Derivatives of Parametric Functions

Home → Calculus → Differentiation of Functions → 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 or tap a problem to see the solution.

Example 1

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

Example 2

\[x = 2t + 1,\;y = 4t - 3.\]

Example 3

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

Example 4

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

Example 5

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

Example 6

\[x = \sinh t,\;y = \cosh t.\]

Example 7

\[x = a\cos t,\;y = b\sin t.\]

Example 8

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

Example 9

\(x = 2{t^2} + t + 1,\) \(y = 8{t^3} + 3{t^2} + 2.\)

Example 10

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

Example 1.

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

Solution.

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

\[{x'_t} = \left( {{t^2}} \right)^\prime = 2t,\;\;{y'_t} = \left( {{t^3}} \right)^\prime = 3{t^2}.\]

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.

\[x = 2t + 1,\;y = 4t - 3.\]

Solution.

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

Consequently,

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

Example 3.

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

Solution.

\[{x'_t} = \left( {{e^{2t}}} \right)^\prime = 2{e^{2t}},\;\;{y'_t} = \left( {{e^{3t}}} \right)^\prime = 3{e^{3t}}.\]

Hence, the derivative \(\frac{{dy}}{{dx}}\) 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

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

Hence,

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

Example 5.

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

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

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

Example 6.

\[x = \sinh t,\;y = \cosh t.\]

Solution.

Calculate the derivatives:

\[{x'_t} = \left( {\sinh t} \right)^\prime = \cosh t,\;\;{y'_t} = \left( {\cosh t} \right)^\prime = \sinh t.\]

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

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

Example 7.

\[{x^2} + {y^2} - 2x - 4y = 4\]

Solution.

These equations describe an ellipse centered at the origin with semi-axes \(a\) and \(b\). Differentiate the variables \(x\) and \(y\) with respect to \(t:\)

\[{x'_t} = \left( {a\cos t} \right)^\prime = - a\sin t,\;\;{y'_t} = \left( {b\sin t} \right)^\prime = b\cos t.\]

The derivative \(\frac{{dy}}{{dx}}\) depends on \(t\) as follows:

\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{b\cos t}}{{\left( { - a\sin t} \right)}} = - \frac{b}{a}\cot t.\]

Here the parameter \(t\) varies from \(-\pi\) to \(\pi\). However the derivative \(\frac{{dy}}{{dx}}\) becomes infinite at the points \(t = 0, \pm \pi .\) Therefore, the domain can be represented as \(0 \lt \left| t \right| \lt \pi .\)

Example 8.

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

Solution.

Compute the derivatives with respect to \(t:\)

\[x_t^\prime = \left( {\sin 2t} \right)^\prime = 2\cos 2t,\;\;y_t^\prime = \left( {-\cos t} \right)^\prime = \sin t.\]

So, the derivative \(\frac{{dy}}{{dx}}\) is given by

\[\frac{{dy}}{{dx}} = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{2\cos 2t}}{{\sin t}}.\]

Compute the derivative at \(t = \frac{\pi }{6}:\)

\[\frac{{dy}}{{dx}}\left( {t = \frac{\pi }{6}} \right) = \frac{{2\cos \left( {2 \cdot \frac{\pi }{6}} \right)}}{{\sin \frac{\pi }{6}}} = \frac{{2\cos \frac{\pi }{3}}}{{\sin \frac{\pi }{6}}} = \frac{{2 \cdot \cancel{\frac{1}{2}}}}{{\cancel{\frac{1}{2}}}} = 2.\]

Example 9.

\(x = 2{t^2} + t + 1,\) \(y = 8{t^3} + 3{t^2} + 2.\)

Solution.

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

\[{x'_t} = \left( {2{t^2} + t + 1} \right)^\prime = 4t + 1,\;\;{y'_t} = \left( {8{t^3} + 3{t^2} + 2} \right)^\prime = 24{t^2} + 6t.\]

Consequently, the derivative \(\frac{{dy}}{{dx}}\) is given by

\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{24{t^2} + 6t}}{{4t + 1}} = \frac{{6t\cancel{\left( {4t + 1} \right)}}}{{\cancel{4t + 1}}} = 6t.\]

Example 10.

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

Solution.

\[{x'_t} = \left( {\sqrt {1 - {t^2}} } \right)^\prime = {\frac{1}{{2\sqrt {1 - {t^2}} }} \cdot {\left( {\sqrt {1 - {t^2}} } \right)^\prime }} = \frac{{ - \cancel{2}t}}{{\cancel{2}\sqrt {1 - {t^2}} }} = - \frac{t}{{\sqrt {1 - {t^2}} }},\;\;{{y'_t} = {\left( {\arcsin t} \right)^\prime } = \frac{1}{{\sqrt {1 - {t^2}} }}.}\]

The derivative \(\frac{{dy}}{{dx}}\) is expressed by the formula

\[\frac{{dy}}{{dx}} = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{\frac{1}{{\sqrt {1 - {t^2}} }}}}{{\frac{{ - t}}{{\sqrt {1 - {t^2}} }}}} = \frac{1}{{\sqrt {1 - {t^2}} }} \cdot \frac{{\sqrt {1 - {t^2}} }}{{\left( { - t} \right)}} = - \frac{1}{t},\]

where the parameter \(t\) can take values satisfying the conditions \(\left| t \right| \lt 1,\;t \ne 0.\)

See more problems on Page 2.