The derivatives of parametric equations are presented with examples and their solutions. Some applications to find slopes of tangent lines and concavity are also included. Explore more questions with solutions below.
In what follows, \( C \) represents the constant of integration where applicable.
First and Second Derivatives of Parametric Equations
Given parametric equations of the form:
\[ \left\{ \begin{aligned} x(t) \\ y(t) \end{aligned} \right. \]what is the derivative \( \dfrac{dy}{dx} \)?
First, find the derivative \( \dfrac{dy}{dt} \) using the chain rule of differentiation:
\[ \dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt} \]Dividing both sides by \( \dfrac{dx}{dt} \) and simplifying yields:
\[ \boxed{\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \qquad \text{if} \; \dfrac{dx}{dt} \ne 0} \qquad (I) \]The second derivative is defined by:
\[ \dfrac{d^2 y}{dx^2} = \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right) \]Substituting \( y \) by \( \dfrac{dy}{dx} \) into formula (I) above gives:
\[ \boxed{ \dfrac{d^2 y}{dx^2} = \dfrac{\dfrac{d}{dt} \left(\dfrac{dy}{dx} \right)}{\dfrac{dx}{dt}} } \qquad (II) \]Examples and Their Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Given parametric equations of the form:
\[ \left\{ \begin{aligned} x(t) &= 2t + 1 \\ y(t) &= t^2 - 2 \end{aligned} \right. \]What is the derivative \( \dfrac{dy}{dx} \)?
Show Solution to Example 1
Find \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \):
\[ \dfrac{dy}{dt} = 2t, \quad \dfrac{dx}{dt} = 2 \]Use formula (I) to obtain:
\[ \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{2t}{2} = t \qquad (I) \]We may also express \( \dfrac{dy}{dx} \) in terms of \( x \). Use the parametric equation \( x = 2t + 1 \) to find \( t \) in terms of \( x \):
\[ t = \dfrac{x - 1}{2} \]Substitute \( t \) into equation (I):
\[ \dfrac{dy}{dx} = \dfrac{x - 1}{2} \]Example 2
a) Find the first derivative \( \dfrac{dy}{dx} \) and the second derivative \( \dfrac{d^2 y}{dx^2} \) given the parametric equations:
\[ \left\{ \begin{aligned} x(t) &= 2(t - \sin t) \\ y(t) &= 2(1 - \cos t) \end{aligned} \right. \]And determine the concavity of the curve.
b) Use a graphing utility to verify the concavity answer from part a).
Show Solution to Example 2
a)
Find \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \):
\[ \dfrac{dy}{dt} = 2\sin t, \quad \dfrac{dx}{dt} = 2(1 - \cos t) \]Use formula (I) to find \( \dfrac{dy}{dx} \):
\[ \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{2\sin t}{2(1 - \cos t)} = \dfrac{\sin t}{1 - \cos t} \]Next, calculate \( \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) \):
\[ \dfrac{d}{dt} \left( \dfrac{\sin t}{1 - \cos t} \right) = \dfrac{\cos t(1 - \cos t) - \sin t(\sin t)}{(1 - \cos t)^2} = \dfrac{\cos t - \cos^2 t - \sin^2 t}{(1 - \cos t)^2} = \dfrac{\cos t - 1}{(1 - \cos t)^2} = -\dfrac{1}{1 - \cos t} = \dfrac{1}{\cos t - 1} \]Use formula (II) to obtain the second derivative:
\[ \dfrac{d^2 y}{dx^2} = \dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}} = \dfrac{\dfrac{1}{\cos t - 1}}{2(1 - \cos t)} = -\dfrac{1}{2(\cos t - 1)^2} \]The second derivative is always negative (except where the denominator equals 0), hence the curve of the parametric equations is concave down.
b)
The plot of the given parametric equations (known as a cycloid) confirms that the curve is entirely concave down.
Example 3
a) Plot the curve defined by the parametric equations:
\[ \left\{ \begin{aligned} x(t) &= \sin t \\ y(t) &= \sin t \cos t \end{aligned} \right., \quad t \in [0, 2\pi] \]b) Find the first derivative \( \dfrac{dy}{dx} \), the equations of the tangent line(s) at the point \( (0, 0) \), and plot them.
Show Solution to Example 3
a)
b)
Find the derivatives \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \):
\[ \dfrac{dy}{dt} = \cos^2 t - \sin^2 t, \quad \dfrac{dx}{dt} = \cos t \]Find \( \dfrac{dy}{dx} \) using formula (I):
\[ \dfrac{dy}{dx} = \dfrac{\cos^2 t - \sin^2 t}{\cos t} \]Find the value(s) of \( t \) for which \( (x, y) = (0, 0) \) by solving \( \sin t = 0 \) and \( \sin t \cos t = 0 \) for \( t \in [0, 2\pi] \):
\[ t = 0 \quad \text{and} \quad t = \pi \]Evaluate \( \dfrac{dy}{dx} \) at \( t = 0 \) for the first slope \( m_1 \):
\[ m_1 = \dfrac{\cos^2 0 - \sin^2 0}{\cos 0} = 1 \]Evaluate \( \dfrac{dy}{dx} \) at \( t = \pi \) for the second slope \( m_2 \):
\[ m_2 = \dfrac{\cos^2 \pi - \sin^2 \pi}{\cos \pi} = -1 \]Equations of the tangents at \( (0, 0) \):
\[ y = x \quad \text{and} \quad y = -x \]
Questions
Practice solving the problems below. Click each solution box to check your answers.
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Question 1
a) Given the parametric equations \( \left\{ \begin{aligned} x(t) &= \cos t + 2 \\ y(t) &= \sin t - 1 \end{aligned} \right., t \in [\pi, 2\pi] \), find the derivative \( \dfrac{dy}{dx} \) at \( t = \dfrac{3\pi}{2} \).
b) What is the concavity of the curve?
c) Plot the curve and check your answers. -
Question 2
A curve is given in polar coordinates by \( r = 3 - 3\cos\theta \).
a) Find the slope of the tangent to the curve for \( \theta = \dfrac{\pi}{4} \).
b) Find \( \theta \) and \( r \) so that the tangent to the curve is horizontal.
c) Graph \( r \) in polar coordinates and check your answers. -
Question 3
In polar coordinates, the equation of a circle of radius \( R \) centered at the origin is \( r = R \). Find the rectangular coordinates of all points on the circle of radius 2 centered at the origin where the tangents have a slope equal to \( \dfrac{1}{2} \).
Solutions to the Above Exercises
Solution to Question 1
Show Solution to Question 1
a) Find \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \):
\[ \dfrac{dy}{dt} = \cos t, \quad \dfrac{dx}{dt} = -\sin t \] \[ \dfrac{dy}{dx} = \dfrac{\cos t}{-\sin t} = -\cot t \]Evaluate at \( t = \dfrac{3\pi}{2} \):
\[ \dfrac{dy}{dx} = -\cot\left(\dfrac{3\pi}{2}\right) = 0 \]b) Calculate the second derivative \( \dfrac{d^2 y}{dx^2} \):
\[ \dfrac{d^2 y}{dx^2} = \dfrac{\dfrac{d}{dt}(-\cot t)}{-\sin t} = \dfrac{\csc^2 t}{-\sin t} = -\csc^3 t \]Since \( \csc t = \dfrac{1}{\sin t} \) is negative over \( (\pi, 2\pi) \), the second derivative \( -\csc^3 t \) is positive, meaning the curve is concave up on \( (\pi, 2\pi) \).
Solution to Question 2
Show Solution to Question 2
a) Using polar-to-rectangular conversion:
\[ x(\theta) = (3 - 3\cos\theta)\cos\theta, \quad y(\theta) = (3 - 3\cos\theta)\sin\theta \]Computing derivatives with respect to \( \theta \):
\[ \dfrac{dy}{d\theta} = 3\sin^2\theta + \cos\theta(3 - 3\cos\theta), \quad \dfrac{dx}{d\theta} = 3\sin(2\theta) - 3\sin\theta \]At \( \theta = \dfrac{\pi}{4} \), substituting values gives slope \( m = \sqrt{2} + 1 \approx 2.41 \).
b) For a horizontal tangent, set \( \dfrac{dy}{d\theta} = 0 \) (with \( \dfrac{dx}{d\theta} \neq 0 \)):
\[ 3\sin^2\theta + 3\cos\theta - 3\cos^2\theta = 0 \implies 2\cos^2\theta - \cos\theta - 1 = 0 \]Solving yields \( \theta = \dfrac{2\pi}{3} \) and \( \theta = \dfrac{4\pi}{3} \) (excluding \( \theta = 0 \) where denominator is zero). The corresponding radii are \( r = 4.5 \).
Solution to Question 3
Show Solution to Question 3
Parametric equations for the circle \( r = 2 \):
\[ x(\theta) = 2\cos\theta, \quad y(\theta) = 2\sin\theta \]The derivative is \( \dfrac{dy}{dx} = -\cot\theta \).
Set the slope equal to \( \dfrac{1}{2} \):
\[ -\cot\theta = \dfrac{1}{2} \implies \tan\theta = -2 \]Solving for angles in \( [0, 2\pi] \):
\[ \theta_1 \approx 2.68 \text{ rad} \, (153.55^\circ), \quad \theta_2 \approx 5.82 \text{ rad} \, (333.46^\circ) \]Evaluating rectangular coordinates:
\[ (2\cos\theta_1, 2\sin\theta_1) \approx (-1.79, 0.89) \] \[ (2\cos\theta_2, 2\sin\theta_2) \approx (1.79, -0.89) \]