If x=3 tant and y=3 sect, then the value of (d^2y)/(dx^2) at t=(π)/4, is:

Correct answer: (B) 1/(6√2). Given x=3tant and y=3sect ⇒dx/dt=3sec^2t and dy/dt=3secttant ⇒dy/dx=dydt/dxdt ⇒dy/dx=sint ⇒(d^2y)/(dx^2)=cost·dt/dx=cost ·(cos^2t)/3=(cos^3t)/3 At t=(π)/4 (d^2y)/(dx^2)=1/(6√2)

Q47 / 90 · JEE Main 2019 (09 Jan Shift 2) · Mathematics

If $x = 3 t a n t$ and $y = 3 s e c t,$ then the value of $\frac{d^{2} y}{d x^{2}}$ at $t = \frac{π}{4},$ is:

A
$\frac{1}{6}$
Correct answer
B
$\frac{1}{6 \sqrt{2}}$
C
$\frac{1}{3 \sqrt{2}}$
D
$\frac{3}{2 \sqrt{2}}$

Solution

Given $x = 3 \tan t$ and $y = 3 \sec t$

$\Rightarrow \frac{d x}{d t} = 3 \sec^{2} t$ and $\frac{d y}{d t} = 3 \sec t \tan t$

$\Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$\Rightarrow \frac{d y}{d x} = \sin t$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \cos t \cdot \frac{d t}{d x} = \cos t \cdot \frac{\cos^{2} t}{3} = \frac{\cos^{3} t}{3}$

At $t = \frac{π}{4}$

$\frac{d^{2} y}{d x^{2}} = \frac{1}{6 \sqrt{2}}$

Hence, the correct answer is option: 'B'.

If x=3 tant and y=3 sect, then the value of d2ydx2 at t=π4, is:

Solution

More Differentiation Questions

If $x = 3 t a n t$ and $y = 3 s e c t,$ then the value of $\frac{d^{2} y}{d x^{2}}$ at $t = \frac{π}{4},$ is: