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# 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