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Tangents and Normals (using Differentiation)
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Tangents and Normals (using Differentiation)
Last updated at May 29, 2023 by Teachoo
Question 14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) π₯=cosβ‘π‘, π¦=sinβ‘π‘ ππ‘ π‘= π/4At π= π /π x = cos π/4 = 1/β2 y = sin π/4 = 1/β2 β΄ At π‘=π/4 , the point is (1/β2 " ," 1/β2) Finding ππ¦/ππ₯ ππ¦/ππ₯=(ππ¦/ππ‘)/(ππ₯/ππ‘) Now, Hence, ππ¦/ππ₯=(ππ¦/ππ‘)/(ππ₯/ππ‘) π=πππβ‘π Differentiating w.r.t.π‘ ππ₯/ππ‘=π(cosβ‘π‘ )/ππ‘ ππ₯/ππ‘=βsinβ‘π‘ π=πππβ‘π Differentiating w.r.t. π‘ ππ¦/ππ‘=π(sinβ‘π‘ )/ππ‘ ππ¦/ππ‘=cosβ‘π‘ ππ¦/ππ₯=cosβ‘π‘/(βsinβ‘π‘ ) ππ¦/ππ₯=βcotβ‘π‘ At π=π /π γππ¦/ππ₯βγ_(π‘=π/4)=βπππ‘(π/4) =β1 β΄ Slope of tangent at (1/β2 " ," 1/β2) is β 1 We know that Slope of tangent Γ Slope of Normal =β1 β1 Γ Slope of Normal =β1 Slope of Normal =(β1)/(β1) Slope of Normal =1 Hence Slope of tangent is β 1 & Slope of Normal is 1 Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (1/β2 " ," 1/β2) & having Slope β1 is π¦ β1/β2 =βπ₯+ 1/β2 π₯+π¦ =1/β2+ 1/β2 π₯+π¦ =2/β2 π₯+π¦ =β2 π+πββπ=π Equation of Normal at (1/β2 " ," 1/β2) & having Slope 1 is (π¦β1/β2)=1(π₯β1/β2) π¦ β1/β2 =π₯β 1/β2 π¦ =π₯β1/β2+ 1/β2 π¦ =π₯ π =π