The derivative of parametric equation

The parametric equation defines a group of quantities as functions of one or more independent variables that are called parameters. Generally the parameteric equations are used to express coordinates of the points that make up a geometric object i.e. curve or surface. In that case the equations are collectively called parametric representation or parametrization of the object(curve, surface).
In pararmetric equations the dependancy of function \(y\) and argument \(x\) is given by the parameter \(t\) and can be written as \begin{cases}x = \phi(t) \\ y = \psi(t) \end{cases} the the first derivative of system of parametric equations can be written as \begin{eqnarray}y_x' &=& \frac{y_t'}{x_t'}\\ \frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\end{eqnarray} To obtain the first derivative of the system of parametric equation first the derivative of argument \(x\) and function \(y\) with respect to \(t\) must be determined and then the ratio between \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) gives the first derivation of function y with respect to argumetn x.

• Example 1 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = a cos t \\ y = a sin t\end{cases}
  Solution - First the \(\frac{dx}{dt}\) must be calculated.\begin{eqnarray}\frac{dx}{dt} &=& (a\cos t)_t' \\ \frac{dx}{dt} &=& -a\sin_t\end{eqnarray} The second step is to calculate \(\frac{dy}{dt}\) as \begin{eqnarray} \frac{dy}{dt} &=& (a\sin t)_t' = a \cos t. \end{eqnarray} Now the \(\frac{dy}{dx}\) can be calculated as \begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx} &=& -\frac{a \cos t }{a\sin t} = \frac{\cos t}{\sin t} = \cot t\end{eqnarray}
• Example 2 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = 2t-1 \\ y = t^3\end{cases}
  Solution The first step is to calculate \(\frac{dx}{dt}\) \begin{eqnarray}\frac{dx}{dt} &=& (2t-1)_t' = 2\end{eqnarray} The second step is to calculate \(\frac{dy}{dt}\) \begin{eqnarray}\frac{dy}{dt} &=& (t^3)_t' = 3t^2\end{eqnarray} The final step is to calcualte \(\frac{dy}{dx}\)\begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ \frac{dy}{dx} &=& \frac{3t^2}{2} = \\frac{3}{2}t^2 \end{eqnarray}
• Example 3 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = \frac{1}{t+1} \\ y = \left(\frac{t}{t+1}\right)^2\end{cases}
  Solution The first step is to calculate the \(\frac{dx}{dt}\) \begin{eqnarray}\frac{dx}{dt} &=& \left(\frac{1}{t+1}\right)_t' = -\frac{1}{(t+1)^2} \end{eqnarray} The second step is to calculate the \(\frac{dy}{dt}\)\begin{eqnarray}\frac{dy}{dt} &=& \left(\left(\frac{t}{t+1}\right)^2\right)_t' = 2\left(\frac{t}{t+1}\right) \frac{t+1-t}{(t+1)^2} = \frac{2t}{(t+1)^3}\end{eqnarray} The final step is to calculate the \(\frac{dy}{dx}\) \begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ \frac{dy}{dx} &=& \frac{\frac{2t}{(t+1)^3}}{-\frac{1}{(t+1)^2}} = -\frac{2t}{t+1}\end{eqnarray}
• Example 4 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = \frac{2at}{1+t^2} \\ y = \frac{a(1-t^2)}{a+t^2}\end{cases}
  Solution The first step is to calculate \(\frac{dx}{dt}\) \begin{eqnarray}\frac{dx}{dt} &=& \left(\frac{2at}{1+t^2}\right)_t' \\\nonumber \frac{dx}{dt} &=& \frac{2a(1+t^2) - 2at\cdot 2t}{(1+t^2)^2} \\\nonumber \frac{dx}{dt} &=& \frac{2a-2at^2}{(1+t^2)^2}\end{eqnarray} The second step is to calcualte \(\frac{dy}{dt}\) \begin{eqnarray}\frac{dy}{dt} &=& \left(\frac{a(1-t^2)}{1+t^2}\right)_t' \\\nonumber \frac{dy}{dt} &=& \frac{-2at(1+t^2) -a(1-t^2)2t}{(1+t^2)^2}\\\nonumber \frac{dy}{dt} &=& \frac{-4at}{(1+t)^2}\end{eqnarray} The third and final step is to calculate \(\frac{dy}{dx}\) \begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\\nonumber \frac{dy}{dx} &=& \frac{\frac{-4at}{(1+t^2)^2}}{\frac{2a-2at^2}{(1+t^2)^2}} = \frac{2t}{(t^2 -1)} \end{eqnarray}
• Example 5 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = \frac{3at}{1+t^3}\\ y = \frac{3at^2}{1+t^3}\end{cases}
  Solution The first step is to calculate \(\frac{dx}{dt}\) \begin{eqnarray}\frac{dx}{dt} &=& \left(\frac{3at}{1+t^3}\right)_t' = \frac{3a(1+t^3)- 3at\cdot 3t^2}{(1+t^3)^2} \\ \frac{dx}{dt} &=& \frac{3a+3at^3 - 9at^3}{(1+t^3)^2} = \frac{3a(1-2t^3)}{(1+t^3)^2}\end{eqnarray} The second step is to calculate \(\frac{dy}{dt}\) \begin{eqnarray}\frac{dy}{dt} &=& \left(\frac{3at^2}{1+t^3}\right)_t' = \frac{6at(1+t^3)-3at^2\cdot 3t^2}{(1+t^3)^2}\\ \frac{dy}{dt} &=& \frac{6at + 6at^4 -9at^4}{(1+t^3)^2} = \frac{3at(2-t^3)}{(1+t^3)^2}\end{eqnarray} The final step is to determine \(\frac{dy}{dx}\) \begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{3at(2-t^3)}{(1+t^3)^2}}{\frac{3a(1-2t^3)}{(1+t^3)^2}}\\ \frac{dy}{dx} &=& \frac{2t-t^4}{(1-2t^3)}\end{eqnarray}
• Example 6 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = \sqrt{t} \\ y = \sqrt[3]{t}\end{cases}
  Solution The first step is to calculate \(\frac{dx}{dt}\)\begin{eqnarray}\frac{dx}{dt} &=& \left(\sqrt{t}\right)_t' = \frac{1}{2}\frac{1}{\sqrt{t}}\end{eqnarray} The second step is to calculate \(\frac{dy}{dt}\) \begin{eqnarray}\frac{dy}{dt} &=& \left(\sqrt[3]{t}\right)_t' = \frac{1}{3}t^{\frac{1}{3}-1} = \frac{1}{3} t^{-\frac{2}{3}} = \frac{1}{3}\frac{1}{\sqrt[3]{t^2}}\end{eqnarray} The final step is to determine \(\frac{dy}{dx}\) \begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{3}\frac{1}{\sqrt[3]{t^2}}}{\frac{1}{2}\frac{1}{\sqrt{t}}} = \frac{2}{3}\frac{\sqrt{t}}{\sqrt[3]{t}}\\\frac{dy}{dx} &=& \frac{2}{3}t^{\frac{1}{2}} t^{-\frac{2}{3}} = \frac{2}{3} t^{-\frac{1}{6}}\end{eqnarray}
• Example 7 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = \sqrt{t^2 +1} \\ y = \frac{t-1}{\sqrt{t^2 +1}}\end{cases}
  Solution The first step is to calculate \(\frac{dx}{dt}\) \begin{eqnarray}\frac{dx}{dt} &=& \frac{1}{2} \frac{2t}{\sqrt{t^2 + 1}} = \frac{t}{\sqrt{t^2 +1}}\end{eqnarray} The second step is to calculate \(\frac{dy}{dt}\)\begin{eqnarray}\frac{dy}{dt} &=& \frac{\sqrt{t^2 + 1} - \frac{1}{2}(t-1)\frac{2t}{\sqrt{t^2 +1}}}{t^2 +1 } \\ \frac{dy}{dt} &=& \frac{1+t}{(t^2 +1)\sqrt{t^2 +1 }}\end{eqnarray} The final step is to determine \(\frac{dy}{dx}\)\begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1+t}{(t^2 +1)\sqrt{t^2 +1 }}}{\frac{t}{\sqrt{t^2 +1}}}\\ \frac{dy}{dx} &=& \frac{1+t}{t(t^2 +1)}\end{eqnarray}
• Example 8 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases}x = a(\cos t + t\sin t) \\ y = a(\sin t - t\cos t) \end{cases}
  Solution The first step is to determine \(\frac{dx}{dt}\) \begin{eqnarray}\frac{dx}{dt} &=& -a\sin t + at\cos t + a \sin t = at\cos t\end{eqnarray} The next step is to determine \(\frac{dy}{dt}\)\begin{eqnarray}\frac{dy}{dt} &=& a\cos t - a \cos t + at\sin t = at\sin t\end{eqnarray} The final step or third step is to determine \(\frac{dy}{dx}\)\begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at\cos t} = \frac{\sin t}{\cos t} = \tan t\end{eqnarray}
• Example 9 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = a \cos^2 t \\ y = b\sin^2 t\end{cases}
  Solution The first step is to determine the \(\frac{dx}{dt}\)\begin{eqnarray}\frac{dx}{dt} &=& 2a\cos t(-\sin t) = -2a\sin t\cos t\end{eqnarray} The second step is to determine \(\frac{dy}{dt}\) \begin{eqnarray}\frac{dy}{dt} &=& 2b\sin t \cos t\end{eqnarray} The final step is to determine the value of \(\frac{dy}{dx}\) \begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -\frac{2b \sin t \cos t}{2a\sin t \cos t} = -\frac{b}{a}\end{eqnarray}
• Example 10 Calculate \(\frac{dy}{dx}\) for a system of parametric equations \begin{cases} x = a \cos^3 t \\ y = b\sin^3 t\end{cases}
  Solution The first step is to determine the \(\frac{dx}{dt}\)\begin{eqnarray}\frac{dx}{dt} &=& 3a\cos^2 t(-\sin t) = -3a\sin t\cos^2 t\end{eqnarray} The second step is to determine \(\frac{dy}{dt}\) \begin{eqnarray}\frac{dy}{dt} &=& 3b\sin^2 t \cos t\end{eqnarray} The final step is to determine the value of \(\frac{dy}{dx}\) \begin{eqnarray}\frac{dy}{dx} &=& \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -\frac{3b\sin^2 t \cos t}{3a\sin t\cos^2 t} = -\frac{b}{a}\frac{\sin t}{\cos t} = - \frac{b}{a} \tan t\end{eqnarray}