Ex 5.6, 8 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Derivatives in parametric form
Derivatives in parametric form
Last updated at April 16, 2024 by Teachoo
Ex 5.6, 8 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = π(cosβ‘γπ‘+logβ‘γ tanγβ‘γπ‘/2γ γ ) , π¦=π sinβ‘π‘Here ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦ = π sinβ‘π‘ ππ¦/ππ‘ " " = π(π sinβ‘π‘)/ππ‘ ππ¦/ππ‘ " " = π π( sinβ‘π‘)/ππ‘ ππ¦/ππ‘ " " = π cosβ‘π‘ Calculating π π/π π π₯ = π(cosβ‘γπ‘+logβ‘γ tanγβ‘γπ‘/2γ γ ) ππ₯/ππ‘ = π(π(cosβ‘γπ‘+logβ‘γ tan γβ‘γπ‘/2γ γ ))/ππ‘ ππ₯/ππ‘ = a ((π(cosβ‘π‘ ) )/ππ‘ +π(logβ‘(γtan γβ‘γπ‘/2γ ) )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +π(logβ‘(tanβ‘γ π‘/2γ ) )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . π(γtan γβ‘γπ‘/2γ )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . (sec^2β‘γπ‘/2γ ).π(π‘/2)/ππ‘ ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . sec^2 (π‘/2). 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +cotβ‘γπ‘/2γ . 1/(cos^2 π‘/2) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +γcos γβ‘γπ‘/2γ/sinβ‘γ π‘/2γ . 1/(cos^2 π‘/2) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/(sinβ‘γ π‘/2γ cosβ‘γ π‘/2γ ) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/(2 sinβ‘γπ‘/2γ cosβ‘γπ‘/2γ )) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/sinβ‘π‘ ) ππ₯/ππ‘ = a ((βsin^2β‘π‘ + 1)/sinβ‘π‘ ) ππ₯/ππ‘ = a ((1 β sin^2β‘π‘)/sinβ‘π‘ ) π π/π π = a (γπππγ^πβ‘π/πππβ‘π ) We know that sin 2ΞΈ =2sin ΞΈ cos ΞΈ Replacing ΞΈ by π‘/2 sin t = 2 sin π‘/2 . cos π‘/2 Therefore, π π/π π " =" (π π/π π)/(π π/π π) ππ¦/ππ₯ = (π cosβ‘π‘)/(γπ cos^2γβ‘π‘/sinβ‘π‘ ) ππ¦/ππ₯ = π cosβ‘π‘ Γ sinβ‘π‘/(π cos^2 π‘) ππ¦/ππ₯ = (π cosβ‘π‘ . sinβ‘π‘)/(π cosβ‘π‘ .cosβ‘π‘ ) ππ¦/ππ₯ = sinβ‘π‘/cosβ‘π‘ π π/π π = πππ§β‘π