Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 202: 89

\[\begin{align} & \text{No horizontal tangent line} \\ & \text{Vertical tangent lines at}\left( 0,0 \right),\left( -\frac{3\sqrt{3}}{2},-\sqrt{3} \right),\left( \frac{3\sqrt{3}}{2},\sqrt{3} \right) \\ \end{align}\]

\[\begin{align} & x\left( 1-{{y}^{2}} \right)+{{y}^{3}}=0 \\ & \text{Differentiate implicitly with respect to }x \\ & \frac{d}{dx}\left[ x\left( 1-{{y}^{2}} \right)+{{y}^{3}} \right]=\frac{d}{dx}\left[ 0 \right] \\ & x\left( -2y\frac{dy}{dx} \right)+1-{{y}^{2}}+3{{y}^{2}}\frac{dy}{dx}=0 \\ & \text{Solve for }\frac{dy}{dx} \\ & 1-2xy\frac{dy}{dx}-{{y}^{2}}+3{{y}^{2}}\frac{dy}{dx}=0 \\ & -2xy\frac{dy}{dx}+3{{y}^{2}}\frac{dy}{dx}={{y}^{2}}-1 \\ & \left( 3{{y}^{2}}-2xy \right)\frac{dy}{dx}={{y}^{2}}-1 \\ & \frac{dy}{dx}=\frac{{{y}^{2}}-1}{3{{y}^{2}}-2xy} \\ & \text{The points on the curve at which the tangent line is horizontal } \\ & \text{are when the slope is 0, then} \\ & \frac{dy}{dx}=\frac{{{y}^{2}}-1}{3{{y}^{2}}-2xy}=0 \\ & {{y}^{2}}-1=0 \\ & y=-1,\text{ }y=1 \\ & \text{Substitute }y=-1\text{ into }x\left( 1-{{y}^{2}} \right)+{{y}^{3}}=0=0 \\ & x\left( 0 \right)+{{\left( -1 \right)}^{3}}=0 \\ & -1\ne 0 \\ & \text{Substitute }y=-1\text{ into }x\left( 1-{{y}^{2}} \right)+{{y}^{3}}=0=0 \\ & x\left( 0 \right)+{{\left( 1 \right)}^{3}}=0 \\ & -1\ne 0 \\ & \text{Therefore, does not exist points at which the curve has slope 0} \\ & \\ & \text{The points on the curve at which the tangent line is vertical} \\ & \text{are when the slope is }\infty \text{, then} \\ & \frac{dy}{dx}=\frac{{{y}^{2}}-1}{3{{y}^{2}}-2xy}=\infty \\ & 3{{y}^{2}}-2xy=0 \\ & y\left( 3y-2x \right)=0 \\ & y=0,\text{ }y=\frac{2x}{3} \\ & \text{Substituting }0\text{ for }y\text{ into }x\left( 1-{{y}^{2}} \right)+{{y}^{3}}=0 \\ & x\left( 1-{{0}^{2}} \right)+{{0}^{3}}=0 \\ & x=0 \\ & \text{We obtain the point }\left( 0,0 \right) \\ & \text{Substituting }\frac{2x}{3}\text{ for }y\text{ into }x\left( 1-{{y}^{2}} \right)+{{y}^{3}}=0 \\ & x\left( 1-{{\left( \frac{2x}{3} \right)}^{2}} \right)+{{\left( \frac{2x}{3} \right)}^{3}}=0 \\ & x\left( 1-\frac{4{{x}^{2}}}{9} \right)+\frac{8{{x}^{3}}}{27}=0 \\ & x-\frac{4{{x}^{3}}}{9}+\frac{8{{x}^{3}}}{27}=0 \\ & x-\frac{4}{27}{{x}^{3}}=0 \\ & x\left( 1-\frac{4}{27}{{x}^{2}} \right)=0 \\ & x=0 \\ & 1-\frac{4}{27}{{x}^{2}}=0 \\ & {{x}^{2}}=\frac{27}{4}= \\ & x=0,\text{ }x=\pm \sqrt{\frac{27}{4}}=\pm \frac{3\sqrt{3}}{2} \\ & y=\frac{2x}{3}\Rightarrow y=0,\text{ }y=-\sqrt{3},\text{ }y=\sqrt{3} \\ & \text{We obtain the points }\left( 0,0 \right),\left( -\frac{3\sqrt{3}}{2},-\sqrt{3} \right),\left( \frac{3\sqrt{3}}{2},\sqrt{3} \right) \\ & \\ & \text{This graph confirms the resul} \\ \end{align}\]