Ex 5.6, 10 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Derivatives in parametric form
Derivatives in parametric form
Last updated at April 16, 2024 by Teachoo
Ex 5.6, 10 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = π (cosβ‘π + π sinβ‘π), π¦ = π (sinβ‘π β π cosβ‘π)Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ π¦ = π (sinβ‘π β π cosβ‘π) ππ¦/ππ = (π(π (sinβ‘π β π cosβ‘π)) )/ππ ππ¦/ππ = a ((π (sinβ‘π β π cosβ‘π ))/ππ) ππ¦/ππ = a ((π (sinβ‘π ) )/ππ β (π (π cosβ‘π ))/ππ) ππ¦/ππ = a (cosβ‘πβ (π (π cosβ‘π ))/ππ) ππ¦/ππ = a (cosβ‘πβ((π (π) )/ππ . cosβ‘π+(π (cosβ‘π ) )/ππ . π)) ππ¦/ππ = a (cosβ‘πβ(cosβ‘π+(βsinβ‘π )) π) Using Product Rule As (π’π£)β = π’βπ£ + π£βπ’ ππ¦/ππ = a (πππ β‘πβcosβ‘π+π sinβ‘π ) ππ¦/ππ = a (π sinβ‘π ) π π/π π½ = π π½. πππβ‘π½ Calculating π π/π π½ π₯=π (cosβ‘π+ π sinβ‘π ) ππ₯/ππ = π(π (cosβ‘π+ π sinβ‘π ))/ππ ππ₯/ππ = π π(cosβ‘π+ π sinβ‘π )/ππ ππ₯/ππ = π (π(cosβ‘π )/ππ + π(π sinβ‘π )/ππ) Using Product Rule As (π’π£)β = π’βπ£ + π£βπ’ ππ₯/ππ = π (βsinβ‘π+ π(π sinβ‘π )/ππ) ππ₯/ππ = π (βsinβ‘π+(ππ/ππ . sinβ‘π+ π(sinβ‘π )/ππ . π)) ππ₯/ππ = π (βsinβ‘π+(sinβ‘π+cosβ‘π. π)) ππ₯/ππ = π (βsinβ‘π+sinβ‘π+π.cosβ‘π ) π π/π π½ = π (π½ πππβ‘π½ ) Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (π (π. sinβ‘π ))/π" " (π cosβ‘π ) ππ¦/ππ₯ = sinβ‘π/cosβ‘π π π/π π = πππβ‘π½