## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.7 - Implicit Differentiation - Exercises - Page 165: 46

#### Answer

a) At $(4,2)$, the slope of the curve is $5/4$ and at $(2,4)$, that is $4/5$. b) Besides the origin, at $(3\sqrt[3]2,3\sqrt[3]4)$, the folium also has a horizontal tangent. c) At the origin and at $(3\sqrt[3]4,3\sqrt[3]2)$, the folium has vertical tangents.

#### Work Step by Step

$$x^3+y^3-9xy=0$$ Find the derivative of the function using implicit differentiation: $$3x^2+3y^2y'-9(y+xy')=0$$ $$x^2+y^2y'-3(y+xy')=0$$ $$x^2+y^2y'-3y-3xy'=0$$ $$y'(y^2-3x)=3y-x^2$$ $$y'=\frac{3y-x^2}{y^2-3x}$$ a) The slope of the folium at $(4,2)$ is $$y'=\frac{3\times2-4^2}{2^2-3\times4}=\frac{-10}{-8}=\frac{5}{4}$$ - The slope of the folium at $(2,4)$ is $$y'=\frac{3\times4-2^2}{4^2-3\times2}=\frac{8}{10}=\frac{4}{5}$$ b) The folium has a horizontal tangent when there are points $A(x,y)$ in the folium for which $y'=0$: $$y'=0$$ $$\frac{3y-x^2}{y^2-3x}=0$$ $$3y-x^2=0$$ $$y=\frac{x^2}{3}$$ Now we substitute $y=(x^2)/3$ back to the formula of the function $x^3+y^3-9xy=0$ to find if there exists any point $A$ besides the origin for which $y'=0$ or not: $$x^3+\Big(\frac{x^2}{3}\Big)^3-9x\times\frac{x^2}{3}=0$$ $$x^3+\frac{x^6}{27}-3x^3=0$$ $$\frac{x^6}{27}-2x^3=0$$ $$x^3\Big(\frac{x^3}{27}-2\Big)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}\frac{x^3}{27}-2=0$$ Since the question asks for points besides the origin, we leave out $x=0$. $$\frac{x^3}{27}-2=0$$ $$x^3=54$$ $$x=\sqrt[3]{54}=\sqrt[3]{27\times2}=3\sqrt[3]2$$ - For $x=3\sqrt[3]2$: $y=\frac{(3\sqrt[3]2)^2}{3}=\frac{9\times\sqrt[3]4}{3}=3\sqrt[3]4$ So at the point $(3\sqrt[3]2,3\sqrt[3]4)$, the folium also has a horizontal tangent. c) Similarly, the folium has a vertical tangent, whose slope is undefined, when there are points $A(x,y)$ in the folium for which $y'$ is also undefined. In this case, it happens when the denominator of $y'$ equals $0$: $$y^2-3x=0$$ $$3x=y^2$$ $$x=\frac{y^2}{3}$$ Now we substitute $x=(y^2)/3$ back to the formula of the function $x^3+y^3-9xy=0$ to find $A$: $$\Big(\frac{y^2}{3}\Big)^3+y^3-9y\times\frac{y^2}{3}=0$$ $$\frac{y^6}{27}+y^3-3y^3=0$$ $$\frac{y^6}{27}-2y^3=0$$ $$y^3\Big(\frac{y^3}{27}-2\Big)=0$$ $$y=0\hspace{1cm}\text{or}\hspace{1cm}\frac{y^3}{27}-2=0$$ $$y=0\hspace{1cm}\text{or}\hspace{1cm}y^3=54$$ $$y=0\hspace{1cm}\text{or}\hspace{1cm}y=\sqrt[3]{54}=3\sqrt[3]2$$ - For $y=0$: this is the origin. - For $y=3\sqrt[3]2$: $x=\frac{(3\sqrt[3]2)^2}{3}=\frac{9\times\sqrt[3]4}{3}=3\sqrt[3]4$ So at the origin and at the point $(3\sqrt[3]4,3\sqrt[3]2)$, the folium has vertical tangents.

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