Question

X=a(cos theta+theta sin theta) y=a(sin theta-theta cos theta) find d2y/dx2

Answer 1

Step-by-step explanation:

x = a(cos Q + Q sin Q)

y = a(sin Q - Q cos Q)

dx/dQ = a(-sinQ + sinQ + QcosQ)

dx/dQ = aQcosQ ------(1)

dy/dQ = a(cosQ - (1*cosQ+Q*-sinQ))

= a(cosQ - cosQ + QsinQ)

= aQsinQ -------(2)

समी.(2) / (1) से

dy/dx = tanQ

d²y/dx² = sec²Q