Question

If \( x = 5 \tan t \) and \( y = 5 \sec t \), then \( \frac{dy}{dx} \) at \( t = \frac{\pi}{3} \) is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( \frac{1}{\sqrt{3}} \)
• E. \( \sqrt{3} \)

Hint:

When given functions involving trigonometric identities, use differentiation rules such as the chain rule and standard trigonometric derivatives to simplify the process. In this case, use \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) and trigonometric values for specific angles to evaluate the final result.

Answer 1

The Correct Option is C

We are given the following: \[ x = 5 \tan t \quad {and} \quad y = 5 \sec t. \] We need to find \( \frac{dy}{dx} \) at \( t = \frac{\pi}{3} \).
Step 1: Differentiate \( x \) and \( y \) with respect to \( t \): \[ \frac{dx}{dt} = \frac{d}{dt} \left( 5 \tan t \right) = 5 \sec^2 t. \] \[ \frac{dy}{dt} = \frac{d}{dt} \left( 5 \sec t \right) = 5 \sec t \tan t. \] Step 2: Use the chain rule to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{5 \sec t \tan t}{5 \sec^2 t} = \frac{\tan t}{\sec t}. \] Since \( \frac{\tan t}{\sec t} = \sin t \), we have: \[ \frac{dy}{dx} = \sin t. \] Step 3: Now, evaluate \( \frac{dy}{dx} \) at \( t = \frac{\pi}{3} \): \[ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}. \] Thus, the correct answer is option (C).