Calculus

Differentiation of Functions

Differentiation of Functions Logo

Derivatives of Parametric Functions

  • The relationship between the variables \(x\) and \(y\) can be defined in parametric form using two equations:

    \[ \left\{ \begin{aligned} x &= x\left( t\right) \\ y &= y\left( t\right) \end{aligned} \right., \]

    where the variable \(t\) is called a parameter. For example, two functions

    \[ \left\{ \begin{aligned} x &= R \cos t \\ y &= R \sin t \end{aligned} \right. \]

    describe in parametric form the equation of a circle centered at the origin with the radius \(R.\) In this case, the parameter \(t\) varies from \(0\) to \(2 \pi.\)

    Find an expression for the derivative of a parametrically defined function. Suppose that the functions \(x = x\left( t \right)\) and \(y = y\left( t \right)\) are differentiable in the interval \(\alpha \lt t \lt \beta \) and \(x’\left( t \right) \ne 0.\) Moreover, we assume that the function \(x = x\left( t \right)\) has an inverse function \(t = \varphi \left( x \right).\)

    By the inverse function theorem we can write:

    \[\frac{{dt}}{{dx}} = {t’_x} = \frac{1}{{{x’_t}}}.\]

    The original function \(y\left( x \right)\) can be considered as a composite function:

    \[y\left( x \right) = y\left( {t\left( x \right)} \right).\]

    Then its derivative is given by

    \[
    {{y’_x} = {y’_t} \cdot {t’_x} }
    = {{y’_t} \cdot \frac{1}{{{x’_t}}} }
    = {\frac{{{y’_t}}}{{{x’_t}}}.}
    \]

    This formula allows to find the derivative of a parametrically defined function without expressing the function \(y\left( x \right)\) in explicit form.

    In the examples below, find the derivative of the parametric function.


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    \[x = {t^2},\;\;y = {t^3}.\]

    Example 2

    \[{x = 2t + 1,\;\;}\kern-0.3pt{y = 4t – 3.}\]

    Example 3

    \[x = {e^{2t}},\;\;y = {e^{3t}}.\]

    Example 4

    \[x = at,\;\;y = b{t^2}.\]

    Example 5

    \[{x = {\sin ^2}t,\;\;}\kern-0.3pt{y = {\cos ^2}t.}\]

    Example 6

    \[{x = \sinh t,\;\;}\kern-0.3pt{y = \cosh t.}\]

    Example 7

    \[{x = a\cos t,\;\;}\kern-0.3pt{y = b\sin t.}\]

    Example 8

    Find the derivative \(\large{\frac{{dy}}{{dx}}}\normalsize\) for the function \(x = \sin 2t,\) \(y = -\cos t\) at the point \(t = \large{\frac{\pi }{6}}\normalsize.\)

    Example 9

    \[ {x = 2{t^2} + t + 1,\;\;}\kern-0.3pt {y = 8{t^3} + 3{t^2} + 2.} \]

    Example 10

    \[{x = \sqrt {1 – {t^2}} ,\;\;}\kern-0.3pt{y = \arcsin t.}\]

    Example 11

    \[{x = {\sin ^3}t,\;\;}\kern-0.3pt{y = {\cos ^3}t.}\]

    Example 12

    \[{x = \frac{{t + 1}}{{t – 1}},\;\;}\kern-0.3pt{y = \frac{{t – 1}}{{t + 1}}.}\]

    Example 13

    Find the derivative \(\large{\frac{{dy}}{{dx}}}\normalsize\) for the function \(x = \large{\frac{1}{t}}\normalsize + t,\) \(y = \large{\frac{1}{{{t^2}}}}\normalsize\) at the point \(t = \large{\frac{1}{2}}\normalsize.\)

    Example 14

    \[{x = \sqrt {{t^2} + 1} ,\;\;}\kern-0.3pt{y = \ln \left( {{t^2} + 1} \right).}\]

    Example 15

    \[{x = {e^t}\sin t,\;\;}\kern-0.3pt{y = {e^{ – t}}\cos t.}\]

    Example 16

    \[{x = t – \sin t,\;\;}\kern-0.3pt{y = 1 – \cos t.}\]

    Example 17

    Find the derivative of the parametric curve \(x = 2 + \cos t,\) \(y = 1 + \sin t\) at the point \(t = \large{\frac{\pi }{3}}\normalsize.\)

    Example 18

    \[ {x = 1 + \sqrt t ,\;\;}\kern-0.3pt {y = t – \frac{1}{{\sqrt t }},\;\;}\kern-0.3pt {\left( {t \gt 0} \right).} \]

    Example 19

    \[{x = {\tan ^2}t,\;\;}\kern-0.3pt{y = {\cos ^2}t.}\]

    Example 20

    \[{x = \arccos \left( {1 – t} \right),\;\;}\kern-0.3pt{y = \sqrt {2t – {t^2}} .}\]

    Example 21

    \[{x = {\sin ^4}2t,\;\;}\kern-0.3pt{y = {\cos ^4}2t.}\]

    Example 22

    \[{x = \arcsin {e^t},\;\;}\kern-0.3pt{y = \sqrt {1 – {e^{2t}}} .}\]

    Example 23

    Find the derivative \(\large{\frac{{dy}}{{dx}}}\normalsize\) of the function \(x = \arctan {e^t},\) \(y = 1 + {e^{2t}}\) at \(t = 0.\)

    Example 24

    Find the derivative \(\large\frac{{dy}}{{dx}}\normalsize\) of the parametrically defined function \(x = t + 2\sin \pi t\), \(y = 3t – \cos \pi t\) at \(t = {\large\frac{1}{2}\normalsize}.\)

    Example 25

    Find the equation of the tangent line to the curve \(x = 4 + 2t,\) \(y = 1 – {t^2}\) at the point \(t = 1.\)

    Example 26

    Find the equation of the tangent line to the curve \(x = {t^2} – 2t,\) \(y = {t^2} + 2t\) at the point \(t = 2.\)

    Example 1.

    \[x = {t^2},\;\;y = {t^3}.\]

    Solution.

    We find the derivatives of \(x\) and \(y\) with respect to \(t:\)

    \[
    {{x’_t} = {\left( {{t^2}} \right)^\prime } = 2t,\;\;}\kern-0.3pt
    {{y’_t} = {\left( {{t^3}} \right)^\prime } = 3{t^2}.}
    \]

    Hence,

    \[
    {\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_t}}}{{{x’_t}}} }
    = {\frac{{3{t^2}}}{{2t}} }
    = {\frac{{3t}}{2}\;\left( {t \ne 0} \right).}
    \]

    Example 2.

    \[{x = 2t + 1,\;\;}\kern-0.3pt{y = 4t – 3.}\]

    Solution.

    \[
    {{x’_t} = \left( {2t + 1} \right) = 2,\;\;}\kern-0.3pt
    {{y’_t} = {\left( {4t – 3} \right)^\prime } = 4.}
    \]

    Consequently,

    \[{\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_t}}}{{{x’_t}}} }={ \frac{4}{2} = 2.}\]

    Example 3.

    \[x = {e^{2t}},\;\;y = {e^{3t}}.\]

    Solution.

    \[
    {{x’_t} = {\left( {{e^{2t}}} \right)^\prime } = 2{e^{2t}},\;\;}\kern-0.3pt
    {{y’_t} = {\left( {{e^{3t}}} \right)^\prime } = 3{e^{3t}}.}
    \]

    Hence, the derivative \(\large\frac{{dy}}{{dx}}\normalsize\) is given by

    \[
    {\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_t}}}{{{x’_t}}} }
    = {\frac{{3{e^{3t}}}}{{2{e^{2t}}}} }
    = {\frac{3}{2}{e^{3t – 2t}} }
    = {\frac{3}{2}{e^t}.}
    \]

    Example 4.

    \[x = at,\;\;y = b{t^2}.\]

    Solution.

    In this example, the derivatives with respect to \(t\) are given by

    \[
    {{x’_t} = {\left( {at} \right)^\prime } = a,\;\;}\kern-0.3pt
    {{y’_t} = {\left( {b{t^2}} \right)^\prime } = 2bt.}
    \]

    Hence,

    \[
    \frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_t}}}{{{x’_t}}}
    = \frac{{2bt}}{a}.
    \]

    Example 5.

    \[{x = {\sin ^2}t,\;\;}\kern-0.3pt{y = {\cos ^2}t.}\]

    Solution.

    Differentiate with respect to the parameter \(t:\)

    \[{{x’_t} = {\left( {{{\sin }^2}t} \right)^\prime } = {2\sin t \cdot \cos t} = {\sin 2t,}}\]

    \[{{y’_t} = {\left( {{{\cos }^2}t} \right)^\prime } = {2\cos t \cdot \left( { – \sin t} \right)} = {- 2\sin t\cos t} = {- \sin 2t.}}\]

    Then

    \[\require{cancel}
    {\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_t}}}{{{x’_t}}} }
    = {\frac{{ – \cancel{\sin 2t}}}{{\cancel{\sin 2t}}} = – 1,\;\;}\kern-0.3pt
    {\text{where}\;\;t \ne \frac{{\pi n}}{2},\;\;}\kern-0.3pt{n \in \mathbb{Z}.}
    \]

    Example 6.

    \[{x = \sinh t,\;\;}\kern-0.3pt{y = \cosh t.}\]

    Solution.

    Calculate the derivatives:

    \[
    {{x’_t} = {\left( {\sinh t} \right)^\prime } = \cosh t,\;\;}\kern-0.3pt
    {{y’_t} = {\left( {\cosh t} \right)^\prime } = \sinh t.}
    \]

    Then the derivative \(\large\frac{{dy}}{{dx}}\normalsize\) is given by

    \[
    {\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_t}}}{{{x’_t}}} }
    = {\frac{{\sinh t}}{{\cosh t}} }
    = {\tanh t.}
    \]

    Page 1
    Problems 1-6
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    Problems 7-26