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

Last updated at March 11, 2021 by

Last updated at March 11, 2021 by

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Ex 5.6, 5 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = cosβ‘π β cosβ‘2π, π¦ = sinβ‘π β sinβ‘2π Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ ππ¦/ππ = π(sinβ‘π β sinβ‘2π)/ππ ππ¦/ππ = π(sin π)/ππ β π(sinβ‘2π)/ππ ππ¦/ππ = cosβ‘πβcosβ‘2π. 2 ππ¦/ππ = πππβ‘π½βπ πππβ‘ππ½ Calculating π π/π π½ ππ₯/ππ = π(cosβ‘π β cosβ‘2π)/ππ ππ₯/ππ " " = π(cosβ‘π)/ππ β π(cosβ‘2π)/ππ ππ₯/ππ " " = βsinβ‘πβ(βsinβ‘2π . 2) ππ₯/ππ = βπππβ‘π½ + π πππβ‘ππ½ Therefore ππ¦/ππ₯ = (ππ¦/(ππ ))/(ππ₯/ππ) ππ¦/ππ₯ = (cosβ‘π β 2 cosβ‘2π)/(βsinβ‘π " +" 2 sinβ‘2π ) π π/π π = (πππβ‘π½ β π πππβ‘ππ½)/(π πππβ‘ππ½ " β" πππβ‘π½ )

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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.