\frac{d}{dt}(\cos t) = -\sin t\), we have:
\[ \frac{dy}{dt} = \alpha \cdot 3 \cos^2 t \cdot (-\sin t) = -3\alpha \cos^2 t \sin t \]

### Step 2: Differentiate \(x\) with respect to \(t\)
Now, we differentiate \(x\):
\[ x = \alpha \sin^3 t \]
Using the chain rule, we differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = \alpha \cdot 3 \sin^2 t \cdot \frac{d}{dt}(\sin t) \]
Since \(\frac{d}{dt}(\sin t) = \cos t\), we have:
\[ \frac{dx}{dt} = \alpha \cdot 3 \sin^2 t \cdot \cos t \]

### Step 3: Apply the chain rule to find \(\frac{dy}{dx}\)
Now, we can find \(\frac{dy}{dx}\) using the relationship:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \frac{-3\alpha \cos^2 t \sin t}{3\alpha \sin^2 t \cos t} \]

### Step 4: Simplify the expression
We can simplify this expression:
\[ \frac{dy}{dx} = \frac{-3\alpha \cos^2 t \sin t}{3\alpha \sin^2 t \cos t} = \frac{-\cos t}{\sin t} = -\cot t \]

### Final Result
Thus, the value of \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\cot t \]

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