Question

If x =3cos⁡θ-cos⁡3θ, y = 3sin⁡θ-sin⁡3θthen dy/dx = ​

Answer 1

GIVEN :–

▪︎ x = 3cos⁡θ - cos⁡3θ & y = 3sin⁡θ - sin⁡3θ

TO FIND :–

• dy/dx = ?

SOLUTION :–

• Differential of 'x' with respect to 'θ' –

=> dx/dθ = 3( -sinθ) - [ - sin(3θ)] (3)

=> dx/dθ = - 3 sinθ + 3 sin(3θ) ____eq.(1)

• Differential of 'y' with respect to 'θ' –

=> dy/dθ = 3(cosθ) - [cos(3θ)] (3)

=> dy/dθ = 3(cosθ) - 3 cos(3θ) ____eq.(2)

[ ∵ dy/dx = (dy/dθ)/(dx/dθ) ]

• Using eq.(1) & eq.(2) –

=> dy/dx = [ 3(cosθ) - 3 cos(3θ) ] / [ - 3 sinθ + 3 sin(3θ) ]

=> dy/dx = 3 [ cos(θ) - cos(3θ)] / 3 [ sin(3θ) - sin(θ) ]

=> dy/dx = [cos(θ) - cos(3θ)] / [ sin(3θ) - sin(θ) ]

USED IDENTITY :–

(1) d(sinθ)/dθ = cosθ

(2) d(cosθ)/dθ = - sinθ

(3) d(kθ)/dθ = k [ k → constant ]

Answer 2

x = 3cos⁡θ - cos⁡3θ & y = 3sin⁡θ - sin⁡3θ

TO FIND :–

• dy/dx = ?

SOLUTION :–

• Differential of 'x' with respect to 'θ' –

=> dx/dθ = 3( -sinθ) - [ - sin(3θ)] (3)

=> dx/dθ = - 3 sinθ + 3 sin(3θ) ____eq.(1)

• Differential of 'y' with respect to 'θ' –

=> dy/dθ = 3(cosθ) - [cos(3θ)] (3)

=> dy/dθ = 3(cosθ) - 3 cos(3θ) ____eq.(2)

[ ∵ dy/dx = (dy/dθ)/(dx/dθ) ]

• Using eq.(1) & eq.(2) –

=> dy/dx = [ 3(cosθ) - 3 cos(3θ) ] / [ - 3 sinθ + 3 sin(3θ) ]

=> dy/dx = 3 [ cos(θ) - cos(3θ)] / 3 [ sin(3θ) - sin(θ) ]

=> dy/dx = [cos(θ) - cos(3θ)] / [ sin(3θ) - sin(θ) ]