Ex 5.6, 11 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Transcript

Ex 5.6, 11 If 𝑥=√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) ) , y=√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) ), show that 𝑑𝑦/𝑑𝑥 = − 𝑦/𝑥Here 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Calculating 𝒅𝒚/𝒅𝒕 𝑦 = √(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) ) 𝑑𝑦/𝑑𝑡 " "= 𝑑(√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) ))/𝑑𝑡 𝑑𝑦/𝑑𝑡 " "= 1/(2√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) )) . 𝑑(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) )/𝑑𝑡 𝑑𝑦/𝑑𝑡 " "= 1/(2√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) )) . 𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡). log⁡𝑎 × 𝑑(〖𝑐𝑜𝑠〗^(−1) 𝑡)/𝑑𝑡 𝑑𝑦/𝑑𝑡 " "= 1/(2√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) )) . 𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡). log⁡𝑎 × (−1)/√(1 − 𝑡^2 ) (𝐴𝑠" " 𝑑(𝑎^𝑥 )/𝑑𝜃=𝑎^𝑥.log⁡𝑎 ) 𝑑𝑦/𝑑𝑡 " "= (− 𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) ". " log⁡𝑎" " )/(2√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) ) . √(1 − 𝑡^2 )) Calculating 𝒅𝒙/𝒅𝒕 𝑥=√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) ) 𝑑𝑥/𝑑𝑡 = 𝑑(√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) ))/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 1/(2√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) )) × 𝑑(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) )/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 1/(2√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) )) × 𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡). log⁡𝑎 . 𝑑(〖𝑠𝑖𝑛〗^(−1) 𝑡)/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 1/(2√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) )) × 𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡). log⁡𝑎 . 1/√(1 − 𝑡^2 ) 𝑑𝑥/𝑑𝑡 = " " 𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡)/(2√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) )) × log⁡𝑎/√(1 − 𝑡^2 ) 𝑑𝑥/𝑑𝑡 = " " (√(𝒂^(〖𝒔𝒊𝒏〗^(−𝟏) 𝒕) ) . 𝒍𝒐𝒈⁡𝒂)/(𝟐√(𝟏 − 𝒕^𝟐 )) Finding 𝒅𝒚/𝒅𝒙 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) 𝑑𝑦/𝑑𝑥 = ((− √(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) ) . log⁡𝑎" " )/(2√(1 − 𝑡^2 )))/((√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) ) . log⁡𝑎)/(2√(1 − 𝑡^2 ))) 𝑑𝑦/𝑑𝑥 = (− √(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) ) . − log⁡𝑎" " )/(2√(1 − 𝑡^2 )) × (2√(1 − 𝑡^2 ))/(√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) ) . log⁡𝑎 ) 𝑑𝑦/𝑑𝑥 = (−√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) )⁡)/√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) ) × (log⁡𝑎 × 2√(1 − 𝑡^2 ))/(log⁡𝑎 × 2√(1 − 𝑡^2 )) 𝑑𝑦/𝑑𝑥 = (−√(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) )⁡)/√(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) ) 𝒅𝒚/𝒅𝒙 = (−𝒚⁡)/𝒙 Hence proved. 𝐴𝑠 √(𝑎^(〖𝑐𝑜𝑠〗^(−1) 𝑡) )=𝑦 √(𝑎^(〖𝑠𝑖𝑛〗^(−1) 𝑡) )=𝑥