Chapter 4.2, Problem 55E

(a)

To determine

To find: The slope of the folium of Descartes at the given points.

Answer to Problem 55E

The slope at the point $\left(4,2\right)$ is $\frac{5}{4}$ .

The slope at the point $\left(2,4\right)$ is $\frac{4}{5}$.

Explanation of Solution

Given Information: The equation of the folium of Descartes is $x^3+y^3-9xy=0$ at the points $\left(4,2\right),\left(2,4\right)$ .

Calculation:

The slope of the folium of Descartes $\frac{dy}{dx}$ is,

  $\begin{array}{c}{x}^3+y^3-9xy=0\\ 3x^2+3y^2\frac{dy}{dx}-9y-9x\frac{dy}{dx}=0\\ \left(3x^2-9y\right)+\left(3y^2-9x\right)\frac{dy}{dx}=0\\ \left(3y^2-9x\right)\frac{dy}{dx}=-\left(3x^2-9y\right)\\ \frac{dy}{dx}=-\frac{\left(3x^2-9y\right)}{3y^2-9x}\\ =-\frac{x^2-3y}{y^2-3x}\end{array}$

The slope at the point $\left(4,2\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{x^2-3y}{y^2-3x}\\ =-\frac{{\left(4\right)}^2-3\left(2\right)}{{\left(2\right)}^2-3\left(4\right)}\\ =-\frac{16-6}{4-12}\\ =-\frac{10}{-8}\\ =\frac{5}{4}\end{array}$

The slope at the point $\left(2,4\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{x^2-3y}{y^2-3x}\\ =-\frac{{\left(2\right)}^2-3\left(4\right)}{{\left(4\right)}^2-3\left(2\right)}\\ =-\frac{4-12}{16-6}\\ =-\frac{-8}{10}\\ =\frac{4}{5}\end{array}$

Conclusion:

The slope at the point $\left(4,2\right)$ is $\frac{5}{4}$ .

The slope at the point $\left(2,4\right)$ is $\frac{4}{5}$.

(b)

To determine

To calculate: Thepoint other than the origin does the folium have a horizontal tangent.

Answer to Problem 55E

The coordinate other than origin is $\left(3\sqrt[3]{2},3\sqrt[3]{4}\right)$ .

Explanation of Solution

Given Information: The equation of the folium of Descartes is $x^3+y^3-9xy=0$ .

Calculation:

The folium of Descartes has a horizontal tangent when $\frac{dy}{dx}=0$ .

The value of $\frac{dy}{dx}$ is,

  $\begin{array}{c}{x}^3+y^3-9xy=0\\ 3x^2+3y^2\frac{dy}{dx}-9y-9x\frac{dy}{dx}=0\\ \left(3x^2-9y\right)+\left(3y^2-9x\right)\frac{dy}{dx}=0\\ \left(3y^2-9x\right)\frac{dy}{dx}=-\left(3x^2-9y\right)\\ \frac{dy}{dx}=-\frac{\left(3x^2-9y\right)}{3y^2-9x}\\ =-\frac{x^2-3y}{y^2-3x}\end{array}$

Equate the derivative to 0,

  $\begin{array}{c}\frac{dy}{dx}=0\\ -\frac{x^2-3y}{y^2-3x}=0\\ x^2-3y=0\\ x^2=3y\\ y=\frac{x^2}{3}\end{array}$

Substitute $y=\frac{x^2}{3}$ in $x^3+y^3-9xy=0$ to obtain the values of x.

  $\begin{array}{c}{x}^3+y^3-9xy=0\\ x^3+{\left(\frac{x^2}{3}\right)}^3-9x\left(\frac{x^2}{3}\right)=0\\ x^3+\frac{x^6}{27}-3x^3=0\\ \frac{x^6}{27}-2x^3=0\\ x^3\left(\frac{x^3}{27}-2\right)=0\\ x=0,\frac{x^3}{27}-2=0\\ x=0,\frac{x^3}{27}=2\\ x=0,x^3=54\\ x=0,\sqrt[3]{54}\\ =0,3\sqrt[3]{2}\end{array}$

Since $y=\frac{x^2}{3}$ ,

  $\begin{array}{l}\text{When }x=0,y=0\\ \text{When }x=3\sqrt[3]{2},y=\frac{{\left(3\sqrt[3]{2}\right)}^2}{3}\\ \text{ =3}\sqrt[3]{4}\end{array}$

Hence, the coordinate other than origin is $\left(3\sqrt[3]{2},3\sqrt[3]{4}\right)$ .

Conclusion:

The coordinate other than origin is $\left(3\sqrt[3]{2},3\sqrt[3]{4}\right)$ .

(c)

To determine

To calculate: Thefolium have a vertical tangent.

Answer to Problem 55E

The coordinate other than origin is $\left(3\sqrt[3]{4},3\sqrt[3]{2}\right)$ .

Explanation of Solution

Given Information: The equation of the folium of Descartes is $x^3+y^3-9xy=0$ .

Calculation:

The folium of Descartes has a vertical tangent when $\frac{dx}{dy}=0$ .

The value of $\frac{dx}{dy}$ is,

  $\begin{array}{c}{x}^3+y^3-9xy=0\\ 3x^2+3y^2\frac{dy}{dx}-9y-9x\frac{dy}{dx}=0\\ \left(3x^2-9y\right)+\left(3y^2-9x\right)\frac{dy}{dx}=0\\ \left(3y^2-9x\right)\frac{dy}{dx}=-\left(3x^2-9y\right)\\ \frac{dy}{dx}=-\frac{\left(3x^2-9y\right)}{3y^2-9x}\\ =-\frac{x^2-3y}{y^2-3x}\\ \frac{dx}{dy}=-\frac{y^2-3x}{x^2-3y}\end{array}$

Equate the derivative to 0,

  $\begin{array}{c}\frac{dx}{dy}=0\\ -\frac{y^2-3x}{x^2-3y}=0\\ y^2-3x=0\\ y^2=3x\\ x=\frac{y^2}{3}\end{array}$

Substitute $x=\frac{y^2}{3}$ in $x^3+y^3-9xy=0$ to obtain the values of x.

  $\begin{array}{c}{x}^3+y^3-9xy=0\\ {\left(\frac{y^2}{3}\right)}^3+y^3-9\left(\frac{y^2}{3}\right)y=0\\ \frac{y^6}{27}+y^3-3y^3=0\\ \frac{y^6}{27}-2y^3=0\\ y^3\left(\frac{y^3}{27}-2\right)=0\\ y=0,\frac{y^3}{27}-2=0\\ y=0,\frac{y^3}{27}=2\\ y=0,y^3=54\\ y=0,\sqrt[3]{54}\\ =0,3\sqrt[3]{2}\end{array}$

Since $x=\frac{y^2}{3}$ ,

  $\begin{array}{l}\text{When }x=0,y=0\\ \text{When }y=3\sqrt[3]{2},x=\frac{{\left(3\sqrt[3]{2}\right)}^2}{3}\\ \text{ =3}\sqrt[3]{4}\end{array}$

Hence, the coordinate other than origin is $\left(3\sqrt[3]{4},3\sqrt[3]{2}\right)$ .

Conclusion:

The coordinate other than origin is $\left(3\sqrt[3]{4},3\sqrt[3]{2}\right)$ .