Derivatives of Parametric Functions
Solved Problems
Example 11.
\[x = {\sin ^3}t,\;y = {\cos ^3}t.\]
Solution.
Calculate the derivatives \(x'_t\), \(y'_t:\)
\[{x'_t} = \left( {{{\sin }^3}t} \right)^\prime = {3\,{\sin ^2}t\cos t,}\;\;{y'_t} = \left( {{{\cos }^3}t} \right)^\prime = - 3\,{\cos ^2}t\sin t.\]
The derivative \(\frac{{dy}}{{dx}}\) is given by
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{ - \cancel{3}{{\cos }^{\cancel{2}}}t\cancel{\sin t}}}{{\cancel{3}{{\sin }^{\cancel{2}}}t\cancel{\cos t}}} = - \frac{{\cos t}}{{\sin t}} = - \cot t,\]
where \(t \ne {\frac{{\pi n}}{2}},\) \(n \in \mathbb{Z}.\) The restrictions on the possible values of \(t\) follow from the condition \({x'_t} \ne 0.\)
Example 12.
\[x = \frac{{t + 1}}{{t - 1}},\;y = \frac{{t - 1}}{{t + 1}}.\]
Solution.
We compute the derivatives of \(x\) and \(y\) with respect to the parameter \(t:\)
\[{x'_t} = \left( {\frac{{t + 1}}{{t - 1}}} \right)^\prime = \frac{{1 \cdot \left( {t - 1} \right) - \left( {t + 1} \right) \cdot 1}}{{{{\left( {t - 1} \right)}^2}}} = \frac{{\cancel{\color{blue}{t}} - \color{red}{1} - \cancel{\color{blue}{t}} - \color{red}{1}}}{{{{\left( {t - 1} \right)}^2}}} = \frac{{ - \color{red}{2}}}{{{{\left( {t - 1} \right)}^2}}};\]
\[{y'_t} = \left( {\frac{{t - 1}}{{t + 1}}} \right)^\prime = \frac{{1 \cdot \left( {t + 1} \right) - \left( {t - 1} \right) \cdot 1}}{{{{\left( {t + 1} \right)}^2}}} = \frac{{\cancel{\color{blue}{t}} + \color{red}{1} - \cancel{\color{blue}{t}} + \color{red}{1}}}{{{{\left( {t + 1} \right)}^2}}} = \frac{\color{red}{2}}{{{{\left( {t + 1} \right)}^2}}}.\]
Then the derivative \(\frac{{dy}}{{dx}}\) is given by
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{\frac{2}{{{{\left( {t + 1} \right)}^2}}}}}{{\frac{{\left( { - 2} \right)}}{{{{\left( {t - 1} \right)}^2}}}}} = - \frac{{{{\left( {t - 1} \right)}^2}}}{{{{\left( {t + 1} \right)}^2}}} = - {\left( {\frac{{t - 1}}{{t + 1}}} \right)^2}.\]
Here the parameter \(t\) can take any values except the points \(t = \pm 1,\) in which the variables \(x\) and \(y\) have a discontinuity.
Example 13.
Find the derivative \(\frac{{dy}}{{dx}}\) for the function \(x = \frac{1}{t} + t,\) \(y = \frac{1}{{{t^2}}}\) at the point \(t = \frac{1}{2}.\)
Solution.
We differentiate both variables \(x\) and \(y\) with respect to \(t:\)
\[x_t^\prime = \left( {\frac{1}{t} + t} \right)^\prime = - \frac{1}{{{t^2}}} + 1 = \frac{{{t^2} - 1}}{{{t^2}}};\]
\[y_t^\prime = \left( {\frac{1}{{{t^2}}}} \right)^\prime = \left( {{t^{ - 2}}} \right)^\prime = - 2{t^{ - 3}} = - \frac{2}{{{t^3}}}.\]
Hence
\[\frac{{dy}}{{dx}} = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\frac{{{t^2} - 1}}{{{t^2}}}}}{{\left( { - \frac{2}{{{t^3}}}} \right)}} = - \frac{{\left( {{t^2} - 1} \right){t^3}}}{{2{t^2}}} = - \frac{{\left( {{t^2} - 1} \right)t}}{2} = \frac{{t - {t^3}}}{2}.\]
Compute the value of \(\frac{{dy}}{{dx}}\) at \(t = \frac{1}{2}:\)
\[\frac{{dy}}{{dx}}\left( {t = \frac{1}{2}} \right) = \frac{{\frac{1}{2} - {{\left( {\frac{1}{2}} \right)}^3}}}{2} = \frac{{\frac{1}{2} - \frac{1}{8}}}{2} = \frac{{\frac{3}{8}}}{2} = \frac{3}{{16}}.\]
Example 14.
\[x = \sqrt {{t^2} + 1} ,\;y = \ln \left( {{t^2} + 1} \right).\]
Solution.
Differentiating with respect to \(t\) yields:
\[{x'_t} = \left( {\sqrt {{t^2} + 1} } \right)^\prime = \frac{1}{{2\sqrt {{t^2} + 1} }} \cdot {\left( {{t^2} + 1} \right)^\prime } = \frac{{\cancel{2}t}}{{\cancel{2}\sqrt {{t^2} + 1} }} = \frac{t}{{\sqrt {{t^2} + 1} }};\]
\[{y'_t} = \left( {\ln \left( {{t^2} + 1} \right)} \right)^\prime = \frac{1}{{{t^2} + 1}} \cdot {\left( {{t^2} + 1} \right)^\prime } = \frac{{2t}}{{{t^2} + 1}}.\]
Consequently,
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{\frac{{2t}}{{{t^2} + 1}}}}{{\frac{t}{{\sqrt {{t^2} + 1} }}}} = \frac{{2t}}{{{t^2} + 1}} \cdot \frac{{\sqrt {{t^2} + 1} }}{t} = \frac{2}{{\sqrt {{t^2} + 1} }}.\]
Example 15.
\[x = {e^t}\sin t,\;y = {e^{ - t}}\cos t.\]
Solution.
Find the derivatives of \(x\) and \(y\) with respect to \(t:\)
\[{x'_t} = \left( {{e^t}\sin t} \right)^\prime = {\left( {{e^t}} \right)^\prime }\sin t + {e^t}{\left( {\sin t} \right)^\prime } = {e^t}\sin t + {e^t}\cos t = {e^t}\left( {\sin t + \cos t} \right);\]
\[{y'_t} = \left( {{e^{ - t}}\cos t} \right)^\prime = {\left( {{e^{ - t}}} \right)^\prime }\cos t + {e^{ - t}}{\left( {\cos t} \right)^\prime } = - {e^{ - t}}\cos t + {e^{ - t}}\left( { - \sin t} \right) = - {e^{ - t}}\left( {\cos t + \sin t} \right).\]
As a result, we have
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{ - {e^{ - t}}\cancel{\left( {\cos t + \sin t} \right)}}}{{{e^t}\cancel{\left( {\sin t + \cos t} \right)}}} = - {e^{ - 2t}}.\]
Note that the derivative exists under the condition
\[\sin t + \cos t \ne 0,\;\; \Rightarrow \tan t + 1 \ne 0,\;\; \Rightarrow \tan t \ne - 1,\;\; \Rightarrow t \ne - \frac{\pi }{4} + \pi n,\;\;n \in \mathbb{Z}.\]
Example 16.
\[x = t - \sin t,\;y = 1 - \cos t.\]
Solution.
The derivatives of \(x\) and \(y\) with respect to \(t\) have the form:
\[{x'_t} = \left( {t - \sin t} \right)^\prime = 1 - \cos t;\;\;{y'_t} = \left( {1 - \cos t} \right)^\prime = \sin t.\]
We write the derivative \(\frac{{dy}}{{dx}}:\)
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{\sin t}}{{1 - \cos t}}.\]
Using the double angle formulas in the numerator and denominator, we obtain:
\[\frac{{dy}}{{dx}} = \frac{{\sin t}}{{1 - \cos t}} = \frac{{\cancel{2}\cancel{\sin \frac{t}{2}}\cos \frac{t}{2}}}{{\cancel{2}{{\sin }^{\cancel{2}}}\frac{t}{2}}} = \frac{{\cos \frac{t}{2}}}{{\sin \frac{t}{2}}} = \cot \frac{t}{2},\]
where \(t \ne 2\pi n,\) \(n \in \mathbb{Z}.\)
Example 17.
Find the derivative of the parametric curve \(x = 2 + \cos t,\) \(y = 1 + \sin t\) at the point \(t = \frac{\pi }{3}.\)
Solution.
Determine the derivatives of \(x\) and \(y\) with respect to \(t:\)
\[x_t^\prime = \left( {2 + \cos t} \right)^\prime = - \sin t,\;\;y_t^\prime = \left( {1 + \sin t} \right)^\prime = \cos t.\]
Then the derivative \(\frac{{dy}}{{dx}}\) is given by
\[\frac{{dy}}{{dx}} = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\cos t}}{{\left( { - \sin t} \right)}} = - \cot t.\]
Calculate the value of the derivative at the point where \(t = \frac{\pi }{3}:\)
\[\frac{{dy}}{{dx}}\left( {t = \frac{\pi }{3}} \right) = - \cot \frac{\pi }{3} = - \frac{1}{{\sqrt 3 }}.\]
Example 18.
\[x = 1 + \sqrt t ,\;y = t - \frac{1}{{\sqrt t }},\] where \({t \gt 0}.\)
Solution.
We differentiate the functions \(x\left( t \right)\) and \(y\left( t \right)\) with respect to \(t:\)
\[{x'_t} = {\left( {1 + \sqrt t } \right)^\prime } = \frac{1}{{2\sqrt t }};\;\;{y'_t} = {\left( {t - \frac{1}{{\sqrt t }}} \right)^\prime } = {\left( {t - {t^{ - \frac{1}{2}}}} \right)^\prime } = 1 + \frac{1}{2}{t^{ - \frac{3}{2}}} = 1 + \frac{1}{{2\sqrt {{t^3}} }}.\]
Then the derivative \(\frac{{dy}}{{dx}}\) is expressed by the formula
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{1 + \frac{1}{{2\sqrt {{t^3}} }}}}{{\frac{1}{{2\sqrt t }}}} = \frac{{\frac{{2\sqrt {{t^3}} + 1}}{{2\sqrt {{t^3}} }}}}{{\frac{1}{{2\sqrt t }}}} = \frac{{\left( {2\sqrt {{t^3}} + 1} \right) \cdot 2\sqrt t }}{{2\sqrt {{t^3}} }} = \frac{{2\sqrt {{t^3}} + 1}}{{\sqrt {{t^2}} }} = \frac{{2\sqrt {{t^3}} + 1}}{{\left| t \right|}} = \frac{{2\sqrt {{t^3}} + 1}}{t},\;\;\text{where}\;\;t \gt 0.\]
Example 19.
\[x = {\tan ^2}t,\;y = {\cos ^2}t.\]
\[{x'_t} = \left( {{{\tan }^2}t} \right)^\prime = 2\tan t \cdot {\left( {\tan t} \right)^\prime } = 2\tan t \cdot \frac{1}{{{{\cos }^2}t}} = \frac{{2\sin t}}{{\cos t}} \cdot \frac{1}{{{{\cos }^2}t}} = \frac{{2\sin t}}{{{{\cos }^3}t}};\]
\[{y'_t} = \left( {{{\cos }^2}t} \right)^\prime = 2\cos t \cdot {\left( {\cos t} \right)^\prime } = 2\cos t \cdot \left( { - \sin t} \right) = - 2\sin t\cos t.\]
As a result, we have
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{ - \cancel{2\sin t}\cos t}}{{\frac{{\cancel{2\sin t}}}{{{{\cos }^3}t}}}} = - \cos t \cdot {\cos ^3}t = - {\cos^4}t.\]
In this case, \(t \ne {\frac{{\pi n}}{2}},\) \(n \in \mathbb{Z}.\)
Example 20.
\[x = \arccos \left( {1 - t} \right), y = \sqrt {2t - {t^2}} .\]
Solution.
Find the derivatives \(x'_t\), \(y'_t:\)
\[{x'_t} = \left( {\arccos \left( {1 - t} \right)} \right)^\prime = - \frac{1}{{\sqrt {1 - {{\left( {1 - t} \right)}^2}} }} \cdot {\left( {1 - t} \right)^\prime } = - \frac{1}{{\sqrt {1 - \left( {1 - 2t + {t^2}} \right)} }} \cdot \left( { - 1} \right) = \frac{1}{{\sqrt {\cancel{1} - \cancel{1} + 2t - {t^2}} }} = \frac{1}{{\sqrt {2t - {t^2}} }};\]
\[{y'_t} = \left( {\sqrt {2t - {t^2}} } \right)^\prime = \frac{1}{{2\sqrt {2t - {t^2}} }} \cdot {\left( {2t - {t^2}} \right)^\prime } = \frac{{2 - 2t}}{{2\sqrt {2t - {t^2}} }} = \frac{{1 - t}}{{\sqrt {2t - {t^2}} }}.\]
Now it's easy to write an expression for the derivative \(\frac{{dy}}{{dx}}:\)
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{\frac{{1 - t}}{{\sqrt {2t - {t^2}} }}}}{{\frac{1}{{\sqrt {2t - {t^2}} }}}} = \frac{{1 - t}}{{\cancel{\sqrt {2t - {t^2}}} }} \cdot \frac{{\cancel{\sqrt {2t - {t^2}}} }}{1} = 1 - t.\]
The valid values of \(t\) are defined by the following system of inequalities:
\[\left\{ \begin{array}{l} - 1 \le 1 - t \le 1\\ 2t - {t^2} \gt 0 \end{array} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {1 - t \ge - 1}\\ {1 - t \le 1}\\ {t\left( {2 - t} \right) \gt 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { - t \ge - 2}\\ { - t \le 0}\\ {0 \lt t \lt 2} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {t \le 2}\\ {t \ge 0}\\ {0 \lt t \lt 2} \end{array}} \right.,\;\; \Rightarrow 0 \lt t \lt 2. \]
Example 21.
\[x = {\sin ^4}2t,\;y = {\cos ^4}2t.\]
Solution.
The derivatives of \(x\) and \(y\) with respect to \(t\) have the form:
\[{x'_t} = \left( {{{\sin }^4}2t} \right)^\prime = 4\,{\sin ^3}2t \cdot {\left( {\sin 2t} \right)^\prime } = 4\,{\sin ^3}2t \cdot 2\cos 2t = 8\,{\sin ^3}2t\cos 2t;\]
\[{y'_t} = \left( {{{\cos }^4}2t} \right)^\prime = 4\,{\cos ^3}2t \cdot {\left( {\cos 2t} \right)^\prime } = 4\,{\cos ^3}2t \cdot \left( { - 2\sin 2t} \right) = - 8\,{\cos ^3}2t\sin 2t.\]
Then the derivative \(\frac{{dy}}{{dx}}\) is given by
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{ - 8\,{{\cos }^3}2t\sin 2t}}{{8\,{{\sin }^3}2t\cos 2t}} = - \frac{{{{\cos }^2}2t}}{{{{\sin }^2}2t}} = - {\cot ^2}2t.\]
In this example, the valid values of \(t\) are limited by the conditions
\[ {x'_t} \ne 0,\;\; \Rightarrow 8\,{\sin ^3}2t\cos 2t \ne 0,\;\; \Rightarrow \left\{ \begin{array}{l} {\sin ^3}2t \ne 0\\ \cos 2t \ne 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} 2t \ne \pi n\\ 2t \ne \frac{\pi }{2} + \pi n \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} t \ne \frac{{\pi n}}{2}\\ t \ne \frac{\pi }{4} + \frac{{\pi n}}{2} \end{array} \right.,\;\; \Rightarrow t \ne \frac{{\pi n}}{4},\;n \in \mathbb{Z}. \]
Example 22.
\[x = \arcsin {e^t},\;y = \sqrt {1 - {e^{2t}}} .\]
Solution.
First we calculate the derivatives of \(x\) and \(y\) with respect to \(t:\)
\[{x'_t} = \left( {\arcsin {e^t}} \right)^\prime = \frac{1}{{\sqrt {1 - {{\left( {{e^t}} \right)}^2}} }} \cdot {\left( {{e^t}} \right)^\prime } = \frac{{{e^t}}}{{\sqrt {1 - {{\left( {{e^t}} \right)}^2}} }};\]
\[{y'_t} = \left( {\sqrt {1 - {e^{2t}}} } \right)^\prime = \frac{1}{{2\sqrt {1 - {e^{2t}}} }} \cdot {\left( {1 - {e^{2t}}} \right)^\prime } = \frac{{ - \cancel{2}{e^{2t}}}}{{\cancel{2}\sqrt {1 - {e^{2t}}} }} = - \frac{{{e^{2t}}}}{{\sqrt {1 - {e^{2t}}} }}.\]
It follows from here that
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{ - \frac{{{e^{2t}}}}{{\sqrt {1 - {e^{2t}}} }}}}{{\frac{{{e^t}}}{{\sqrt {1 - {e^{2t}}} }}}} = - \frac{{{e^{2t}}}}{{\cancel{\sqrt {1 - {e^{2t}}}} }} \cdot \frac{{\cancel{\sqrt {1 - {e^{2t}}}} }}{{{e^t}}} = - \frac{{{e^{2t}}}}{{{e^t}}} = - {e^t}.\]
Note that the parameter \(t\) can take the values defined by the inequality
\[1 - {e^{2t}} \gt 0,\;\; \Rightarrow {e^{2t}} \lt 1,\;\; \Rightarrow {e^{2t}} \lt {e^0},\;\; \Rightarrow 2t \lt 0,\;\; \Rightarrow t \lt 0,\]
that is only negative values of \(t\) are allowed.
Example 23.
Find the derivative \(\frac{{dy}}{{dx}}\) of the function \(x = \arctan {e^t},\) \(y = 1 + {e^{2t}}\) at \(t = 0.\)
Solution.
We differentiate the variables \(x\) and \(y\) with respect to \(t:\)
\[x_t^\prime = \left( {\arctan {e^t}} \right)^\prime = \frac{1}{{1 + {{\left( {{e^t}} \right)}^2}}} \cdot \left( {{e^t}} \right)^\prime = \frac{{{e^t}}}{{1 + {e^{2t}}}};\]
\[y_t^\prime = \left( {1 + {e^{2t}}} \right)^\prime = 2{e^{2t}}.\]
Then
\[\frac{{dy}}{{dx}} = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{2{e^{2t}}}}{{\frac{{{e^t}}}{{1 + {e^{2t}}}}}} = \frac{{2{e^{2t}}\left( {1 + {e^{2t}}} \right)}}{{{e^t}}} = 2{e^t}\left( {1 + {e^{2t}}} \right).\]
Calculate the value of the derivative \(\frac{{dy}}{{dx}}\) at \(t = 0:\)
\[\frac{{dy}}{{dx}}\left( {t = 0} \right) = 2{e^0}\left( {1 + {e^{2 \cdot 0}}} \right) = 2 \cdot 1 \cdot \left( {1 + 1} \right) = 4.\]
Example 24.
Find the derivative \(\frac{{dy}}{{dx}}\) of the parametrically defined function \(x = t + 2\sin \pi t\), \(y = 3t - \cos \pi t\) at \(t = {\frac{1}{2}}.\)
Solution.
First we find expressions for the derivatives \(x'_t\), \(y'_t:\)
\[{x'_t} = \left( {t + 2\sin \pi t} \right)^\prime = {1 + 2\pi \cos \pi t,}\;\;{y'_t} = {\left( {3t - \cos \pi t} \right)^\prime } = {3 + \pi \sin \pi t.}\]
Then the derivative \(\frac{{dy}}{{dx}}\) is described by the formula
\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{3 + \pi \sin \pi t}}{{1 + 2\pi \cos \pi t}}.\]
Substituting \(t = {\frac{1}{2}}\) we calculate the value of the derivative at the given point:
\[\frac{{dy}}{{dx}}\left( {t = \frac{1}{2}} \right) = \frac{{3 + \pi \sin \frac{\pi }{2}}}{{1 + 2\pi \cos \frac{\pi }{2}}} = \frac{{3 + \pi \cdot 1}}{{1 + 2\pi \cdot 0}} = 3 + \pi .\]
Example 25.
Find the equation of the tangent line to the curve \(x = 4 + 2t,\) \(y = 1 - {t^2}\) at the point \(t = 1.\)
Solution.
The equation of the tangent line is given by
\[y - {y_0} = \frac{{dy}}{{dx}}\left( {x - {x_0}} \right),\]
where \({{x_0},{y_0}}\) are the coordinates of the point of tangency.
Calculate the coordinates \({{x_0},{y_0}}:\)
\[{x_0} = x\left( {t = 1} \right) = 4 + 2 \cdot 1 = 6;\]
\[{y_0} = y\left( {t = 1} \right) = 1 - {1^2} = 0.\]
Now we determine the derivative \(\frac{{dy}}{{dx}}.\) As this is a parametrically defined curve, we find the derivatives of \(x\) and \(y\) with respect to \(t:\)
\[x_t^\prime = \left( {4 + 2t} \right)^\prime = 2;\]
\[y_t^\prime = \left( {1 - {t^2}} \right)^\prime = - 2t.\]
Then
\[\frac{{dy}}{{dx}} = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{ - 2t}}{2} = - t,\]
so
\[\frac{{dy}}{{dx}}\left( {t = 1} \right) = - 1.\]
Substitute these values into the equation of the tangent line:
\[y - {y_0} = \frac{{dy}}{{dx}}\left( {x - {x_0}} \right),\;\; \Rightarrow y - 0 = \left( { - 1} \right) \cdot \left( {x - 6} \right),\;\; \Rightarrow y = - x + 6,\;\; \Rightarrow y = 6 - x.\]
Example 26.
Find the equation of the tangent line to the curve \(x = {t^2} - 2t,\) \(y = {t^2} + 2t\) at the point \(t = 2.\)
Solution.
The equation of the tangent line is written in the form
\[y - {y_0} = \frac{{dy}}{{dx}}\left( {x - {x_0}} \right),\]
where \({{x_0},{y_0}}\) are the coordinates of the point of tangency.
Calculate these coordinates:
\[{x_0} = x\left( {t = 2} \right) = {2^2} - 2 \cdot 2 = 0;\]
\[{y_0} = y\left( {t = 2} \right) = {2^2} + 2 \cdot 2 = 8.\]
Now we compute the derivative \(\frac{{dy}}{{dx}}.\) As this is a parametric curve, we find the derivatives of \(x\) and \(y\) with respect to \(t:\)
\[x_t^\prime = \left( {{t^2} - 2t} \right)^\prime = 2t - 2;\]
\[y_t^\prime = \left( {{t^2} + 2t} \right)^\prime = 2t + 2.\]
Then
\[\frac{{dy}}{{dx}} = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{2t + 2}}{{2t - 2}} = \frac{{t + 1}}{{t - 1}}.\]
At the point where \(t = 2,\) the derivative is equal to
\[\frac{{dy}}{{dx}}\left( {t = 2} \right) = \frac{{2 + 1}}{{2 - 1}} = 3.\]
Substitute these values into the equation of the tangent line:
\[y - {y_0} = \frac{{dy}}{{dx}}\left( {x - {x_0}} \right),\;\; \Rightarrow y - 8 = 3\left( {x - 0} \right),\;\; \Rightarrow y - 8 = 3x,\;\; \Rightarrow y = 3x + 8.\]
