Calculus

Differentiation of Functions

Derivatives of Parametric Functions

Page 1
Problems 1-4
Page 2
Problems 5-20

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 on problem description to see 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

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

 Example 9

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

 Example 10

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

 Example 11

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

 Example 12

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

 Example 13

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

 Example 14

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

 Example 15

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

 Example 16

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

 Example 17

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

 Example 18

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

 Example 19

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

 Example 20

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

Page 1
Problems 1-4
Page 2
Problems 5-20