# Ex 5.6, 10 - 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, 10 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = π (cosβ‘π + π sinβ‘π), π¦ = π (sinβ‘π β π cosβ‘π)Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ π¦ = π (sinβ‘π β π cosβ‘π) ππ¦/ππ = (π(π (sinβ‘π β π cosβ‘π)) )/ππ ππ¦/ππ = a ((π (sinβ‘π β π cosβ‘π ))/ππ) ππ¦/ππ = a ((π (sinβ‘π ) )/ππ β (π (π cosβ‘π ))/ππ) ππ¦/ππ = a (cosβ‘πβ (π (π cosβ‘π ))/ππ) ππ¦/ππ = a (cosβ‘πβ((π (π) )/ππ . cosβ‘π+(π (cosβ‘π ) )/ππ . π)) ππ¦/ππ = a (cosβ‘πβ(cosβ‘π+(βsinβ‘π )) π) Using Product Rule As (π’π£)β = π’βπ£ + π£βπ’ ππ¦/ππ = a (πππ β‘πβcosβ‘π+π sinβ‘π ) ππ¦/ππ = a (π sinβ‘π ) π π/π π½ = π π½. πππβ‘π½ Calculating π π/π π½ π₯=π (cosβ‘π+ π sinβ‘π ) ππ₯/ππ = π(π (cosβ‘π+ π sinβ‘π ))/ππ ππ₯/ππ = π π(cosβ‘π+ π sinβ‘π )/ππ ππ₯/ππ = π (π(cosβ‘π )/ππ + π(π sinβ‘π )/ππ) Using Product Rule As (π’π£)β = π’βπ£ + π£βπ’ ππ₯/ππ = π (βsinβ‘π+ π(π sinβ‘π )/ππ) ππ₯/ππ = π (βsinβ‘π+(ππ/ππ . sinβ‘π+ π(sinβ‘π )/ππ . π)) ππ₯/ππ = π (βsinβ‘π+(sinβ‘π+cosβ‘π. π)) ππ₯/ππ = π (βsinβ‘π+sinβ‘π+π.cosβ‘π ) π π/π π½ = π (π½ πππβ‘π½ ) Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (π (π. sinβ‘π ))/π" " (π cosβ‘π ) ππ¦/ππ₯ = sinβ‘π/cosβ‘π π π/π π = πππβ‘π½

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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.