Question

If x=3sint-sin3t and y=3cost-cos3t find d^2y/dx^2 at t=π/3

Answer 1

Answer:

$\frac{d^2y}{dx^2}=\frac{-16}{27}$

Step-by-step explanation:

$x=3sint-sin3t\\\\\frac{dx}{dt}=3cost-3cos3t\\\\\frac{dx}{dt}=3[cost-cos3t]\\\\\frac{dx}{dt}=3[-2.sin\frac{4t}{2}.sin\frac{-2t}{2}]\\\\\frac{dx}{dt}=3[-2.sin2t.sin(-t)]\\\\\frac{dx}{dt}=6sin2t.sint$

$y=3cost-cos3t\\\\\frac{dy}{dt}=-3sint+3sin3t\\\\\frac{dy}{dt}=3[sin3t-sint]\\\\\frac{dy}{dt}=3[2.cos\frac{4t}{2}.sin\frac{2t}{2}]\\\\\frac{dy}{dt}=6.cos2t.sint$

$Now,\\\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\\\\frac{dy}{dx}=\frac{6.cos2t.sint}{6sin2t.sint}=cot2t\\\\\frac{d^2y}{dx^2}=\frac{d(cot2t)}{dx}\\\\\frac{d^2y}{dx^2}=(-2cosec^22t).\frac{dt}{dx}\\\\\frac{d^2y}{dx^2}=(-2cosec^22t).\frac{1}{6.sin2t.sint}\\\\\frac{d^2y}{dx^2}=\frac{-1}{3.sin^{3}2t.sint}$

$when \:t=\frac{\pi}{3}\\\frac{d^2y}{dx^2}=\frac{-1}{3.sin^{3}\frac{2\pi}{3}.sin\frac{\pi}{3}}\\\\\frac{d^2y}{dx^2}=\frac{-1}{3.(\frac{\sqrt3}{2})^3.\frac{\sqrt3}{2}}\\\\\frac{d^2y}{dx^2}=\frac{-1}{3.(\frac{3\sqrt3}{8}).\frac{\sqrt3}{2}}\\\\\frac{d^2y}{dx^2}=\frac{-1}{3.(\frac{3\sqrt3}{8}).\frac{\sqrt3}{2}}\\\\\frac{d^2y}{dx^2}=\frac{-16}{27}$

Answer 2

-16/27

is the answer od the question