# Misc 20 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Misc 20 The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2,– 1) is (A) 22/7 (B) 6/7 (C) 7/6 (D) (− 6)/7 We need to find slope of tangent at (2, −1) We know that slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥= (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Finding 𝒅𝒙/𝒅𝒕 Given 𝑥 = t2 + 3t – 8 Differentiating w.r.t t 𝑑𝑥/𝑑𝑡= (𝑑(𝑡^2 + 3𝑡 −8))/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 2t + 3 Finding 𝒅𝒚/𝒅𝒕 Given 𝑦 = 2t2 − 2t − 5 Differentiating w.r.t t 𝑑𝑦/𝑑𝑡= (𝑑 (2𝑡2 − 2𝑡 − 5))/𝑑𝑡 𝑑𝑦/𝑑𝑡=4t −2 Thus, 𝑑𝑦/𝑑𝑥= (𝑑𝑦∕𝑑𝑡)/(𝑑𝑥∕𝑑𝑡) 𝑑𝑦/𝑑𝑥= (4𝑡 − 2)/(2𝑡 + 3) Now we need to find value of t, Given point is (2, –1) Putting 𝑥 = 2 & 𝑦 = –1 in the curve x = t2 + 3t – 8 2 = t2 + 3t − 8 t2 + 3t – 8 – 2 =0 t2 + 3t − 10 = 0 t2 + 5t – 2t −10 = 0 t (t + 5) – 2 (t − 5) = 0 (t −2) (t + 5) = 0 So, t = 2 & t = − 5 y = 2t2 – 2t – 5 – 1 = 2t2 – 2t – 5 2t2 – 2t – 5 + 1 = 0 2t2 – 2t – 4 = 0 2(t2 – t – 2 ) = 0 t2 – t – 2 = 0 t2 – 2t + t – 2 = 0 t (t − 2) + 1(t − 2) = 0 (t + 1) (t – 2) = 0 So, t = −1 & t = 2 Since t = 2 is common in both parts So, we will calculate 𝑑𝑦/𝑑𝑥= (4𝑡 −2)/(2𝑡 −3) at t = 2 At t = 2 𝑑𝑦/𝑑𝑥= (4𝑡 −2)/(2𝑡 + 3) 𝑑𝑦/𝑑𝑥= (4 (2) − 2)/(2 (2) + 3)= (8 − 2)/(4 +3)= 6/7 Hence the correct answer is (B)

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.