The parametric equations below describe a curve \(C\):
\[ \begin{cases} x(t) = t^2\\ y(t) = t^3 \end{cases} , \textrm { for } t \geq 0 \]

Suppose that we want to find the equation of a tangent line to the curve at a given point \((x,y)\).

Consider the point \((x,y) = (4,8)\).

First, how can we verify if this point lies on the curve generated by our parametric equations?

We need to see if \(x=4\) for some \(t\)-value and \(y=8\) for some potentially different \(t\)-value. We need to determine if there is a common \(t\)-value for which both \(x(t) = 4\) and \(y(t)=8\) simultaneously.

In this case, we find when \(t=\answer {2}\), \(x(t) = 4\) and \(y(t) =8\).

Now that we now that \((4,8)\) lies on our curve, let’s find the equation of the tangent line at this point. We can proceed two ways:

We can work directly with the parametric description of the curve.

Recall that point-slope form of the equation of a line is given by \(y-y_{0}=m_{tan}(x-x_{0})\). Here \(m\) is the slope of the line and \((x_{0}, y_{0})\) is a point that lies on the line.

We need to find the slope of the line through the point \((4,8)\). Since our curve is given parametrically, we use \(\dd [y]{x}=\frac {dy/dt}{dx/dt}\) to determine the slope:

\[ \dd [y]{x}=\answer {\frac {3}{2}t} \]

When \(t=2\), we find \(\dd [y]{x} \bigg |_{t=2}= 3\).

Thus, using the point-slope form of a line:

\begin{align*} y-y(2)& = m_{tan}\left (x-x(2)\right ) \\ y-\left (\answer {8}\right )& =\answer {3} \cdot \left (x-\answer {4}\right ) \\ \end{align*}

Thus the tangent line to our curve at the point \((4, 8)\) is \(y=\answer {3x-4}\).

Follow the instructions in the earlier exercise to draw a picture of this on Desmos. In an unused line, type “\(y=3x-4\)” and verify that this is indeed the tangent line at \((4,8)\).
Of course, we can also find a description of our curve in Cartesian coordinates, then find the tangent line. Notice that given our parametric equations that \(x^3=\answer {t^6}\) and that \(y^2=\answer {t^6}\).

(give answers in terms of \(t\))

Thus the the \(x\)-coordinate and \(y\)-coordinate of any point on our given curve satisfies \(y^{\answer {2}}=x^{\answer {3}}\).

We will use implicit differentiation since the above equation does not represent a function. Differentiating both sides with respect to \(x\) gives:

\begin{align*} \left (\answer {2y}\right )\dd [y]{x} &= \answer {3x^2} \\ \dd [y]{x} &= \answer {\frac {3x^2}{2y}} \end{align*}
Now we can calculate the derivative at the point \((4,8)\) to find:
\[ \dd [y]{x}\bigg |_{(x,y)=(4,8)} = \answer {3}. \]

Thus, using the point-slope form of a line:

\begin{align*} y-y_0& = m_{tan}\left (x-x_0\right ) \\ y-\left (\answer {8}\right )& =\answer {3} \cdot \left (x-\answer {4}\right ). \\ \end{align*}

Thus, the tangent line to our curve at the point \((4, 8)\) is \(y=\answer {3x-4}\).

Do your answers using both methods agree?

Yes No