Ex 5.6, 6 - Find dy/dx, x = a (theta - sin), y = a (1 + cos)

Ex 5.6, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.6, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 5.6, 6 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯ = π‘Ž (πœƒ – sinβ‘πœƒ), 𝑦 = π‘Ž (1 + cosβ‘πœƒ)Here 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑦 = π‘Ž (1+cosβ‘πœƒ) 𝑑𝑦/π‘‘πœƒ = 𝑑(π‘Ž (1+cosβ‘πœƒ))/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = π‘Ž (𝑑(1 + cosβ‘πœƒ )/π‘‘πœƒ) 𝑑𝑦/π‘‘πœƒ = π‘Ž (0βˆ’sinβ‘πœƒ ) 𝑑𝑦/π‘‘πœƒ = π‘Ž (βˆ’sinβ‘πœƒ ) π’…π’š/π’…πœ½ = βˆ’π’‚ π’”π’Šπ’β‘πœ½ Calculating 𝒅𝒙/π’…πœ½ π‘₯=π‘Ž (πœƒ –sinβ‘πœƒ ) 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž πœƒ βˆ’π‘Ž sinβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž πœƒ)/π‘‘πœƒ βˆ’ 𝑑(π‘Ž sinβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Žβˆ’γ€–π‘Ž cosγ€—β‘πœƒ 𝒅𝒙/π’…πœ½ = 𝒂(γ€–πŸβˆ’π’„π’π’”γ€—β‘πœ½ ) Therefore, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = (βˆ’π‘Ž sinβ‘πœƒ)/π‘Ž(γ€–1 βˆ’ cosγ€—β‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = (βˆ’sinβ‘πœƒ)/γ€–1 βˆ’ cosγ€—β‘πœƒ 𝑑𝑦/𝑑π‘₯ = (βˆ’2 γ€–sin γ€—β‘γ€–πœƒ/2γ€— γ€–cos γ€—β‘γ€–πœƒ/2γ€—)/(2 γ€–sin^2 γ€—β‘γ€–πœƒ/2γ€— ) 𝑑𝑦/𝑑π‘₯ = (βˆ’γ€–cos γ€—β‘γ€–πœƒ/2γ€—)/(sin πœƒ/2) π’…π’š/𝒅𝒙 = βˆ’π’„π’π’•β‘γ€–πœ½/πŸγ€— Rough We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by πœƒ/2 sin ΞΈ = 2 π’”π’Šπ’β‘γ€–πœ½/πŸγ€— π’„π’π’”β‘γ€–πœ½/πŸγ€— and cos 2ΞΈ = 2cos2 ΞΈ – 1 Replacing ΞΈ by πœƒ/2 cos ΞΈ = 2cos2 𝜽/𝟐 – 1

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.