Finding an Equation of the Tangent Line to a Smooth Curve

(a) Find an equation of the tangent line to the plane curve with parametric equations \(x(t) =3t^{2},\quad y(t) =2t\), when \(t=1\).

(b) Find all the points on the plane curve at which the tangent line is vertical.

Solution (a) The curve is smooth \(\bigg(\dfrac{{\it dy}}{{\it dt}}=2 {\rm \, is \, never \, zero} \bigg)\). Since \(\dfrac{dx}{dt}=6t\) is not 0 at \(t=1\), the slope of the tangent line to the curve is \[ \begin{equation*} \frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{\dfrac{d}{dt} ( 2t) }{\dfrac{d}{dt}\left( 3t^{2}\right) }=\dfrac{2}{6t}=\frac{1 }{3t} \end{equation*} \]

When \(t=1\), the slope of the tangent line is \(\dfrac{1}{3}\). Since \(x=3\) and \(y=2\) when \(t=1\), an equation of the tangent line is \[ \begin{array}{rcl@{\qquad\qquad}l} y-2 &=&\dfrac{1}{3} ( x-3 ) & {\color{#0066A7}{{y-y_{1} =m( x-x_{1});}}\;{\color{#0066A7}{ x_{1} =3, y_{1} =2, m= \dfrac{1}{3}}}} \\ y &=&\dfrac{1}{3}x+1 \end{array} \]

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(b) A vertical tangent line occurs when \(\dfrac{dx}{dt}=0\) and \( \dfrac{dy}{dt}\neq 0.\) Since \(\dfrac{dx}{dt}=0\) and \(\dfrac{dy}{ dt}=2\) when \(t=0,\) there is a vertical tangent line to the curve when \(t=0,\) namely at the point \((0,0)\).