1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise

Transcript

Ex 6.3,6 Find the slope of the normal to the curve ð¥=1âð sinâ¡ð , ð¦ =ð cosï·®2ï·¯ ð at ð= ðï·®2ï·¯ Slope of tangent is ðð¦ï·®ðð¥ï·¯ ðð¦ï·®ðð¥ï·¯= ðð¦ï·®ððï·¯ï·® ðð¥ï·®ððï·¯ï·¯ Now, ðð¦ï·®ðð¥ï·¯= ðð¦ï·®ððï·¯ï·® ððï·®ððï·¯ï·¯ ðð¦ï·®ðð¥ï·¯= â 2ð sinï·®ð cosï·®ðï·¯ï·¯ï·®â ð cosï·®ðï·¯ï·¯ ðð¦ï·®ðð¥ï·¯= 2ð sinï·®ðï·¯ï·®ðï·¯ We need to find Slope of tangent at ð= ðï·®2ï·¯ Putting ð= ðï·®2ï·¯ ðð¦ï·®ðð¥ï·¯ï·¯ï·®ð = ðï·®2ï·¯ï·¯= 2ðï·®ðï·¯ ð ðð ðï·®2ï·¯ï·¯ = 2ðï·®ðï·¯ 1ï·¯ We know that Tangent is Perpendicular to Normal Hence Slope of tangent Ã Slope of Normal =â1 Slope of Normal = â 1ï·®ððððð ðð ð¡ðððððð¡ ï·¯ Slope of Normal = â 1ï·® ðð¦ï·®ðð¥ï·¯ï·¯ Slope of Normal =â1 Ã ðï·®2ðï·¯ Slope of Normal = âðï·®2ðï·¯ Hence Slope of Normal at is âðï·®ððï·¯