If x = 3 tan t and y = 3 sec t, then the value of at is

Engineering

Mathematics

Derivative of Standard Functions

Parametric Differentiation

Question

If x = 3 tan t and y = 3 sec t, then the value of at $t = \frac{d^{2} y}{{dx}^{2}}, \frac{π}{4}$ is

$\frac{1}{6 \sqrt{2}}$

$\frac{1}{6}$

$\frac{1}{3 \sqrt{2}}$

$\frac{3}{2 \sqrt{2}}$

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\frac{dx}{dt} = 3 \sec^{3} t$

$\frac{dy}{dt} = 3 sect tant$

$\frac{dy}{dx} = \frac{tant}{sect} = sint$

$= \frac{cost}{3 \sec^{2} t} = \frac{\cos^{3} t}{3} = \frac{1}{3 .2 \sqrt{2}} = \frac{1}{6 \sqrt{2}}$