Figure 15 $x(t)=t^3-4t,\quad y(t)=t^2$

The plane curve represented by the parametric equations $x(t)=t^{3}-4t,\quad y(t) =t^{2}$, crosses itself at the point $(0,4)$, and so has two tangent lines there, as shown in Figure 15. Find equations for these tangent lines.

Solution We begin by finding the numbers $t$ that correspond to the point $( 0,4) .$ $$\begin{array}{rcl@{\quad\qquad}lrll} x &=&0 & y&=&4 \\ t^{3}-4t &=&0 & t^{2}&=&4 \\ t( t^{2}-4) &=& 0 & t&=&-2\hbox{ or }2 \\ t &=&0,-2,\hbox{or }2 & \end{array}$$

We exclude $t=0,$ since if $t=0,$ then $y\neq 4.$ So we investigate only $t=-2$ and $t=2.$ Since $\dfrac{dx}{dt}=3t^{2}-4\neq 0$ for $t=-2$ and $t=2,$ the slope of the tangent lines is given by $$\begin{equation*} \frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{\dfrac{d}{dt} ( t^{2}) }{\dfrac{d}{dt}( t^{3}-4t) }=\dfrac{2t}{ 3t^{2}-4} \end{equation*}$$

650

When $t=-2$, the slope of the tangent line is $$\begin{equation*} \dfrac{dy}{dx}=\dfrac{2( -2) }{3( -2) ^{2}-4}=\dfrac{-4 }{8}=-\dfrac{1}{2} \end{equation*}$$

and an equation of the tangent line at the point $( 0,4)$ is $$\begin{eqnarray*} y-4 &=&-\dfrac{1}{2}( x-0) \\ y &=&-\dfrac{1}{2}x+4 \end{eqnarray*}$$

When $t=2$, the slope of the tangent line at the point $( 0,4)$ is $$\begin{equation*} \dfrac{dy}{dx}=\dfrac{2( 2) }{3( 2) ^{2}-4}=\dfrac{4}{8 }=\dfrac{1}{2} \end{equation*}$$

and an equation of the tangent line at $( 0,4)$ is $$\begin{eqnarray*} y-4 &=&\dfrac{1}{2}( x-0) \\ y &=&\dfrac{1}{2}x+4 \end{eqnarray*}$$