Chapter 3.2, Problem 55E

(a)

To determine

To find : slope of the folium of Descartes at indicated points.

Answer to Problem 55E

Slope at $\left(4,2\right)$ is $\frac{5}{4}$ and Slope at $\left(2,4\right)$ is $\frac{4}{5}$ .

Explanation of Solution

Given information :

The equation of the folium of Descartes is $x^3+y^3-9xy=0$ and point $\left(4,2\right)$ and, $\left(2,4\right)$ .

Formula needed :

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Slope of curve is $m={\frac{dy}{dx}}_{\left(a,b\right)}$ .

Differentiate $x^3+y^3-9xy=0$ with respect to $x$ .

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

Solve further,

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

Substitute $\left(4,2\right)$ in the $\frac{dy}{dx}=-\frac{\left(x^2-3y\right)}{\left(y^2-3x\right)}$ in order to find slope at $\left(4,2\right)$ .

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

Substitute $\left(2,4\right)$ in the $\frac{dy}{dx}=-\frac{\left(x^2-3y\right)}{\left(y^2-3x\right)}$ in order to find slope at $\left(2,4\right)$ .

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

(b)

To determine

To find : point other than origin that have horizontal tangent.

Answer to Problem 55E

The other coordinate is $\left(3{\left(2\right)}^{1/3},3{\left(2\right)}^{2/3}\right)$ .

Explanation of Solution

Given information :

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

Formula needed :

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

For horizontal tangents, $\frac{dy}{dx}=0$ .

Differentiate $x^3+y^3-9xy=0$ with respect to $x$ .

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

Solve further,

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

For horizontal tangents, $\frac{dy}{dx}=0$ .

  $\begin{array}{c}\frac{dy}{dx}=0\\ -\frac{\left(x^2-3y\right)}{\left(y^2-3x\right)}=0\\ \left(x^2-3y\right)=0\left[\left(y^2-3x\right)\ne 0\right]\\ y=x^2/3\end{array}$

Substitute $y=x^2/3$ in the curve $x^3+y^3-9xy=0$ .

  $\begin{array}{c}{x}^3+{\left(\frac{x^2}{3}\right)}^3-9x\left(\frac{x^2}{3}\right)=0\\ x^3+\frac{x^6}{27}-3x^3=0\\ \frac{x^6}{27}-2x^3=0\\ x^3\left(\frac{x^3}{27}-2\right)=0\\ x=3{\left(2\right)}^{1/3},0\end{array}$

Substitute $x=3{\left(2\right)}^{1/3}$ in $y=x^2/3$ .

  $\begin{array}{c}y={\left(3{\left(2\right)}^{1/3}\right)}^2/3\\ =\left(9{\left(2\right)}^{2/3}\right)/3\\ =3{\left(2\right)}^{2/3}\end{array}$

The coordinate is $\left(3{\left(2\right)}^{1/3},3{\left(2\right)}^{2/3}\right)$ .

(c)

To determine

To find : point that have vertical tangent.

Answer to Problem 55E

The coordinate is $\left(3{\left(2\right)}^{2/3},3{\left(2\right)}^{1/3}\right)$ where the curve has vertical tangent.

Explanation of Solution

Given information :

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

Formula needed :

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

For vertical tangents, $\frac{dx}{dy}=0$ .

Differentiate $x^3+y^3-9xy=0$ with respect to $x$ .

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

Solve further,

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

For vertical tangents, $\frac{dx}{dy}=0$ .

  $\begin{array}{c}\frac{dx}{dy}=0\\ -\frac{\left(y^2-3x\right)}{\left(x^2-3y\right)}=0\\ \left(y^2-3x\right)=0\left[\left(x^2-3y\right)\ne 0\right]\\ x=y^2/3\end{array}$

Substitute $x=y^2/3$ in the curve $x^3+y^3-9xy=0$ .

  $\begin{array}{c}{y}^3+{\left(\frac{y^2}{3}\right)}^3-9\left(\frac{y^2}{3}\right)y=0\\ y^3+\frac{y^6}{27}-3y^3=0\\ \frac{y^6}{27}-2y^3=0\\ y^3\left(\frac{y^3}{27}-2\right)=0\\ y=3{\left(2\right)}^{1/3},0\end{array}$

Substitute $y=3{\left(2\right)}^{1/3}$ in $x=y^2/3$ .

  $\begin{array}{c}x={\left(3{\left(2\right)}^{1/3}\right)}^2/3\\ =\left(9{\left(2\right)}^{2/3}\right)/3\\ =3{\left(2\right)}^{2/3}\end{array}$

The coordinate is $\left(3{\left(2\right)}^{2/3},3{\left(2\right)}^{1/3}\right)$ where the curve has vertical tangent.