Question

X=a(t-sint), y=a(1+cost), prove that d2y/dx2 = (1/4a)cosec^4(t/2)

Answer 1

Answer:

$\frac{d^2y}{dx^2}=\frac{1}{4a}cosec^4\frac{t}{2}}$

Step-by-step explanation:

Formula used:

$cosA=1-2sin^2\frac{A}{2}$

$sinA=2\:sin\frac{A}{2}\:cos\frac{A}{2}$

$x=a(t-sint)$

$\frac{dx}{dt}=a(1-cost)$

$\frac{dx}{dt}=a(2\:sin^2\frac{t}{2})$

$\frac{dx}{dt}=2a\:sin^2\frac{t}{2}$

$y=a(1+cost)$

$\frac{dy}{dt}=a(-sint)$

$\frac{dy}{dt}=-asint$

$\frac{dy}{dt}=-2a\:sin\frac{t}{2}\:cos\frac{t}{2}$

Now,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\frac{dy}{dx}=\frac{-2a\:sin\frac{t}{2}\:cos\frac{t}{2}}{2a\:sin^2\frac{t}{2}}$

$\frac{dy}{dx}=\frac{-cos\frac{t}{2}}{sin\frac{t}{2}}$

$\frac{dy}{dx}=-cot\frac{t}{2}}$

Differentiate with respect to x

$\frac{d^2y}{dx^2}=\frac{1}{2}cosec^2\frac{t}{2}}\:\frac{dt}{dx}$

$\frac{d^2y}{dx^2}=\frac{1}{2}cosec^2\frac{t}{2}}\:\frac{1}{\frac{dx}{dt}}$

$\frac{d^2y}{dx^2}=\frac{1}{2}cosec^2\frac{t}{2}}\:\frac{1}{2a\:sin^2\frac{t}{2}}$

$\frac{d^2y}{dx^2}=\frac{1}{2}cosec^4\frac{t}{2}}\:\frac{1}{2a}$

$\frac{d^2y}{dx^2}=\frac{1}{4a}cosec^4\frac{t}{2}}$