Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability (Important Question)

Transcript

Ex 5.6, 7 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑𝑥, 𝑥 =(〖𝑠𝑖𝑛〗^3 𝑡)/√(cos⁡2𝑡 ) , 𝑦 = (〖𝑐𝑜𝑠〗^3 𝑡)/√(cos⁡2𝑡 )Here, 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Calculating 𝒅𝒚/𝒅𝒕 𝑦 = (〖𝑐𝑜𝑠〗^3 𝑡)/√(cos⁡2𝑡 ) 𝑑𝑦/𝑑𝑡 " " = 𝑑/𝑑𝑡 ((〖𝑐𝑜𝑠〗^3 𝑡)/√(cos⁡2𝑡 )) 𝑑𝑦/𝑑𝑡 " " = (𝑑(〖𝑐𝑜𝑠〗^3 𝑡)/𝑑𝑡 . √(cos⁡2 𝑡) − 𝑑(√(cos⁡2𝑡 ))/𝑑𝑡 .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " = " (3 cos^2⁡〖𝑡 〗. 𝑑(cos⁡𝑡 )/𝑑𝑡. √(cos⁡2 𝑡) − 1/(2√(cos⁡2𝑡 )) . 𝑑(cos⁡2𝑡 )/𝑑𝑡 .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 Using quotient rule As (𝑢/𝑣)^′ = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 𝑑𝑦/𝑑𝑡 " = " (3 cos^2⁡〖𝑡 〗. (−sin⁡𝑡 ) . √(cos⁡2 𝑡) − 1/(2√(cos⁡2𝑡 )) . (−2 sin⁡2𝑡) .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " =" (−3 cos^2⁡〖𝑡 〗 sin⁡𝑡 √(cos⁡2 𝑡) + 1/√(cos⁡2𝑡 ) . sin⁡2𝑡 .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " =" ((−3 cos^2⁡〖𝑡 〗 sin⁡𝑡 √(cos⁡2 𝑡) × √(cos⁡2𝑡 ) + sin⁡2𝑡 .〖 cos^3〗⁡𝑡)/√(cos⁡2𝑡 ))/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " =" (−3 cos^2⁡〖𝑡 〗 sin⁡𝑡 (cos⁡2 𝑡) + sin⁡2𝑡 .〖 cos^3〗⁡𝑡)/((√(cos⁡2 𝑡))^2 (√(cos⁡2 𝑡)) ) 𝑑𝑦/𝑑𝑡 " =" ( cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/((cos⁡2 𝑡)^(3/2) ) Calculating 𝒅𝒙/𝒅𝒕 𝑥 = (〖𝑠𝑖𝑛〗^3 𝑡)/√(cos⁡2𝑡 ) 𝑑𝑥/𝑑𝑡 = 𝑑/𝑑𝑥 ((〖𝑠𝑖𝑛〗^3 𝑡)/√(cos⁡2𝑡 )) 𝑑𝑥/𝑑𝑡 = (𝑑(〖𝑠𝑖𝑛〗^3 𝑡)/𝑑𝑡 . √(cos⁡2𝑡 ) − (𝑑(√(cos⁡2𝑡 )) )/𝑑𝑥 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/(√(cos⁡2𝑡 ))^2 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . (𝑑(sin⁡𝑡 ) )/𝑑𝑡 . √(cos⁡2𝑡 ) − 1/(2√(cos⁡2𝑡 )) . (𝑑(cos⁡2𝑡 ) )/𝑑𝑥 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/(√(cos⁡2𝑡 ))^2 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . cos⁡𝑡 . √(cos⁡〖2 𝑡〗 ) − 1/(2√(cos⁡〖2 𝑡〗 )) . (−sin⁡2𝑡 ) . 2 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/((cos⁡〖2 𝑡〗 ) ) 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . cos⁡𝑡 . (√(cos⁡2𝑡 )) . (√(cos⁡2𝑡 )) + sin⁡2𝑡 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/((√(cos⁡2𝑡 )) (cos⁡2𝑡 ) ) 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/(cos⁡2𝑡 )^(3/2) 𝑑𝑥/𝑑𝑡 = (〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) )/(cos⁡2𝑡 )^(3/2) Finding 𝒅𝒚/𝒅𝒙 𝒅𝒚/𝒅𝒙 = ((cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/((cos⁡2 𝑡)^(3/2) ))/((〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) )/(cos⁡2𝑡 )^(3/2) ) 𝑑𝑦/𝑑𝑥 = (cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/(〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = (cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/(〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡)/(3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 )) Taking cos 2t common 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((cos⁡2𝑡 (−3 sin⁡𝑡 + cos⁡𝑡 sin⁡2𝑡/cos⁡2𝑡 ))/(cos⁡2𝑡 (3 cos⁡〖𝑡 〗+sin⁡𝑡 . sin⁡2𝑡/cos⁡2𝑡 ) )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 sin⁡𝑡 + cos⁡𝑡 sin⁡2𝑡/cos⁡2𝑡 )/(3 cos⁡〖𝑡 〗+〖 sin〗⁡𝑡 . sin⁡2𝑡/cos⁡2𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 sin⁡𝑡 + cos⁡𝑡 tan⁡2𝑡)/(3 cos⁡〖𝑡 〗+〖 sin〗⁡𝑡 . tan⁡2𝑡 )) Taking cos t common 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((cos⁡𝑡 (−3 sin⁡𝑡/cos⁡𝑡 + tan⁡2𝑡))/(cos⁡𝑡 (3 + sin⁡𝑡/cos⁡𝑡 . tan⁡2𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 tan⁡𝑡 + tan⁡2𝑡)/(3 +〖 tan〗⁡𝑡 . tan⁡2𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((tan⁡2𝑡 − 3 tan⁡𝑡)/(3 +〖 tan〗⁡𝑡 . tan⁡2𝑡 )) Using tan 2𝜃 = (2 𝑡𝑎𝑛⁡𝜃)/(1 〖𝑡𝑎𝑛〗^2⁡𝜃 ) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 (((2 tan⁡𝑡)/(1 − tan^2⁡𝑡 ) − 3 tan⁡𝑡)/(3 + (tan⁡𝑡 ) ((2 tan⁡𝑡)/(1 −tan^2⁡𝑡 )) )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 (((2 tan⁡𝑡 − 3 tan⁡𝑡 (1 − tan^2⁡𝑡 ))/((1 − tan^2⁡𝑡)))/((3 (1− tan^2⁡𝑡 ) + tan⁡𝑡 (2 tan⁡𝑡 ))/((1 − tan^2⁡𝑡)))) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((2 tan⁡𝑡 −3 tan⁡𝑡 (1 − tan^2⁡𝑡 ))/(3 (1− tan^2⁡𝑡 ) + tan⁡𝑡 (2 tan⁡𝑡 ) )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((2 tan⁡𝑡 −3 tan⁡𝑡 + 3 tan^3⁡𝑡)/(3 − 3 tan^2⁡𝑡 + 2 tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−tan⁡𝑡 + 3 tan^3⁡𝑡 ) )/((3 −tan^2⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 (−(tan⁡𝑡 −3 tan^3⁡𝑡 ) )/((3 −tan^2⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = 〖−cot〗^2 𝑡 ((tan⁡𝑡 −3 tan^3⁡𝑡 ) )/((3 −tan^2⁡𝑡 ) ) Multiplying cot2 t to numerator 𝑑𝑦/𝑑𝑥 = −((cot^2⁡𝑡 × tan⁡𝑡 − 3 cot^2⁡𝑡 tan^3⁡𝑡)/(3 −tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − ((1/tan^2⁡𝑡 × tan⁡𝑡 − 3 × 1/tan^2⁡𝑡 ×tan^3⁡𝑡)/(3 −tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − ((1/tan⁡𝑡 .− 3 tan⁡𝑡 )/(3 − tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − (((1 −3 tan⁡𝑡 (tan⁡〖𝑡)〗)/tan⁡𝑡 )/(3 − tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − ((1 − 3 tan^2⁡𝑡 )/(tan⁡𝑡 (3 − tan^2⁡𝑡 ) )) 𝑑𝑦/𝑑𝑥 = − ((1 − 3 tan^2⁡𝑡 )/(3 tan⁡𝑡 −tan^3⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = (−1)/(((3 tan⁡𝑡 −〖 tan〗^3⁡𝑡)/(1 −3 tan^2⁡𝑡 )) ) 𝑑𝑦/𝑑𝑥 = − ((1 − 3 tan^2⁡𝑡 )/(3 tan⁡𝑡 −tan^3⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = (−1)/(((3 tan⁡𝑡 −〖 tan〗^3⁡𝑡)/(1 −3 tan^2⁡𝑡 )) ) 𝐴𝑠 tan⁡3𝑥=(3 tan⁡𝑥 − tan^3⁡𝑥)/(1 − 3 tan^2⁡𝑥 ) 𝑑𝑦/𝑑𝑥 = (−1)/tan⁡3𝑡 𝒅𝒚/𝒅𝒙 = −𝒄𝒐𝒕⁡𝟑𝒕