Explanation Formula Used: Quotient rule d d t ( y x ) = x d y d t − y d x d t x 2 Area enclosed by C = ∮ C x d y = ∮ C y d x = 1 2 ∫ C x d y − y d x Given: x 3 + y 3 = 3 x y Calculation: (a) We have to show that the folium has a parameterization in terms of t = y x , given by x = 3 t 1 + t 3 and y = 3 t 2 1 + t 3 ( − ∞ < t < ∞ ) ( t ≠ 1 ) We have been given, x 3 + y 3 = 3 x y In order to prove the parameterization, we substitute the x and y values in the given equation. Left hand-side = x 3 + y 3 = ( 3 t 1 + t 3 ) 3 + ( 3 t 2 1 + t 3 ) 3 x 3 + y 3 = ( 27 t 3 ( 1 + t 3 ) 3 ) + ( 27 t 6 ( 1 + t 3 ) 3 ) x 3 + y 3 = 27 t 3 + 27 t 6 ( 1 + t 3 ) 3 x 3 + y 3 = 27 t 3 ( 1 + t 3 ) ( 1 + t 3 ) 3 x 3 + y 3 = 27 t 3 ( 1 + t 3 ) 2 ------------------------(1) Right hand-side = 3 x y = 3 ( 3 t 1 + t 3 ) ( 3 t 2 1 + t 3 ) 3 x y = 27 t 3 ( 1 + t 3 ) 2 ------------------------------(2) From (1) and (2) we can see that Left hand-side = Right hand-side That means x = 3 t 1 + t 3 and y = 3 t 2 1 + t 3 satisfies the given equation, hence it is proved that the curve parameterized by x = 3 t 1 + t 3 and y = 3 t 2 1 + t 3 lies on the folium of Descartes. We can also show that, this parameterization can be solved in terms of t as t = y x Therefore, y x = 3 t 2 1 + t 3 3 t 1 + t 3 y x = 3 t 2 1 + t 3 ⋅ 1 + t 3 3 t y x = t Hence Proved...

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Chapter 18.1, Problem 23E

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To derive: the parameters and area of the folium of Descartes.

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Chapter 18.1, Problem 22E

Chapter 18.1, Problem 24E

Chapter 18 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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Finding Local Maxima and Minima by Differentiation; Author: Professor Dave Explains;https://www.youtube.com/watch?v=pvLj1s7SOtk;License: Standard YouTube License, CC-BY

To derive: the parameters and area of the folium of Descartes.

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To derive: the parameters and area of the folium of Descartes.

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Chapter 18 Solutions