Misc 5 - Show that normal at any point is at constant distance - Finding equation of tangent/normal when point and curve is given

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  1. Chapter 6 Class 12 Application of Derivatives
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Misc 5 Show that the normal at any point ฮธ to the curve x = a cos ๐œƒ + a ๐œƒ sin ๐œƒ, y = a sin ๐œƒ โ€“ a ๐œƒ cos ๐œƒ is at a constant distance from the origin. Given curve ๐‘ฅ=๐‘Ž cosโก๐œƒ+๐‘Ž ๐œƒ , ๐‘ฆ=๐‘Ž sinโก๐œƒ โ€“ ๐‘Ž cosโก๐œƒ 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 ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ= ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ๏ทฎ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ๏ทฏ Given ๐‘ฅ=๐‘Ž cosโก๐œƒ+๐‘Ž ๐œƒ sinโก๐œƒ Diff. w.r.t ฮธ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ= ๐‘‘ ๐‘Ž cos๏ทฎ๐œƒ๏ทฏ + ๐‘Ž ๐œƒ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ๏ทฎ๐‘‘๐œƒ๏ทฏ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ= ๐‘‘(๐‘Ž cos๏ทฎ๐œƒ๏ทฏ)๏ทฎ๐‘‘๐œƒ๏ทฏ + ๐‘‘(๐‘Ž๐œƒ sin๏ทฎ๐œƒ๏ทฏ)๏ทฎ๐‘‘๐œƒ๏ทฏ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ = ๐‘Ž โˆ’ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ+๐‘Ž ๐‘‘(๐œƒ sin๏ทฎ๐œƒ๏ทฏ)๏ทฎ๐‘‘๐œƒ๏ทฏ Using product rule (u v)โ€™ = uโ€™ v + vโ€™ u ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ=โˆ’ a sin ฮธ+a ๐‘‘๐œƒ๏ทฎ๐‘‘๐œƒ๏ทฏ sin ฮธ+ ๐‘‘(๐‘ ๐‘–๐‘› ๐œƒ)๏ทฎ๐‘‘๐œƒ๏ทฏ ฮธ๏ทฏ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ=โˆ’ ๐‘Ž ๐‘ ๐‘–๐‘› ๐œƒ+๐‘Ž ๐‘ ๐‘–๐‘› ๐œƒ+ ๐‘๐‘œ๐‘  ๐œƒ .๏ทฎ๐œƒ๏ทฏ๏ทฏ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ=โˆ’ ๐‘Ž sinโก๐œƒ+๐‘Ž sin๏ทฎ๐œƒ๏ทฏ+๐œƒ cos๏ทฎ๐œƒ๏ทฏ๏ทฏ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ=๐‘Ž ๐œƒ cos๏ทฎ๐œƒ๏ทฏ Also, ๐‘ฆ = ๐‘Ž sinโก๐œƒ โ€“ ๐‘Ž ๐œƒ cosโกฮธ diff w.r.t. ฮธ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ= ๐‘‘(๐‘Ž sin๏ทฎ๐œƒ โˆ’ ๐‘Ž ๐œƒ cos๏ทฎ๐œƒ)๏ทฏ๏ทฏ๏ทฎ๐‘‘ ๐œƒ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ= ๐‘‘(๐‘Ž sin๏ทฎ๐œƒ)๏ทฏ๏ทฎ๐‘‘๐œƒ๏ทฏโˆ’ ๐‘‘(๐‘Ž ๐œƒ cos๏ทฎ๐œƒ)๏ทฏ๏ทฎ๐‘‘๐œƒ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ=๐‘Ž cosโก๐œƒโˆ’๐‘Ž ๐‘‘(๐œƒ cos๏ทฎ๐œƒ)๏ทฏ๏ทฎ๐‘‘๐œƒ๏ทฏ Using product rule (u v)โ€™ = uโ€™v + vโ€™u ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ=๐‘Ž cosโก๐œƒโˆ’๐‘Ž ๐‘‘๐œƒ ๏ทฎ๐‘‘๐œƒ๏ทฏ . cos๏ทฎ๐œƒ๏ทฏ+ ๐‘‘ cos๏ทฎ๐œƒ๏ทฏ๏ทฏ๏ทฎ๐‘‘๐œƒ๏ทฏ. ๐œƒ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ=๐‘Ž cosโก๐œƒโˆ’๐‘Ž cos๏ทฎ๐œƒ๏ทฏ+ โˆ’ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ๐œƒ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ=๐‘Ž cos๏ทฎ๐œƒ๏ทฏโˆ’ ๐‘Ž cos๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž ๐œƒ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ=๐‘Ž cos ๐œƒโˆ’๐‘Ž cosโก๐œƒ + ๐‘Ž . ๐œƒ sinโก๐œƒ. ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ= ๐‘Ž. ๐œƒ .sin. ๐œƒ Now, ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ= ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐œƒ๏ทฏ๏ทฎ ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐œƒ๏ทฏ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ= ๐‘Ž ๐œƒ sin๏ทฎ๐œƒ๏ทฏ๏ทฎ๐‘Ž ๐œƒ cos๏ทฎ๐œƒ๏ทฏ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ= sin๏ทฎ๐œƒ๏ทฏ๏ทฎ cos๏ทฎ๐œƒ๏ทฏ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ= tanโก๐œƒ We know that Slope of tangent of ร— Slope of normal = โˆ’1 tan ฮธ ร— Slope of normal = โˆ’1 Slope of normal = โˆ’1๏ทฎ tan๏ทฎ๐œƒ๏ทฏ๏ทฏ Slope of normal =โˆ’cotโก๐œƒ Equation of normal which passes through the curve ๐‘ฅ = a cos ฮธ + a ฮธ sin ฮธ & ๐‘ฆ = a sin ฮธ - a ฮธ cos ฮธ & has slope โˆ’cotโก๐œƒ is ๐‘ฆโˆ’ ๐‘Ž sin๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž cos๏ทฎ๐œƒ๏ทฏ๏ทฏ๏ทฏ=โˆ’ cot๏ทฎ๐œƒ ๐‘ฅโˆ’ ๐‘Ž cos๏ทฎ๐œƒ๏ทฏ+๐‘Ž ๐œƒ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ๏ทฏ๏ทฏ ๐‘ฆโˆ’๐‘Ž sin๏ทฎ๐œƒ๏ทฏ+๐‘Ž ๐œƒ cos๏ทฎ๐œƒ๏ทฏ๏ทฏ= โˆ’ cos๏ทฎ๐œƒ๏ทฏ๏ทฎ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ ๐‘ฅโˆ’๐‘Ž cos๏ทฎ๐œƒ๏ทฏ+๐‘Ž ๐œƒ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ sin๏ทฎ๐œƒ ๐‘ฆโˆ’๐‘Ž sin๏ทฎ๐œƒ๏ทฏ+๐‘Ž ๐œƒ cos๏ทฎ๐œƒ๏ทฏ๏ทฏ๏ทฏ=โˆ’ cos๏ทฎ๐œƒ ๐‘ฅโˆ’๐‘Ž cos๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž ๐œƒ sin๏ทฎ๐œƒ๏ทฏ๏ทฏ๏ทฏ ๐‘ฆ sin๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž sin2 ๐œƒ+๐‘Ž ๐œƒ . cos๏ทฎ๐œƒ๏ทฏ sin๏ทฎ๐œƒ๏ทฏ=โˆ’๐‘ฅ cos๏ทฎ๐œƒ๏ทฏ+๐‘Ž cos2 ๐œƒ+๐‘Ž ๐œƒsinโก๐œƒ cosโก๐œƒ ๐‘ฆ sin๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž sin2 ๐œƒ+๐‘ฅ cos๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž cos2 ๐œƒ=๐‘Ž ๐œƒ sin๏ทฎ๐œƒ๏ทฏ cos๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž ๐œƒ sinโก๐œƒ cosโก๐œƒ ๐‘ฆ sin๏ทฎ๐œƒ๏ทฏ+๐‘ฅ cos๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž sin2 ๐œƒโˆ’๐‘Ž cos2 ๐œƒ=0 ๐‘ฆ sin๏ทฎ๐œƒ๏ทฏ+ ๐‘ฅ cos๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž sin2 ๐œƒ+cos2 ๐œƒ๏ทฏ=0 ๐‘ฆ sin๏ทฎ๐œƒ๏ทฏ+๐‘ฅ cos๏ทฎ๐œƒ๏ทฏโˆ’๐‘Ž 1๏ทฏ=0 cosโก๐œƒ . ๐‘ฅ+sinโก๐œƒ . ๐‘ฆโˆ’๐‘Ž = 0 Thus, equation of normal is cos ฮธ . ๐‘ฅ + sin ฮธ . ๐‘ฆ โ€“ a = 0 Above equation is of the form a๐‘ฅ + b๐‘ฆ + c = 0 where a = cos ฮธ, b = sin ฮธ & c = โˆ’ a We need to find distance of normal from origin i.e. ๐‘ฅ1 = 0 & ๐‘ฆ1 = 0 โ‡’ ๐‘‘= cos๏ทฎ๐œƒ 0๏ทฏ + sin๏ทฎ๐œƒ 0๏ทฏ โˆ’ ๐‘Ž๏ทฏ๏ทฏ๏ทฏ๏ทฎ ๏ทฎ cos๏ทฎ2๏ทฏ๏ทฎ๐œƒ๏ทฏ + sin๏ทฎ2๏ทฏ๏ทฎ๐œƒ๏ทฏ๏ทฏ๏ทฏ d = 0 + 0 โˆ’ ๐‘Ž๏ทฏ๏ทฎ ๏ทฎ1๏ทฏ๏ทฏ d = โˆ’๐‘Ž๏ทฏ๏ทฎ1๏ทฏ d = ๐‘Ž๏ทฎ1๏ทฏ d = a (constant) Hence distance of normal from origin is a constant. Hence proved.

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  • Abhinav Shukla's image
    Abhinav Shukla
    May 7, 2018, 7:20 p.m.

    In misc exam class 12ch6 q 5X=acostita+atitasintita and the other one are the equation of curve so how can you consider it to be a point for x1andy1 while making the equation of normal

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