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
Last updated at Oct. 13, 2021 by Teachoo
Transcript
Question 3 The tangent to the curve given by x = et . cost, y = et . sint at t = π/4 makes with x-axis an angle: 0 (B) π/4 (C) π/3 (D) π/2 Given curve π=π^π.ππππ and π=π^π.ππππ Finding slope of tangent π π/π π=(π π/π π)/(π π/π π) π=π^π.ππππ Differentiating π₯ π€.π.π‘ π‘ ππ₯/ππ‘=π^π‘ πππ π‘βπ^π‘ π πππ‘ π π/π π=π^π‘ (πππ π‘βπ πππ‘) π=π^π.ππππ Differentiating y π€.π.π‘ π‘ ππ¦/ππ‘=π^π‘ π πππ‘+π^π‘ πππ π‘ π π/π π=π^π‘ (π πππ‘\+πππ π‘) Now, π π/π π=(π π/π π)/(π π/π π) ππ¦/ππ₯=(π^π‘ (π πππ‘\+πππ π‘)" " )/(π^π‘ (πππ π‘βπ πππ‘)" " ) ππ¦/ππ₯=( (ππππ\+ππππ)" " )/( (ππππβππππ)" " ) Putting π=π/π β π π/π πβ€|_(π=π /π)=( (π ππ π/4 \+ πππ π/4)" " )/( (πππ π/4 β π ππ π/4)" " ) = (1/β2+1/β2)/(1/β2β1/β2) = (2/(β2))/0 = β Let π½ be the angle made by tangent with the π₯- axis. Slope = πππ§β‘π½ β = tanβ‘π tanβ‘γπ=βγ β΄ π½= π /π So, the correct answer is (D)
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