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Misc 20 - Slope of tangent to x = t2 + 3t - 8, y = 2t2 - 2t - 5

Misc 20 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 20 - Chapter 6 Class 12 Application of Derivatives - Part 3
Misc 20 - Chapter 6 Class 12 Application of Derivatives - Part 4

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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)/7We 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.