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

Answer: (B)

$\frac {4√2}{3a}$

Answer 2

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
$\frac {dx}{dt}$ = $(\frac {d^2y}{dx^2})$tπ/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) $\frac {4√2}{3a}$