If x=sinθ,y=sin^3θ then (d^2y)/(dx^2) at θ = (π)/2 is ….

Correct answer: (B) 6. We have , x=sinθ,y=sin^3θ ∴dy/dx=dy/(dθ)dx/(dθ)=(3sin^2θcosθ)/(cosθ) ⇒dy/dx=3sin^2θ Now, (d^2y)/(dx^2)=32sinθcosθ(dθ)/dx =32sinθcosθ1/(cosθ)=6sinθ at θ=(π)/2,(d^2y)/(dx^2)=6sin⁡(π)/2=6×1=6

Q76 / 150 · MHT CET 2019 (Shift 2) · Mathematics

If $x = s i n θ, y = {s i n}^{3} θ$ then $\frac{d^{2} y}{d x^{2}}$ at $θ$ = $\frac{π}{2}$ is ….

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

Solution

We have ,

x = s i n θ, y = {s i n}^{3} θ

∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 {s i n}^{2} θ (c o s θ)}{c o s θ}

\Rightarrow \frac{d y}{d x} = 3 {s i n}^{2} θ

Now,

\frac{d^{2} y}{d x^{2}} = 3 (2 s i n θ c o s θ) \frac{d θ}{d x}

= 3 (2 s i n θ c o s θ) \frac{1}{c o s θ} = 6 s i n θ

θ = \frac{π}{2}, \frac{d^{2} y}{d x^{2}} = 6 \sin (\frac{π}{2}) = 6 \times 1 = 6

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

If x=sinθ,y=sin3θ then d2ydx2 at θ = π2 is ….

Solution

Relevant Questions

If $x = s i n θ, y = {s i n}^{3} θ$ then $\frac{d^{2} y}{d x^{2}}$ at $θ$ = $\frac{π}{2}$ is ….