Misc 5

Transcript

Misc 5 Show that the normal at any point θ to the curve x = a cos 𝜃 + a 𝜃 sin 𝜃, y = a sin 𝜃 – a 𝜃 cos 𝜃 is at a constant distance from the origin. Given curve 𝑥=𝑎 cos⁡𝜃+𝑎 𝜃 , 𝑦=𝑎 sin⁡𝜃 – 𝑎 cos⁡𝜃 We need to show distance of a normal from (0,0) is constant First , calculating equation of normal We know that slope of tangent is 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑𝑦﷮𝑑𝜃﷯﷮ 𝑑𝑥﷮𝑑𝜃﷯﷯ Given 𝑥=𝑎 cos⁡𝜃+𝑎 𝜃 sin⁡𝜃 Diff. w.r.t θ 𝑑𝑥﷮𝑑𝜃﷯= 𝑑 𝑎 cos﷮𝜃﷯ + 𝑎 𝜃 sin﷮𝜃﷯﷯﷮𝑑𝜃﷯ 𝑑𝑥﷮𝑑𝜃﷯= 𝑑(𝑎 cos﷮𝜃﷯)﷮𝑑𝜃﷯ + 𝑑(𝑎𝜃 sin﷮𝜃﷯)﷮𝑑𝜃﷯ 𝑑𝑥﷮𝑑𝜃﷯ = 𝑎 − sin﷮𝜃﷯﷯+𝑎 𝑑(𝜃 sin﷮𝜃﷯)﷮𝑑𝜃﷯ Using product rule (u v)’ = u’ v + v’ u 𝑑𝑥﷮𝑑𝜃﷯=− a sin θ+a 𝑑𝜃﷮𝑑𝜃﷯ sin θ+ 𝑑(𝑠𝑖𝑛 𝜃)﷮𝑑𝜃﷯ θ﷯ 𝑑𝑥﷮𝑑𝜃﷯=− 𝑎 𝑠𝑖𝑛 𝜃+𝑎 𝑠𝑖𝑛 𝜃+ 𝑐𝑜𝑠 𝜃 .﷮𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯=− 𝑎 sin⁡𝜃+𝑎 sin﷮𝜃﷯+𝜃 cos﷮𝜃﷯﷯ 𝑑𝑥﷮𝑑𝜃﷯=𝑎 𝜃 cos﷮𝜃﷯ Also, 𝑦 = 𝑎 sin⁡𝜃 – 𝑎 𝜃 cos⁡θ diff w.r.t. θ 𝑑𝑦﷮𝑑𝜃﷯= 𝑑(𝑎 sin﷮𝜃 − 𝑎 𝜃 cos﷮𝜃)﷯﷯﷮𝑑 𝜃﷯ 𝑑𝑦﷮𝑑𝜃﷯= 𝑑(𝑎 sin﷮𝜃)﷯﷮𝑑𝜃﷯− 𝑑(𝑎 𝜃 cos﷮𝜃)﷯﷮𝑑𝜃﷯ 𝑑𝑦﷮𝑑𝜃﷯=𝑎 cos⁡𝜃−𝑎 𝑑(𝜃 cos﷮𝜃)﷯﷮𝑑𝜃﷯ Using product rule (u v)’ = u’v + v’u 𝑑𝑦﷮𝑑𝜃﷯=𝑎 cos⁡𝜃−𝑎 𝑑𝜃 ﷮𝑑𝜃﷯ . cos﷮𝜃﷯+ 𝑑 cos﷮𝜃﷯﷯﷮𝑑𝜃﷯. 𝜃﷯ 𝑑𝑦﷮𝑑𝜃﷯=𝑎 cos⁡𝜃−𝑎 cos﷮𝜃﷯+ − sin﷮𝜃﷯﷯𝜃﷯ 𝑑𝑦﷮𝑑𝜃﷯=𝑎 cos﷮𝜃﷯− 𝑎 cos﷮𝜃﷯−𝑎 𝜃 sin﷮𝜃﷯﷯ 𝑑𝑦﷮𝑑𝜃﷯=𝑎 cos 𝜃−𝑎 cos⁡𝜃 + 𝑎 . 𝜃 sin⁡𝜃. 𝑑𝑦﷮𝑑𝜃﷯= 𝑎. 𝜃 .sin. 𝜃 Now, 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 We know that Slope of tangent of × Slope of normal = −1 tan θ × Slope of normal = −1 Slope of normal = −1﷮ tan﷮𝜃﷯﷯ Slope of normal =−cot⁡𝜃 Equation of normal which passes through the curve 𝑥 = a cos θ + a θ sin θ & 𝑦 = a sin θ - a θ cos θ & has slope −cot⁡𝜃 is 𝑦− 𝑎 sin﷮𝜃﷯−𝑎 cos﷮𝜃﷯﷯﷯=− cot﷮𝜃 𝑥− 𝑎 cos﷮𝜃﷯+𝑎 𝜃 sin﷮𝜃﷯﷯﷯﷯ 𝑦−𝑎 sin﷮𝜃﷯+𝑎 𝜃 cos﷮𝜃﷯﷯= − cos﷮𝜃﷯﷮ sin﷮𝜃﷯﷯ 𝑥−𝑎 cos﷮𝜃﷯+𝑎 𝜃 sin﷮𝜃﷯﷯ sin﷮𝜃 𝑦−𝑎 sin﷮𝜃﷯+𝑎 𝜃 cos﷮𝜃﷯﷯﷯=− cos﷮𝜃 𝑥−𝑎 cos﷮𝜃﷯−𝑎 𝜃 sin﷮𝜃﷯﷯﷯ 𝑦 sin﷮𝜃﷯−𝑎 sin2 𝜃+𝑎 𝜃 . cos﷮𝜃﷯ sin﷮𝜃﷯=−𝑥 cos﷮𝜃﷯+𝑎 cos2 𝜃+𝑎 𝜃sin⁡𝜃 cos⁡𝜃 𝑦 sin﷮𝜃﷯−𝑎 sin2 𝜃+𝑥 cos﷮𝜃﷯−𝑎 cos2 𝜃=𝑎 𝜃 sin﷮𝜃﷯ cos﷮𝜃﷯−𝑎 𝜃 sin⁡𝜃 cos⁡𝜃 𝑦 sin﷮𝜃﷯+𝑥 cos﷮𝜃﷯−𝑎 sin2 𝜃−𝑎 cos2 𝜃=0 𝑦 sin﷮𝜃﷯+ 𝑥 cos﷮𝜃﷯−𝑎 sin2 𝜃+cos2 𝜃﷯=0 𝑦 sin﷮𝜃﷯+𝑥 cos﷮𝜃﷯−𝑎 1﷯=0 cos⁡𝜃 . 𝑥+sin⁡𝜃 . 𝑦−𝑎 = 0 Thus, equation of normal is cos θ . 𝑥 + sin θ . 𝑦 – a = 0 Above equation is of the form a𝑥 + b𝑦 + c = 0 where a = cos θ, b = sin θ & c = − a We need to find distance of normal from origin i.e. 𝑥1 = 0 & 𝑦1 = 0 ⇒ 𝑑= cos﷮𝜃 0﷯ + sin﷮𝜃 0﷯ − 𝑎﷯﷯﷯﷮ ﷮ cos﷮2﷯﷮𝜃﷯ + sin﷮2﷯﷮𝜃﷯﷯﷯ d = 0 + 0 − 𝑎﷯﷮ ﷮1﷯﷯ d = −𝑎﷯﷮1﷯ d = 𝑎﷮1﷯ d = a (constant) Hence distance of normal from origin is a constant. Hence proved.