# 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.