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The tangent to the curve given by x = e t . cost, y = e t . sint at t =

Ο€/4 makes with x-axis an angle:


(B) Ο€/4Β 

(C) Ο€/3Β 

(D) Ο€/2


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Question 3 The tangent to the curve given by x = et . cost, y = et . sint at t = πœ‹/4 makes with x-axis an angle: 0 (B) πœ‹/4 (C) πœ‹/3 (D) πœ‹/2 Given curve 𝒙=𝒆^𝒕.𝒄𝒐𝒔𝒕 and π’š=𝒆^𝒕.π’”π’Šπ’π’• Finding slope of tangent π’…π’š/𝒅𝒙=(π’…π’š/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) 𝒙=𝒆^𝒕.𝒄𝒐𝒔𝒕 Differentiating π‘₯ 𝑀.π‘Ÿ.𝑑 𝑑 𝑑π‘₯/𝑑𝑑=𝑒^𝑑 π‘π‘œπ‘ π‘‘βˆ’π‘’^𝑑 𝑠𝑖𝑛𝑑 𝒅𝒙/𝒅𝒕=𝑒^𝑑 (π‘π‘œπ‘ π‘‘βˆ’π‘ π‘–π‘›π‘‘) π’š=𝒆^𝒕.π’”π’Šπ’π’• Differentiating y 𝑀.π‘Ÿ.𝑑 𝑑𝑦/𝑑𝑑=𝑒^𝑑 𝑠𝑖𝑛𝑑+𝑒^𝑑 π‘π‘œπ‘ π‘‘ π’…π’š/𝒅𝒕=𝑒^𝑑 (𝑠𝑖𝑛𝑑\+π‘π‘œπ‘ π‘‘) Now, π’…π’š/𝒅𝒙=(π’…π’š/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) 𝑑𝑦/𝑑π‘₯=(𝑒^𝑑 (𝑠𝑖𝑛𝑑\+π‘π‘œπ‘ π‘‘)" " )/(𝑒^𝑑 (π‘π‘œπ‘ π‘‘βˆ’π‘ π‘–π‘›π‘‘)" " ) 𝑑𝑦/𝑑π‘₯=( (π’”π’Šπ’π’•\+𝒄𝒐𝒔𝒕)" " )/( (π’„π’π’”π’•βˆ’π’”π’Šπ’π’•)" " ) Putting 𝐭=𝛑/πŸ’ β”œ π’…π’š/𝒅𝒙─|_(𝒕=𝝅/πŸ’)=( (𝑠𝑖𝑛 πœ‹/4 \+ π‘π‘œπ‘  πœ‹/4)" " )/( (π‘π‘œπ‘  πœ‹/4 βˆ’ 𝑠𝑖𝑛 πœ‹/4)" " ) = (1/√2+1/√2)/(1/√2βˆ’1/√2) = (2/(√2))/0 = ∞ Let 𝜽 be the angle made by tangent with the π‘₯- axis. Slope = 𝐭𝐚𝐧⁑𝜽 ∞ = tanβ‘πœƒ tanβ‘γ€–πœƒ=βˆžγ€— ∴ 𝜽= 𝝅/𝟐 So, the correct answer is (D)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.