# Ex 5.6, 3 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

Ex 5.6, 3 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯=sinβ‘π‘, π¦=cosβ‘2π‘Here, ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π ππ¦/ππ‘ " " = π(cosβ‘2π‘)/ππ‘ ππ¦/ππ‘ = βsinβ‘2π‘ . 2 ππ¦/ππ‘ = β2 sinβ‘2π‘ Calculating π π/π π ππ₯/ππ‘ " " = π(sinβ‘π‘ )/ππ‘ ππ₯/ππ‘ " " = cosβ‘π‘ Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = (β2 sinβ‘2π‘" " )/cosβ‘π‘ ππ¦/ππ₯ = (β2(2 sinβ‘π‘ .γ cosγβ‘π‘ ))/cosβ‘π‘ π π/π π = βπ πππβ‘π (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.