Ex 5.6, 4 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 5.6, 4 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = 4π‘, π¦ = 4/π‘Here ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π ππ¦/ππ‘ " " = π/ππ‘ (4/π‘) ππ¦/ππ‘ = 4 π/ππ‘ (1/π‘) ππ¦/ππ‘ = β π/π^π Calculating π π/π π ππ₯/ππ‘ = π(4π‘)/ππ‘ ππ₯/ππ‘ " " = 4 Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = ("β " 4/π‘^2 )/4 π π/π π = ("β" π)/π^π
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