6. x=a(9-sin θ), y = a (1+cos θ)
It is given that x and y are parametrically connected by the equations,
x = a ( θ − sin θ ) (1)
And,
y = a ( 1 + cos θ ) (2)
Differentiate equation (2) with respect to θ .
d y d θ = d [ a ( 1 + cos θ ) ] d θ d y d θ = a ( d ( 1 ) d θ + d ( cos θ ) d θ ) d y d θ = a ( 0 − sin θ ) d y d θ = − a sin θ
Differentiate equation (1) with respect to θ .
d x d θ = a ( θ − sin θ ) d x d θ = d ( a θ ) d θ − d ( a sin θ ) d θ d x d θ = a − a cos θ d x d θ = a ( 1 − cos θ )
We know that,
d y d x = d y d θ d x d θ
Substitute the value of d y d θ and d x d θ .
d y d x = − a sin θ a ( 1 − cos θ ) d y d x = − sin θ ( 1 − cos θ ) d y d x = − 2 sin θ 2 cos θ 2 2 sin 2 θ 2 d y d x = − cot θ 2
Thus, the solution is d y d x = − cot θ 2 .
It is given that
And,
Differentiate equation (2) with respect to
Differentiate equation (1) with respect to
We know that,
Substitute the value of
Thus, the solution is
