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

π/4 makes with x-axis an angle:

(A)0 

(B) π/4 

(C) π/3 

(D) π/2

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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

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 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.