Ex 5.6

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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### Transcript

Ex 5.6, 8 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = π(cosβ‘γπ‘+logβ‘γ tanγβ‘γπ‘/2γ γ ) , π¦=π sinβ‘π‘Here ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating ππ/ππ π¦ = π sinβ‘π‘ ππ¦/ππ‘ " " = π(π sinβ‘π‘)/ππ‘ ππ¦/ππ‘ " " = π π( sinβ‘π‘)/ππ‘ ππ¦/ππ‘ " " = π cosβ‘π‘ Calculating ππ/ππ π₯ = π(cosβ‘γπ‘+logβ‘γ tanγβ‘γπ‘/2γ γ ) ππ₯/ππ‘ = π(π(cosβ‘γπ‘+logβ‘γ tan γβ‘γπ‘/2γ γ ))/ππ‘ ππ₯/ππ‘ = a ((π(cosβ‘π‘ ) )/ππ‘ +π(logβ‘(γtan γβ‘γπ‘/2γ ) )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +π(logβ‘(tanβ‘γ π‘/2γ ) )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . π(γtan γβ‘γπ‘/2γ )/ππ‘) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . (sec^2β‘γπ‘/2γ ).π(π‘/2)/ππ‘ ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/tanβ‘γ π‘/2γ . sec^2 (π‘/2). 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +cotβ‘γπ‘/2γ . 1/(cos^2 π‘/2) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +γcos γβ‘γπ‘/2γ/sinβ‘γ π‘/2γ . 1/(cos^2 π‘/2) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/(sinβ‘γ π‘/2γ cosβ‘γ π‘/2γ ) . 1/2 ) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/(2 sinβ‘γπ‘/2γ cosβ‘γπ‘/2γ )) ππ₯/ππ‘ = a (βsinβ‘π‘ +1/sinβ‘π‘ ) ππ₯/ππ‘ = a ((βsin^2β‘π‘ + 1)/sinβ‘π‘ ) ππ₯/ππ‘ = a ((1 β sin^2β‘π‘)/sinβ‘π‘ ) ππ/ππ = a (γπππγ^πβ‘π/πππβ‘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) ππ¦/ππ₯ = (π cosβ‘π‘)/(γπ cos^2γβ‘π‘/sinβ‘π‘ ) ππ¦/ππ₯ = π cosβ‘π‘ Γ sinβ‘π‘/(π cos^2 π‘) ππ¦/ππ₯ = (π cosβ‘π‘ . sinβ‘π‘)/(π cosβ‘π‘ .cosβ‘π‘ ) ππ¦/ππ₯ = sinβ‘π‘/cosβ‘π‘ ππ/ππ = π­ππ§β‘π

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.