Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 202: 87

\[\begin{align} & \text{No horizontal tangent line} \\ & \text{Vertical tangent lines at}\left( 2,1 \right),\left( -2,1 \right) \\ \end{align}\]

\[\begin{align} & {{x}^{2}}\left( 3{{y}^{2}}-2{{y}^{3}} \right)=4 \\ & \text{Differentiate implicitly with respect to }x \\ & \frac{d}{dx}\left[ {{x}^{2}}\left( 3{{y}^{2}}-2{{y}^{3}} \right) \right]=\frac{d}{dx}\left[ 4 \right] \\ & {{x}^{2}}\left( 6y\frac{dy}{dx}-6{{y}^{2}}\frac{dy}{dx} \right)+2x\left( 3{{y}^{2}}-2{{y}^{3}} \right)=0 \\ & \text{Solve for }\frac{dy}{dx} \\ & 6{{x}^{2}}y\left( 1-y \right)\frac{dy}{dx}=-2x\left( 3{{y}^{2}}-2{{y}^{3}} \right) \\ & \frac{dy}{dx}=\frac{-2x\left( 3{{y}^{2}}-2{{y}^{3}} \right)}{6{{x}^{2}}y\left( 1-y \right)} \\ & \frac{dy}{dx}=\frac{-\left( 3{{y}^{2}}-2{{y}^{3}} \right)}{3xy\left( 1-y \right)} \\ & \frac{dy}{dx}=\frac{2{{y}^{3}}-3{{y}^{2}}}{3xy-3x{{y}^{2}}} \\ & \frac{dy}{dx}=\frac{2{{y}^{2}}-3y}{3x-3xy} \\ & \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{2{{y}^{2}}-3y}{3x-3xy}=0 \\ & 2{{y}^{2}}-3y=0 \\ & y\left( 2y-3 \right)=0 \\ & y=0,\text{ }y=\frac{3}{2} \\ & \text{Substitute }y=0\text{ and }y=\frac{3}{2}\text{ into }{{x}^{2}}\left( 3{{y}^{2}}-2{{y}^{3}} \right)=4 \\ & {{x}^{2}}\left( 3{{\left( 0 \right)}^{2}}-2{{\left( 0 \right)}^{3}} \right)=4 \\ & 0\ne 4 \\ & {{x}^{2}}\left( 3{{\left( \frac{3}{2} \right)}^{2}}-2{{\left( \frac{3}{2} \right)}^{3}} \right)=4 \\ & {{x}^{2}}\left( \frac{27}{4}-\frac{27}{4} \right)=4 \\ & 0\ne 4 \\ & \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{2{{y}^{2}}-3y}{3x-3xy}=\infty \\ & 3x-3xy=\infty \\ & 3x\left( 1-y \right)=\infty \\ & x=0,\text{ }y=1 \\ & \text{Substituting }x=0\text{ into }{{x}^{2}}\left( 3{{y}^{2}}-2{{y}^{3}} \right)=4 \\ & {{\left( 0 \right)}^{2}}\left( 3{{y}^{2}}-2{{y}^{3}} \right)=4 \\ & 0\ne 4 \\ & \text{Substituting }y=1\text{ into }{{x}^{2}}\left( 3{{y}^{2}}-2{{y}^{3}} \right)=4 \\ & {{x}^{2}}\left( 3{{\left( 1 \right)}^{2}}-2{{\left( 1 \right)}^{3}} \right)=4 \\ & {{x}^{2}}\left( 1 \right)=4 \\ & {{x}^{2}}=4 \\ & x=\pm 2 \\ & \text{We obtain the points }\left( 2,1 \right)\text{ and }\left( -2,1 \right) \\ & \text{The points at which the curve has vertical tangent lines } \text{ are }\left( 2,1 \right),\left( -2,1 \right) \\ & \\ & \text{This graph confirms the result} \\ \end{align}\]