Question

if x=a(θ + sinθ) y= a(1+ cosθ) prove d2y/dx2 = -a/y^2

Answer 1

Step-by-step explanation:

x = a(Q + sinQ)

dx/dQ = a(1 + cosQ)

y = a(1 + cosQ)

dy/dQ = -asinQ

dy/dx = -asinQ/a(1+cosQ) (समी.(1)से)

= -sinQ/1+cosQ

= -asinQ/y

Answer 2

Step-by-step explanation:

given

$x = a( \theta + sin \theta) \\ y = a(1 + cos \theta) \\ \\ \frac{dx}{d \theta} = a(1 + cos \theta) \\ \\ \frac{dy}{d \theta} = a( - sin \theta) \\ \\ \frac{dx}{dx} = \frac{a( - sin \theta)}{a(1 + cos \theta)} \\ \\ = \frac{ - sin \theta}{1 + cos \theta}$

$\\ \\ \frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} ( \frac{dy}{dx} ) \\ \\ = - \frac{d}{dx} ( \frac{sin \theta}{1 + cos \theta} ) \\ \\ = - ( \frac{(1 + cos \theta)cos \theta \frac{d \theta}{dx} - sin \theta( - sin \theta) \frac{d \theta}{dx} }{( {1 + cos \theta)}^{2} } ) \\ \\ = - ( \frac{cos \theta \frac{d \theta}{dx } + {cos}^{2} \theta \frac{d \theta}{dx} + {sin}^{2} \theta \frac{d \theta}{dx} }{( {1 + cos \theta)}^{2} } ) \\ \\ = - ( \frac{cos \theta \frac{d \theta}{dx} + \frac{d \theta}{dx} }{( {1 + cos \theta)}^{2} } )$

$= - ( \frac{(1 + cos \theta) \frac{d \theta}{dx} }{( {1 + cos \theta)}^{2} } ) \\ \\ = - ( \frac{ \frac{d \theta}{dx} }{1 + cos \theta} ) \\ \\ = - ( \frac{ \frac{1}{a(1 + cos \theta)} }{1 + cos \theta} ) \\ \\ = \frac{ - 1}{a(1 + cos \theta)} \\ \\ = \frac{ - {a}^{} }{ {a}^{2}( {1 + cos \theta)}^{2} } \\ \\ = \frac{ - a}{ {y}^{2} }$