Question

The second derivative of a sin 3t w.r.t. a cos 3t at t =π/4 is

• A. \(- \frac {4√2}{3a}\)
• B. \(\frac {4√2}{3a}\)
• C. \(\frac {4√3}{3a}\)
• D. 12a

Answer 1

The Correct Option is B

Let y = asin 3t , x=acos 3t;
\(\frac {dy}{dx}\) = 3a sin 2t cos t
\(\frac {dx}{dt}\) = −3a cos 2t sin t
∴ \(\frac {dx}{dy}\) = −\(\frac {3a\  cos^2\ t\  sin\  t}{3a\ sin^2\ t\  cos\ t}\)
\(\frac {dx}{dy}\) = − \(\frac {cos\  t}{sin \ t}\)=−tan t
Differentiating with respect to x
\(\frac {d^2y}{dx^2}\) = −sec 2t
\(\frac {dx}{dt}\)= -\(\frac {sec^2\ t}{-3a\ cos^2\ t sin \ t}\)
\(\frac {dx}{dt}\) = \(\frac {1}{3}\)a cos⁴t sin t
\(\frac {dx}{dt}\) = \((\frac {d^2y}{dx^2})\)_t^π/4
\(\frac {dx}{dt}\) = 3a.\(\frac {1}{3}\)a\((\frac {1}{\sqrt 2})^4\).\(\frac {1}{\sqrt 2}\)
\(\frac {dx}{dt}\) = \(\frac {(\sqrt2)^5}{3a}\)
\(\frac {dx}{dt}\) =\(\frac {4√2}{3a}\)
Therefore the correct option is (B) \(\frac {4√2}{3a}\)