Question

x=a(cosθ+θsinθ), y=a(sinθ-θcosθ),Find dy/dx :(Wherever y is defined as a function of x and dx/dt or dx/dθ≠0)

Answer 1

x = a(cosθ + θsinθ)

differentiate with respect to θ,

dx/dθ = d{a(cosθ + θsinθ)}/dθ

= a[d(cosθ)/dθ + d(θsinθ)/dθ]

= a[ -sinθ + θ.cosθ + sinθ]

= aθ.cosθ ........(1)

again, y = a(sinθ - θ.cosθ)

differentiate with respect to θ,

dy/dθ = d{a(sinθ - θcosθ)}/dθ

= a[dsinθ/dθ - d(θ.cosθ)/dθ ]

= a[cosθ - θ(-sinθ) - cosθ ]

= a[ cosθ + θsinθ - cosθ]

= aθ.sinθ ...........(2)

now, dy/dx = {dy/dθ}/{dx/dθ}

= aθ.sinθ/aθ.cosθ

= tanθ

hence, dy/dx = tanθ

Answer 2

HELLO DEAR,



GIVEN:-

x = a(cosθ + θsinθ)

differentiate x w.r.t θ,

dx/dθ = d{a(cosθ + θsinθ)}/dθ

= a[d(cosθ)/dθ + d(θsinθ)/dθ]

= a[ -sinθ + θcosθ + sinθ]

= aθcosθ -----------( 1 )

AND,
y = a(sinθ - θcosθ)

differentiate with respect to θ,

dy/dθ = d{a(sinθ - θcosθ)}/dθ

= a[dsinθ/dθ - d(θcosθ)/dθ ]

= a[cosθ - θ(-sinθ) - cosθ ]

= a[ cosθ + θsinθ - cosθ]

= aθsinθ ----------( 2 )


now,
divide-------( 1 ) by -----( 2 )

dy/dx = {dy/dθ}/{dx/dθ}

= {aθsinθ}/{aθcosθ}

= tanθ

therefore,dy/dx = tanθ


I HOPE ITS HELP YOU DEAR,
THANKS