Misc 5 - Chapter 6 Class 12 Application of Derivatives

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 sin⁡𝜃 , 𝑦=𝑎 sin⁡𝜃– 𝑎 𝜃 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) 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 )) 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 )) We know that Equation of line passing through (𝑥1 , 𝑦1) & having slope m Is (𝑦−𝑦1) = m(𝑥−𝑥1) (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 𝑥 cos⁡𝜃+𝑦 sin⁡𝜃 −𝑎 = 0 Thus, Equation of normal is 𝑥 cos⁡𝜃+𝑦 sin⁡𝜃 −𝑎 = 0 Finding Distance of Normal from Origin Equation of normal is 𝑥 cos⁡𝜃+𝑦 sin⁡𝜃 −𝑎 = 0 Comparing with 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 We know that Distance of line ax + by + c = 0 from point (x1, y1) is d = |𝑎𝑥1 + 𝑏𝑦1 +𝑐|/√(𝐴^2 + 𝐵^2 ) d = |≺𝑎|/1 d = 𝑎/1 d = a (constant) Hence, distance of normal from origin is a constant. Hence proved.