Question

If \( y = a \sin^3 t \), \( x = a \cos^3 t \), then \( \frac{dy}{dx} \) at \( t = \frac{3\pi}{4} \) is:

• A. \( \frac{1}{\sqrt{3}} \)
• B. \( -\sqrt{3} \)
• C. \( 1 \)
• D. \( -1 \)

Hint:

When using the chain rule to differentiate implicit functions, always make sure to simplify the derivatives before substituting the given value of \( t \). In trigonometric problems, recall the key values of the trigonometric functions at standard angles.

Answer 1

The Correct Option is C

We are given: \[ y = a \sin^3 t, \quad x = a \cos^3 t \] To find \( \frac{dy}{dx} \), we use the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] First, calculate \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \): 1. **Differentiating \( y = a \sin^3 t \) with respect to \( t \):** \[ \frac{dy}{dt} = 3a \sin^2 t \cdot \cos t \] 2. **Differentiating \( x = a \cos^3 t \) with respect to \( t \):** \[ \frac{dx}{dt} = -3a \cos^2 t \cdot \sin t \] Thus, we have: \[ \frac{dy}{dx} = \frac{3a \sin^2 t \cdot \cos t}{-3a \cos^2 t \cdot \sin t} = -\frac{\sin t}{\cos t} = -\tan t \] Now, substitute \( t = \frac{3\pi}{4} \) into the equation: \[ \frac{dy}{dx} = -\tan\left( \frac{3\pi}{4} \right) \] Since \( \tan\left( \frac{3\pi}{4} \right) = -1 \), we get: \[ \frac{dy}{dx} = -(-1) = 1 \] Thus, the correct answer is \( 1 \), which corresponds to option (3).