Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.3,16 Show that the tangents to the curve 𝑦=7𝑥3+11 at the points where 𝑥=2 and 𝑥 =−2 are parallel. We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line 𝑚1=𝑚2 We know that Slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ Given Curve is 𝑦=7﷐𝑥﷮3﷯+11 Differentiating w.r.t.𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐𝑑﷐7𝑥3 + 11﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=21﷐𝑥﷮2﷯ We need to show that tangent at 𝑥=2 & tangent at 𝑥=−2 are parallel i.e. we need to show (Slope of tangent at 𝑥=2) = (Slope of tangent at 𝑥=−2) Now, Slope of tangent = ﷐𝑑𝑦﷮𝑑𝑥﷯=21﷐𝑥﷮2﷯ Slope of tangent at 𝑥=2 ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮𝑥 = 2﷯=21﷐﷐2﷯﷮2﷯=21 ×4=84 & Slope of tangent at 𝑥=−2 ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮𝑥 =− 2﷯=21﷐﷐−2﷯﷮2﷯=21 ×4=84 Since, (Slope of tangent at 𝑥=2) = (Slope of tangent at 𝑥=−2) Thus , tangent at 𝑥=2 & tangent at 𝑥 are parallel Hence Proved