Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12






Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Misc 5 Show that the normal at any point ΞΈ to the curve x = a cos π + a π sin π, y = a sin π β a π cos π is at a constant distance from the origin.Given curve π₯=π cosβ‘π+π π sinβ‘π , π¦=π sinβ‘πβ π π cosβ‘π 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 ππ¦/ππ₯ ππ¦/ππ₯= (ππ¦/ππ)/(ππ₯/ππ) Given π₯=π cosβ‘π+π π sinβ‘π Diff. w.r.t ΞΈ ππ₯/ππ= π(π cosβ‘π + π π sinβ‘π )/ππ ππ₯/ππ= (π(π cosβ‘π))/ππ + (π(ππ sinβ‘π))/ππ ππ₯/ππ = π (βsinβ‘π )+π (π(π sinβ‘π))/ππ Using product rule (u v)β = uβ v + vβ u ππ₯/ππ=β a sin ΞΈ+a (ππ/ππ sin ΞΈ+ (π(π ππ π))/ππ ΞΈ) ππ¦/ππ=π cosβ‘πβπ(cosβ‘π+(βsinβ‘π )π) ππ¦/ππ=π cosβ‘πβ(π cosβ‘πβπ π sinβ‘π ) ππ¦/ππ=π cos πβπ cosβ‘π + π . π sinβ‘π. ππ¦/ππ= π. π .sin. π Now, ππ¦/ππ₯= (ππ¦βππ)/(ππ₯βππ) ππ¦/ππ₯=(π π sinβ‘π)/(π π cosβ‘π ) ππ¦/ππ₯=sinβ‘π/cosβ‘π ππ¦/ππ₯= tanβ‘π We know that Slope of tangent of Γ Slope of normal = β1 tan ΞΈ Γ Slope of normal = β1 Slope of normal = (β1)/tanβ‘π Slope of normal =βcotβ‘π Equation of normal which passes through the curve π₯ = a cos ΞΈ + a ΞΈ sin ΞΈ & π¦ = a sin ΞΈ - a ΞΈ cos ΞΈ & has slope βcotβ‘π is (π¦β(π sinβ‘πβπ cosβ‘π ))=βcotβ‘π(π₯β(π cosβ‘π+π π sinβ‘π )) We know that Slope of tangent of Γ Slope of normal = β1 tan ΞΈ Γ Slope of normal = β1 Slope of normal = (β1)/tanβ‘π Slope of normal =βcotβ‘π Equation of normal which passes through the curve π₯ = a cos ΞΈ + a ΞΈ sin ΞΈ & π¦ = a sin ΞΈ - a ΞΈ cos ΞΈ & has slope βcotβ‘π is (π¦β(π sinβ‘πβπ cosβ‘π ))=βcotβ‘π(π₯β(π cosβ‘π+π π sinβ‘π )) We know that Equation of line passing through (π₯1 , π¦1) & having slope m Is (π¦βπ¦1) = m(π₯βπ₯1) (π¦βπ sinβ‘π+π π cosβ‘π )=(βcosβ‘π)/sinβ‘π (π₯βπ cosβ‘π+π π sinβ‘π ) sinβ‘π(π¦βπ sinβ‘π+π π cosβ‘π )=βcosβ‘π(π₯βπ cosβ‘πβπ π sinβ‘π ) π¦ sinβ‘πβπ sin2 π+π π .cosβ‘π sinβ‘π=βπ₯ cosβ‘π+π cos2 π+π πsinβ‘π cosβ‘π π¦ sinβ‘πβπ sin2 π+π₯ cosβ‘πβπ cos2 π=π π sinβ‘π cosβ‘πβπ π sinβ‘π cosβ‘π π¦ sinβ‘π+π₯ cosβ‘πβπ sin2 πβπ cos2 π=0 π¦ sinβ‘π+ π₯ cosβ‘πβπ (sin2 π+cos2 π)=0 π¦ sinβ‘π+π₯ cosβ‘πβπ (1)=0 π₯ cosβ‘π+π¦ sinβ‘π βπ = 0 Thus, Equation of normal is π₯ cosβ‘π+π¦ sinβ‘π βπ = 0 Finding Distance of Normal from Origin Equation of normal is π₯ cosβ‘π+π¦ sinβ‘π βπ = 0 Comparing with aπ₯ + bπ¦ + c = 0 where a = cos ΞΈ, b = sin ΞΈ & c = β a We need to find distance of normal from origin i.e. π₯1 = 0 & π¦1 = 0 π= |cosβ‘γπ(0) + sinβ‘γπ(0) β πγ γ |/β(cos^2β‘π + sin^2β‘π ) d = |0 + 0 β π|/β1 We know that Distance of line ax + by + c = 0 from point (x1, y1) is d = |ππ₯1 + ππ¦1 +π|/β(π΄^2 + π΅^2 ) d = |βΊπ|/1 d = π/1 d = a (constant) Hence, distance of normal from origin is a constant. Hence proved.
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