Derivatives in parametric form
Derivatives in parametric form
Last updated at Dec. 16, 2024 by Teachoo
Example 31 Find ππ¦/ππ₯ , if π₯ = π cosβ‘π, π¦ = π sin ΞΈ.Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ π¦ = π sin ΞΈ ππ¦/ππ = π(π sin π)/ππ ππ¦/ππ = π πππβ‘π½ Calculating π π/π π½ π¦ = π cos ΞΈ ππ¦/ππ = π(π cos π)/ππ ππ¦/ππ = βπ πππβ‘π½ Now, ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (π cosβ‘ΞΈ)/(βπ sinβ‘ΞΈ ) π π/π π = βππ¨πβ‘π