You are here
Question 16 - Finding point when tangent is parallel/ perpendicular - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Finding point when tangent is parallel/ perpendicular
Chapter 6 Class 12 Application of Derivatives
Concept wise
- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
-
Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
- Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima
Transcript
Question 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=7𝑥^3+11 Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=𝑑(7𝑥3 + 11)/𝑑𝑥 𝑑𝑦/𝑑𝑥=21𝑥^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=21𝑥^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
