# Ex 5.6, 2 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Sept. 25, 2018 by Teachoo

Last updated at Sept. 25, 2018 by Teachoo

Transcript

Ex 5.6, 2 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯=π cosβ‘π, π¦=π cosβ‘π π₯=π cosβ‘π, π¦=π cosβ‘π ππ¦/ππ₯ = ππ¦/ππ₯ Γ ππ/ππ (β("Multiplying Numerator" @"& Denominator by d" π)) ππ¦/ππ₯ = ππ¦/ππ Γ ππ/ππ₯ ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ π¦ = π cosβ‘π ππ¦/ππ = π(π πππ β‘π)/ππ ππ¦/ππ = b . (βsinβ‘π ) ππ¦/ππ = β b sin π Calculating π π/π π½ π₯=π cosβ‘π ππ₯/ππ = π(π cosβ‘π )/ππ ππ₯/ππ = π . (βsinβ‘π ) ππ₯/ππ = βπ sinβ‘π Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (β b sin π)/(β π sinβ‘π ) π π/π π = π/π

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.