Ex 5.6, 7 - 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, 7 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ =(γπ ππγ^3 π‘)/β(cosβ‘2π‘ ) , π¦ = (γπππ γ^3 π‘)/β(cosβ‘2π‘ )Here, ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦ = (γπππ γ^3 π‘)/β(cosβ‘2π‘ ) ππ¦/ππ‘ " " = π/ππ‘ ((γπππ γ^3 π‘)/β(cosβ‘2π‘ )) ππ¦/ππ‘ " " = (π(γπππ γ^3 π‘)/ππ‘ . β(cosβ‘2 π‘) β π(β(cosβ‘2π‘ ))/ππ‘ .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " = " (3 cos^2β‘γπ‘ γ. π(cosβ‘π‘ )/ππ‘. β(cosβ‘2 π‘) β 1/(2β(cosβ‘2π‘ )) . π(cosβ‘2π‘ )/ππ‘ .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 Using quotient rule As (π’/π£)^β² = (π’^β² π£ β π£^β² π’)/π£^2 ππ¦/ππ‘ " = " (3 cos^2β‘γπ‘ γ. (βsinβ‘π‘ ) . β(cosβ‘2 π‘) β 1/(2β(cosβ‘2π‘ )) . (β2 sinβ‘2π‘) .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " =" (β3 cos^2β‘γπ‘ γ sinβ‘π‘ β(cosβ‘2 π‘) + 1/β(cosβ‘2π‘ ) . sinβ‘2π‘ .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " =" ((β3 cos^2β‘γπ‘ γ sinβ‘π‘ β(cosβ‘2 π‘) Γ β(cosβ‘2π‘ ) + sinβ‘2π‘ .γ cos^3γβ‘π‘)/β(cosβ‘2π‘ ))/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " =" (β3 cos^2β‘γπ‘ γ sinβ‘π‘ (cosβ‘2 π‘) + sinβ‘2π‘ .γ cos^3γβ‘π‘)/((β(cosβ‘2 π‘))^2 (β(cosβ‘2 π‘)) ) ππ¦/ππ‘ " =" ( cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/((cosβ‘2 π‘)^(3/2) ) Calculating π π/π π π₯ = (γπ ππγ^3 π‘)/β(cosβ‘2π‘ ) ππ₯/ππ‘ = π/ππ₯ ((γπ ππγ^3 π‘)/β(cosβ‘2π‘ )) ππ₯/ππ‘ = (π(γπ ππγ^3 π‘)/ππ‘ . β(cosβ‘2π‘ ) β (π(β(cosβ‘2π‘ )) )/ππ₯ . γ π ππγ^3 π‘ )/(β(cosβ‘2π‘ ))^2 ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . (π(sinβ‘π‘ ) )/ππ‘ . β(cosβ‘2π‘ ) β 1/(2β(cosβ‘2π‘ )) . (π(cosβ‘2π‘ ) )/ππ₯ . γ π ππγ^3 π‘ )/(β(cosβ‘2π‘ ))^2 ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . cosβ‘π‘ . β(cosβ‘γ2 π‘γ ) β 1/(2β(cosβ‘γ2 π‘γ )) . (βsinβ‘2π‘ ) . 2 . γ π ππγ^3 π‘ )/((cosβ‘γ2 π‘γ ) ) ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . cosβ‘π‘ . (β(cosβ‘2π‘ )) . (β(cosβ‘2π‘ )) + sinβ‘2π‘ . γ π ππγ^3 π‘ )/((β(cosβ‘2π‘ )) (cosβ‘2π‘ ) ) ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . γ π ππγ^3 π‘ )/(cosβ‘2π‘ )^(3/2) ππ₯/ππ‘ = (γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) )/(cosβ‘2π‘ )^(3/2) Finding π π/π π π π/π π = ((cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/((cosβ‘2 π‘)^(3/2) ))/((γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) )/(cosβ‘2π‘ )^(3/2) ) ππ¦/ππ₯ = (cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/(γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) ) ππ¦/ππ₯ = (cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/(γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) ) ππ¦/ππ₯ = cot^2 π‘ ((β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘)/(3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ )) Taking cos 2t common ππ¦/ππ₯ = cot^2 π‘ ((cosβ‘2π‘ (β3 sinβ‘π‘ + cosβ‘π‘ sinβ‘2π‘/cosβ‘2π‘ ))/(cosβ‘2π‘ (3 cosβ‘γπ‘ γ+sinβ‘π‘ . sinβ‘2π‘/cosβ‘2π‘ ) )) ππ¦/ππ₯ = cot^2 π‘ ((β3 sinβ‘π‘ + cosβ‘π‘ sinβ‘2π‘/cosβ‘2π‘ )/(3 cosβ‘γπ‘ γ+γ sinγβ‘π‘ . sinβ‘2π‘/cosβ‘2π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((β3 sinβ‘π‘ + cosβ‘π‘ tanβ‘2π‘)/(3 cosβ‘γπ‘ γ+γ sinγβ‘π‘ . tanβ‘2π‘ )) Taking cos t common ππ¦/ππ₯ = cot^2 π‘ ((cosβ‘π‘ (β3 sinβ‘π‘/cosβ‘π‘ + tanβ‘2π‘))/(cosβ‘π‘ (3 + sinβ‘π‘/cosβ‘π‘ . tanβ‘2π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((β3 tanβ‘π‘ + tanβ‘2π‘)/(3 +γ tanγβ‘π‘ . tanβ‘2π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((tanβ‘2π‘ β 3 tanβ‘π‘)/(3 +γ tanγβ‘π‘ . tanβ‘2π‘ )) Using tan 2π = (2 π‘ππβ‘π)/(1 γπ‘ππγ^2β‘π ) ππ¦/ππ₯ = cot^2 π‘ (((2 tanβ‘π‘)/(1 β tan^2β‘π‘ ) β 3 tanβ‘π‘)/(3 + (tanβ‘π‘ ) ((2 tanβ‘π‘)/(1 βtan^2β‘π‘ )) )) ππ¦/ππ₯ = cot^2 π‘ (((2 tanβ‘π‘ β 3 tanβ‘π‘ (1 β tan^2β‘π‘ ))/((1 β tan^2β‘π‘)))/((3 (1β tan^2β‘π‘ ) + tanβ‘π‘ (2 tanβ‘π‘ ))/((1 β tan^2β‘π‘)))) ππ¦/ππ₯ = cot^2 π‘ ((2 tanβ‘π‘ β3 tanβ‘π‘ (1 β tan^2β‘π‘ ))/(3 (1β tan^2β‘π‘ ) + tanβ‘π‘ (2 tanβ‘π‘ ) )) ππ¦/ππ₯ = cot^2 π‘ ((2 tanβ‘π‘ β3 tanβ‘π‘ + 3 tan^3β‘π‘)/(3 β 3 tan^2β‘π‘ + 2 tan^2β‘π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((βtanβ‘π‘ + 3 tan^3β‘π‘ ) )/((3 βtan^2β‘π‘ ) ) ππ¦/ππ₯ = cot^2 π‘ (β(tanβ‘π‘ β3 tan^3β‘π‘ ) )/((3 βtan^2β‘π‘ ) ) ππ¦/ππ₯ = γβcotγ^2 π‘ ((tanβ‘π‘ β3 tan^3β‘π‘ ) )/((3 βtan^2β‘π‘ ) ) Multiplying cot2 t to numerator ππ¦/ππ₯ = β((cot^2β‘π‘ Γ tanβ‘π‘ β 3 cot^2β‘π‘ tan^3β‘π‘)/(3 βtan^2β‘π‘ )) ππ¦/ππ₯ = β ((1/tan^2β‘π‘ Γ tanβ‘π‘ β 3 Γ 1/tan^2β‘π‘ Γtan^3β‘π‘)/(3 βtan^2β‘π‘ )) ππ¦/ππ₯ = β ((1/tanβ‘π‘ .β 3 tanβ‘π‘ )/(3 β tan^2β‘π‘ )) ππ¦/ππ₯ = β (((1 β3 tanβ‘π‘ (tanβ‘γπ‘)γ)/tanβ‘π‘ )/(3 β tan^2β‘π‘ )) ππ¦/ππ₯ = β ((1 β 3 tan^2β‘π‘ )/(tanβ‘π‘ (3 β tan^2β‘π‘ ) )) ππ¦/ππ₯ = β ((1 β 3 tan^2β‘π‘ )/(3 tanβ‘π‘ βtan^3β‘π‘ )) ππ¦/ππ₯ = (β1)/(((3 tanβ‘π‘ βγ tanγ^3β‘π‘)/(1 β3 tan^2β‘π‘ )) ) ππ¦/ππ₯ = β ((1 β 3 tan^2β‘π‘ )/(3 tanβ‘π‘ βtan^3β‘π‘ )) ππ¦/ππ₯ = (β1)/(((3 tanβ‘π‘ βγ tanγ^3β‘π‘)/(1 β3 tan^2β‘π‘ )) ) π΄π tanβ‘3π₯=(3 tanβ‘π₯ β tan^3β‘π₯)/(1 β 3 tan^2β‘π₯ ) ππ¦/ππ₯ = (β1)/tanβ‘3π‘ π π/π π = βπππβ‘ππ