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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.6, 10 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦𝑑𝑥, 𝑥 = 𝑎 (cos𝜃 + 𝜃 sin𝜃), 𝑦 = 𝑎 (sin𝜃 – 𝜃 cos𝜃) 𝑥 = 𝑎 cos𝜃+ 𝜃 sin𝜃, 𝑦 = 𝑎 (sin𝜃 – 𝜃 cos𝜃) 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑥 × 𝑑𝜃𝑑𝜃 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝜃 × 𝑑𝜃𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝜃 𝑑𝑥𝑑𝜃 Calculating 𝒅𝒚𝒅𝜽 𝑦 = 𝑎 (sin𝜃 – 𝜃 cos𝜃) 𝑑𝑦𝑑𝜃 = 𝑑 𝑎 ( sin𝜃 − 𝜃 cos𝜃) 𝑑𝜃 𝑑𝑦𝑑𝜃 = a 𝑑 sin𝜃 − 𝜃 cos𝜃𝑑𝜃 𝑑𝑦𝑑𝜃 = a 𝑑 sin𝜃 𝑑𝜃 − 𝑑 𝜃 cos𝜃𝑑𝜃 𝑑𝑦𝑑𝜃 = a cos𝜃− 𝑑 𝜃 cos𝜃𝑑𝜃 𝑑𝑦𝑑𝜃 = a cos𝜃− 𝑑 𝜃 𝑑𝜃 . cos𝜃+ 𝑑 cos𝜃 𝑑𝜃 . 𝜃 𝑑𝑦𝑑𝜃 = a cos𝜃− cos𝜃+ − sin𝜃 𝜃 𝑑𝑦𝑑𝜃 = a 𝑐𝑜𝑠𝜃− cos𝜃+𝜃 sin𝜃 𝑑𝑦𝑑𝜃 = a 𝜃 sin𝜃 𝑑𝑦𝑑𝜃 = 𝑎 𝜃. sin𝜃 Calculating 𝒅𝒙𝒅𝜽 𝑥=𝑎 cos𝜃+ 𝜃 sin𝜃 𝑑𝑥𝑑𝜃 = 𝑑 𝑎 cos𝜃+ 𝜃 sin𝜃𝑑𝜃 𝑑𝑥𝑑𝜃 = 𝑎 𝑑 cos𝜃+ 𝜃 sin𝜃𝑑𝜃 𝑑𝑥𝑑𝜃 = 𝑎 𝑑 cos𝜃𝑑𝜃 + 𝑑 𝜃 sin𝜃𝑑𝜃 𝑑𝑥𝑑𝜃 = 𝑎 − sin𝜃+ 𝑑 𝜃 sin𝜃𝑑𝜃 𝑑𝑥𝑑𝜃 = 𝑎 − sin𝜃+ 𝑑𝜃𝑑𝜃 . sin𝜃+ 𝑑 sin𝜃𝑑𝜃 . 𝜃 𝑑𝑥𝑑𝜃 = 𝑎 − sin𝜃+ sin𝜃+ cos𝜃. 𝜃 𝑑𝑥𝑑𝜃 = 𝑎 − sin𝜃+ sin𝜃+𝜃. cos𝜃 𝑑𝑥𝑑𝜃 = 𝑎 𝜃 cos𝜃 Therefore 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝜃 𝑑𝑥𝑑𝜃 𝑑𝑦𝑑𝑥 = 𝑎 𝜃. sin𝜃𝑎 𝜃 cos𝜃 𝑑𝑦𝑑𝑥 = sin𝜃 cos𝜃 𝒅𝒚𝒅𝒙 = 𝒕𝒂𝒏𝜽

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