Question

If x=a cos theta, y=b sin theta , then d³y/dx³ is :-

Answer 1

Answer:

differentiate y/x one by three times

and u will got the answer

Answer 2

$\boxed{\textbf{\large{Given}}}$

x = x=a cos θ,

y=b sin θ

$\boxed{\textbf{\large{To find}}}$

$\frac{ {d}^{3} y}{d {x}^{3} }$

$\boxed{\textbf{\large{Explanation}}}$

x = a cos θ, y = b sinθ

◾first differentiate x = a cos θ

wrt to parameter θ

dx / d θ

= d / d theta ( a cos θ)

= a ( -sin θ)

= - a sin θ

◾Now, differentiate y = b sin θ

wrt θ

dy / d θ

= d / d θ ( b sin θ)

= b ( cos θ)

= b cos θ

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★ we know ,

(dy / d x )

=( dy / dθ ) / (dx / dθ)

therefor ,

( dy / dx )

= ( b cos θ ) / (- a sin θ)

=- ( (b / a ) x ( cot θ))

( dy /dx ) = - ( (b / a ) x ( cot θ))

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◾Now, again differentiate wrt x

(d^2 y / d x^2 )

= -( ( b / a ) (d / dx ( cot θ)))

= - (( b / a) ( - cosec^2 θ)

x ( d θ / dx)

= ( b / a) ( cosec ^2 θ)

x ( d θ / dx)

= ( b / a) ( cosec ^2 θ)

x ( 1 / ( dx / dθ) )

we have, ( dx / dθ) = - a sinθ

= ( b / a) ( cosec ^2 θ )

x( 1 / ( - a sin θ)

= -( b / a^2 ) x ( cosec^2 θ x cosec θ)

= ( - b / a^2 ) x ( cosec^3 θ)

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◾Again differentiate wrt x

( d^3 y / d x^3 )

= d / dx ( ( - b / a^2 ) x ( cosec^3 θ) )

=( -b / a^2 ) x ( 3 cosec^2 θ x ( - cosec θ x cot θ) x ( d θ / d x)

= ( -b / a^2 ) x ( 3 cosec^2 θ x ( - cosec θ x cot θ) x( 1 / (dx / d θ) )

put the value of ( dx / d θ ) = - a sin θ

= ( -b / a^2 ) x ( 3 cosec^2 θ x ( - cosec θ x cot θ) x( 1 / (- a sin θ ) )

= ( b / a^3 ) ( 3 cosec^2 θ x ( - cosec θ x cot θ) x ( cosec θ)

( d^3 y / d x^3 )

= ( b / a^3 ) ( 3 cosec^3 θ x ( - cosec θ x cot θ)

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