Ex 5.6, 10 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Ex 5.6, 10 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦﷮𝑑𝑥﷯, 𝑥 = 𝑎 (cos⁡𝜃 + 𝜃 sin⁡𝜃), 𝑦 = 𝑎 (sin⁡𝜃 – 𝜃 cos⁡𝜃) 𝑥 = 𝑎 cos﷮𝜃﷯+ 𝜃 sin﷮𝜃﷯﷯, 𝑦 = 𝑎 (sin⁡𝜃 – 𝜃 cos⁡𝜃) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑥﷯ × 𝑑𝜃﷮𝑑𝜃﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝜃﷯ × 𝑑𝜃﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝜃﷯﷮ 𝑑𝑥﷮𝑑𝜃﷯﷯ Calculating 𝒅𝒚﷮𝒅𝜽﷯ 𝑦 = 𝑎 (sin⁡𝜃 – 𝜃 cos⁡𝜃) 𝑑𝑦﷮𝑑𝜃﷯ = 𝑑 𝑎 ( sin﷮𝜃﷯ − 𝜃 cos⁡𝜃)﷯ ﷮𝑑𝜃﷯ 𝑑𝑦﷮𝑑𝜃﷯ = a 𝑑 sin﷮𝜃﷯ − 𝜃 cos﷮𝜃﷯﷯﷮𝑑𝜃﷯﷯ 𝑑𝑦﷮𝑑𝜃﷯ = a 𝑑 sin﷮𝜃﷯﷯ ﷮𝑑𝜃﷯ − 𝑑 𝜃 cos﷮𝜃﷯﷯﷮𝑑𝜃﷯﷯ 𝑑𝑦﷮𝑑𝜃﷯ = a cos﷮𝜃﷯− 𝑑 𝜃 cos﷮𝜃﷯﷯﷮𝑑𝜃﷯﷯ 𝑑𝑦﷮𝑑𝜃﷯ = a cos﷮𝜃﷯− 𝑑 𝜃﷯ ﷮𝑑𝜃﷯ . cos﷮𝜃﷯+ 𝑑 cos﷮𝜃﷯﷯ ﷮𝑑𝜃﷯ . 𝜃﷯﷯ 𝑑𝑦﷮𝑑𝜃﷯ = a cos﷮𝜃﷯− cos﷮𝜃﷯+ − sin﷮𝜃﷯﷯﷯ 𝜃﷯ 𝑑𝑦﷮𝑑𝜃﷯ = a 𝑐𝑜𝑠﷮𝜃﷯− cos﷮𝜃﷯+𝜃 sin﷮𝜃﷯﷯ 𝑑𝑦﷮𝑑𝜃﷯ = a 𝜃 sin﷮𝜃﷯﷯ 𝑑𝑦﷮𝑑𝜃﷯ = 𝑎 𝜃. sin﷮𝜃﷯ Calculating 𝒅𝒙﷮𝒅𝜽﷯ 𝑥=𝑎 cos﷮𝜃﷯+ 𝜃 sin﷮𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑑 𝑎 cos﷮𝜃﷯+ 𝜃 sin﷮𝜃﷯﷯﷯﷮𝑑𝜃﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 𝑑 cos﷮𝜃﷯+ 𝜃 sin﷮𝜃﷯﷯﷮𝑑𝜃﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 𝑑 cos﷮𝜃﷯﷯﷮𝑑𝜃﷯ + 𝑑 𝜃 sin﷮𝜃﷯﷯﷮𝑑𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 − sin﷮𝜃﷯+ 𝑑 𝜃 sin﷮𝜃﷯﷯﷮𝑑𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 − sin﷮𝜃﷯+ 𝑑𝜃﷮𝑑𝜃﷯ . sin﷮𝜃﷯+ 𝑑 sin﷮𝜃﷯﷯﷮𝑑𝜃﷯ . 𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 − sin﷮𝜃﷯+ sin﷮𝜃﷯+ cos﷮𝜃﷯. 𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 − sin﷮𝜃﷯+ sin﷮𝜃﷯+𝜃. cos﷮𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 𝜃 cos﷮𝜃﷯﷯ Therefore 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝜃﷯﷮ 𝑑𝑥﷮𝑑𝜃﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎 𝜃. sin﷮𝜃﷯﷯﷮𝑎 𝜃 cos﷮𝜃﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = sin﷮𝜃﷯﷮ cos﷮𝜃﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒕𝒂𝒏﷮𝜽﷯