The curvex^3+y^3−3xy=0is known as the Folium of Decsartes. In 1638 the famous philosopher, diplomat, soldier and mathematician René Descartes challenged the lawyer and rival mathematician Pierre de Fermat to find the tangent to this curve at an arbitrary point. Fermat easily found the tangent, which is quite impressive since his method pre-dates Calculus by about 40 years! Now, about 400 years later we can find the tangent easily using implicit differentiation:dydx= .Evaluated at the point [a,b]=[2/3,4/3]=[2/3,4/3] the gradient ism=dydx(2/3,4/3)(2/3,4/3)= .The formula for the equation of the tangent at [a,b] with gradient m is:y−b=m(x−a)At the point [a,b]=[2/3,4/3]=[2/3,4/3], this can be rewritten asy== .The Folium of Descartes (red) and the tangent at [2/3,4/3][2/3,4/3] (blue) are displayed below The curveis known as the Folium of Decsartes. In 1638 the famous philosopher, diplomat, soldier and mathematician RenéDescartes challenged the lawyer and rival mathematician Pierre de Fermat to find the tangent to this curve at anarbitrary point.2³+y³ - 3xy=03Pierre de Fermat (1601-1665)Fermat easily found the tangent, which is quite impressive since his method pre-dates Calculus by about 40 years!Now, about 400 years later we can find the tangent easily using implicit differentiation:René Descartes (1596-1650)dydx= (y-x^2)/(y^2-x)Evaluated at the point [a, b] = [2/3,4/3] the gradient isdy (2/3,4/3) = 0dzThe formula for the equation of the tangent at [a, b] with gradient m is:y-b=m(x-a).m =At the point [a, b] = [2/3,4/3], this can be rewritten asy =The Folium of Descartes (red) and the tangent at [2/3,4/3] (blue) are displayed below.3

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Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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The curve

x^3+y^3−3xy=0

is known as the Folium of Decsartes. In 1638 the famous philosopher, diplomat, soldier and mathematician René Descartes challenged the lawyer and rival mathematician Pierre de Fermat to find the tangent to this curve at an arbitrary point.

Fermat easily found the tangent, which is quite impressive since his method pre-dates Calculus by about 40 years! Now, about 400 years later we can find the tangent easily using implicit differentiation:

dydx= .

Evaluated at the point [a,b]=[2/3,4/3]=[2/3,4/3] the gradient is

m=dydx(2/3,4/3)(2/3,4/3)= .

The formula for the equation of the tangent at [a,b] with gradient m is:

y−b=m(x−a)

At the point [a,b]=[2/3,4/3]=[2/3,4/3], this can be rewritten as

y== .

The Folium of Descartes (red) and the tangent at [2/3,4/3][2/3,4/3] (blue) are displayed below