Derivatives of Parametric Equations and Applications
The derivative of the parametric equations are presented with examples and their solutions. Some applications to find slopes of tangent line are also included. More questions with solutions are included.
First and Second Derivatives of Parametric Equations
Given parametric equations of the form
\[
\begin{equation}
\left\{ \begin{aligned}
x(t) & \\\\
y(t) &
\end{aligned} \right.
\end{equation}
\] what is the derivative \( \dfrac{dy}{dx} \)?
First find the derivative \( \dfrac{dy}{dt} \) using the the chain rule of differentiation as follows
\[ \dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt} \]
Divide left and right side by \( \dfrac{dx}{dt} \) and simplify to obtain
\[ \boxed {\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \qquad \text{if} \; \dfrac{dx}{dt} \ne 0} \qquad (I)\]
The second derivative is defined by
\[ \dfrac{d^2 y}{dx^2} = \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right) \]
The right side in the above is obtained by substituting \( y \) by \( \dfrac{dy}{dx} \) in formula (I) above, hence
\[ \boxed{ \dfrac{d^2 y}{dx^2} = \dfrac{\dfrac{d}{dt} \left(\dfrac{dy}{dx} \right)}{\dfrac{dx}{dt}} } \qquad (II)\]
Examples and Their Solutions
Example 1
Given parametric equations of the form
\(
\begin{equation}
\left\{ \begin{aligned}
x(t) & = 2 t + 1 \\\\
y(t) & = t^2 2
\end{aligned} \right.
\end{equation}
\) , what is the derivative \( \dfrac{dy}{dx} \)?
Solution to Example 1
Find \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \)
\[ \dfrac{dy}{dt} = 2 t \]
\[ \dfrac{dx}{dt} = 2 \]
Use the above formula in (I) to obtain
\[ \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \\[30pt] = \dfrac{2 t}{2} \]
Simplify
\[ \dfrac{dy}{dx} = t \qquad (I) \]
We may also express \( \dfrac{dy}{dx} \) in terms of \( x \). Use the parametric equation \( x = 2 t + 1 \) to find \( t \) in terms of \( x \)
\[ t = \dfrac{x  1}{2} \]
Substitute \( t \) by \( \dfrac{x  1}{2} \) in (I) above
\[ \dfrac{dy}{dx} = \dfrac{x  1}{2} \]
Example 2
a) Find the first derivative \( \dfrac{dy}{dx} \) and the second derivative \( \dfrac{d^2 y}{dx^2} \) given parametric equations
\(
\begin{equation}
\left\{ \begin{aligned}
x(t) & = 2(t  \sin t) \\\\
y(t) & = 2(1  \cos t)
\end{aligned} \right.
\end{equation}
\)
and determine the concavity of the curve.
b) Use a graphing calculator to verify the answer about the concavity in part a).
Solution to Example 2
a)
Find \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \)
\[ \dfrac{dy}{dt} = 2 \sin t \]
\[ \dfrac{dx}{dt} = 2 (1  \cos t) \]
Use formula (I) above to find \( \dfrac{dy}{dx} \)
\[ \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \\[35 pt] = \dfrac{\sin t}{ (1  \cos t) } \]
We now calculate \( \dfrac{d}{dt} \left(\dfrac{dy}{dx} \right) \)
\[ \dfrac{d}{dt} \left(\dfrac{dy}{dx} \right) = \dfrac{d}{dt} \left( \dfrac{ \sin t}{ 1  \cos t } \right) \\
= \dfrac{1}{\cos t  1}
\]
Use formula (II) and the above derivative to obtain the second derivative
\[ \dfrac{d^2 y}{dx^2} = \dfrac{\dfrac{d}{dt} \left(\dfrac{dy}{dx} \right)}{\dfrac{dx}{dt}} \\[30 pt] = \dfrac{ \dfrac{1}{\cos t  1} }{ 2 (1  \cos t) } \]
Simplify
\[ \dfrac{d^2 y}{dx^2} =  \; \frac{1}{2\left(\cos \left(t\right)1\right)^2} \]
The second derivative is always negative, except at values of \( t \) that makes the denominator equal to \( 0 \), hence the curve of the given parametric equations is concave down .
b)
The plot of the given parametric equations is shown below and it can be seen that the curve is concave down as shown in part a) above. The plot is called a cycloid.
Example 3
a) Use a graphing calculator to plot the curve defined by the parametric equation
\(
\begin{equation}
\left\{ \begin{aligned}
x(t) & = \sin t \\\\
y(t) & = \sin t \cos t
\end{aligned} \right. , t \in [0,2\pi]
\end{equation}
\)
b) Find the first derivative \( \dfrac{dy}{dx} \) and the equation of the tangent(s) at the point \( (0,0) \) and plot them.
Solution to Example 3
a)
The plot of the given parametric equations is shown below.
b)
Find the derivatives \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \). \[ \dfrac{dy}{dt} = \cos^2 t  \sin^2 t \] \[ \dfrac{dx}{dt} = \cos t \] Find the derivative \( \dfrac{dy}{dx} \) using formula (I) above. \[ \dfrac{dy}{dx} = \dfrac{\cos^2 t  \sin^2 t}{\cos t} \] Find the value(s) of \( t \) for which \( (x,y) = (0,0) \) by solving the equations \[ \sin t = 0 \] and \[ \sin t \cos t = 0 \] which gives \[ \sin t = 0 \] and the solutions within the interval \( [0,2\pi ] \) \[ t = 0 \] and \[ t = \pi \] Evaluate \( \dfrac{dy}{dx} \) at \( t = 0 \) to find the slope \(m_1\) of the first tangent \[ m_1 = \dfrac{\cos^2 0  \sin^2 0}{\cos 0} \\[15 pt] = 1 \] Evaluate \( \dfrac{dy}{dx} \) at \( t = \pi \) to find the slope \(m_2\) of the first tangent \[ m_2 = \dfrac{\cos^2 \pi  \sin^2 \pi}{\cos \pi} \\[15 pt] =  1\] Equations of the tangent at \( (0,0) \) with slope \( m_1 = 1 \) \[ y = x \] Equations of the tangent at \( (0,0) \) with slope \( m_2 = 1 \) \[ y =  x \]
Questions
 Question 1
a) Given the parametric equations \( \begin{equation} \left\{ \begin{aligned} x(t) & = \cos t + 2 \\\\ y(t) & = \sin t  1 \end{aligned} \right. \end{equation} t \in [ \pi , 2 \pi ] \), find the derivative \( \dfrac{dy}{dx} \) at \( t = \dfrac{3 \pi}{2} \).
b) What is the concavity of the curve whose parametric equations are given in part a)?
c) Plot the curve of the given parametric equations and check your answers to part a) and b) above.
 Question 2
A curve is given by its equation in polar coordinates as \( r = 33\cos\theta \).
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) Use any graphing calculator to graph \( r \) in polar coordinates and check answers to part b).
 Question 3
In polar coordinates, the equation of a circle with radius \( R \) and centered at the origin is given by \( r = R \).
Find the rectangular coordinates of all points on the circle of radius 2 and centered at the origin so that the tangents at this points have a slope equal to \( \dfrac{1}{2} \)
Solutions to the Above Exercises

Solution to Question 1
a)
Find the derivatives \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \). \[ \dfrac{dy}{dt} = \cos t \] \[ \dfrac{dx}{dt} =  \sin t \] Find the derivative \( \dfrac{dy}{dx} \) using formula (I) above. \[ \dfrac{dy}{dx} =  \dfrac{\cos t}{\sin t} =  \cot t\] Evaluate \( \dfrac{dy}{dx} \) at \( t = \dfrac{3 \pi }{2} \) \[ \dfrac{dy}{dx} =  \cot \left(\dfrac{3 \pi }{2} \right) = 0\]
b)
The concavity is given by the second derivative \( \dfrac{d^2 y}{dx^2} \) given by formula (II) above.
\[ \begin{aligned} & \dfrac{d^2 y}{dx^2} = \dfrac{\dfrac{d}{dt} \left(\dfrac{dy}{dx} \right)}{\dfrac{dx}{dt}} \\[15pt] & \color{red}{\text{Substitute \( \dfrac{dy}{dx} \) and \( \dfrac{dx}{dt} \) }} \\[12pt] & = \dfrac{\dfrac{d}{dt} \left(  \cot t \right)}{ \sin t} \\[15pt] & \color{red}{\text{Evaluate the above to obtain}} \\[8pt] & \dfrac{d^2 y}{dx^2} = \csc^3 (x) \\[15pt] \end{aligned} \]
\( \csc x = \dfrac{1}{\sin x} \) is negative over the interval \( (\pi , 2\pi) \) and therefore the second derivative \( \dfrac{d^2 y}{dx^2} =  \csc^3 (x) \) is positive and therefore the curve is concave up on the interval \( (\pi , 2\pi) \).
c)
The plot of the given parametric equations is shown below and we can see that the tangent at \( t=\dfrac{3\pi}{2} \) is horizontal as predicted in part a) where the slope found is equal to zero.
The curve is concave up as predicted in part b) above.

Solution to Question 2
a)
The parametric equations of the curve given in polar coordinates as \( r = 33\cos\theta \).
Using the convertion from polar to rectangular coordinates , we write the parametric equations as
\( x(\theta) = r \cos \theta = (33\cos\theta)\cos \theta \)
and
\( y(\theta) = r \sin \theta = (33\cos\theta)\sin \theta \)
where \( \theta \) is the polar angle.
Use formula (I) to find the derivative and hence the slope of the tangent. \[ \begin{aligned} & \dfrac{dy}{d\theta} = 3\sin^2 \theta + \cos \theta ( 33\cos \theta) \\[15pt] & \dfrac{dx}{d\theta} = 3\sin (2\theta)3\sin \theta \end{aligned} \] and hence \[ \begin{aligned} &\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} \\[15pt] & \color{red}{\text{Substitute \( \dfrac{dy}{d\theta} \) and \( \dfrac{dx}{d\theta} \) by their expressions found above} } \\[12pt] & \dfrac{dy}{dx} = \dfrac{3\sin^2 \theta + \cos \theta ( 33\cos \theta) }{3\sin (2\theta)3\sin \theta } \\[15pt] & \color{red}{\text{The slope \( m \) at \( \theta = \dfrac{\pi}{4} \) is given by the value of the first derivative \( \dfrac{dy}{d\theta} \) at \( \theta = \dfrac{\pi}{4} \). Hence}} \\[10pt] & m = \dfrac{3\sin^2 \left(\dfrac{\pi}{4}\right) + \cos \left(\dfrac{\pi}{4}\right) ( 33\cos \left(\dfrac{\pi}{4}\right)) }{3\sin (2\left(\dfrac{\pi}{4}\right))3\sin \left(\dfrac{\pi}{4}\right) } \\[15pt] & = \sqrt{2}+1 \\[15pt] & \approx 2.41 \end{aligned} \]
b)
For the tangent to be horizontal the slope, given by the first derivative \( \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} \), must be equal to zero.
Hence we need to solve \( \dfrac{dy}{d\theta} = 0 \) such that \( \dfrac{dx}{d\theta} \ne 0 \). \[ 3\sin^2 \theta + \cos \theta ( 33\cos \theta) = 0 \] Use the identity \( \sin^2 \theta = 1  \cos^2 \theta \) in the above equation and rewrite as \[ 3 (1  \cos^2 \theta ) + \cos \theta ( 33\cos \theta) = 0 \] Group like terms, simplify and rewrite as \[ 2 \cos^2 \theta  \cos \theta  1 = 0\] Solve the above quadartic like equations to find
\( \cos \theta =1 \) and \( \cos \theta = \frac{1}{2} \)
which gives the solutions angles
\( \theta_1 = 0 \qquad \) , \( \qquad \theta_2 = \dfrac{2\pi}{3} \qquad \) and \( \qquad \theta_3 = \dfrac{4\pi}{3} \)
Note that \( \theta_1 = 0 \) is not accepted as a solution because it will make the denomiantor \( \dfrac{dx}{d\theta} \) equal to zero.
Find \( r \) at \( \qquad \theta = \dfrac{2\pi}{3} \) , \( \qquad r = 33 \cos \left(\dfrac{2\pi}{3} \right) = 4.5 \)
Find \( r \) at \( \qquad \theta = \dfrac{4\pi}{3} \) , \( \qquad r = 33 \cos \left(\dfrac{4\pi}{3} \right) = 4.5 \)
c)
The plot of the given polar equations is shown below and we can see that the tangent at the polar points \( \left(4.5 , \left(\dfrac{2\pi}{3} \right) \right) \) and \( \left(4.5 , \left(\dfrac{4\pi}{3} \right) \right)\) are horizontal as predicted in part b).

Solution to Question 3
Convert from polar to rectangular coordinates , we write the parametric equations as
\( x(\theta) = r \cos \theta = 2 \cos \theta \)
and
\( y(\theta) = r \sin \theta = 2 \sin \theta \)
where \( \theta \) is the polar angle.
Calculate \( \dfrac{dy}{d\theta} \) and \( \dfrac{dx}{d\theta} \) \[ \dfrac{dy}{d\theta} = 2 \cos \theta \] \[ \dfrac{dx}{d\theta} =  2 \sin \theta \] Find \( \dfrac{dy}{dx} \) using forrmula (I) above \[ \begin{aligned} &\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} \\[15pt] & \color{red}{\text{Substitute \( \dfrac{dy}{d\theta} \) and \( \dfrac{dx}{d\theta} \) by their expressions found above} } \\[12pt] & = \dfrac{ 2 \cos \theta }{ 2 \sin \theta} \\\\[15pt] & \color{red}{\text{Simplify the above} } \\[12pt] & =  \cot \theta \end{aligned} \]
The slope at a given point of the circle is equal to the value of the derivative found above. Hence we need to solve the equation \[  \cot \theta = 2 \] which gives the general solutions \( \theta = 2.68 +\pi n \) where \( n = 0, \pm 1 , \pm 2, .... \) We need two solutions in the interval \( [0, 2\pi ] \)
In radians
\( \theta_1 \approx 2.68 \qquad \) and \( \qquad \theta_2 \approx 2.68 + \pi = 5.82 \)
In degrees
\( \theta_1 \approx 153.55^{\circ} \qquad \) and \( \qquad \theta_2 = 333.46^{\circ} \)
The coordinates of point A corresponding to the solution \( \theta_1 \) are given by \[ (2 \cos \theta_1 , 2 \sin \theta_1) = (1.79 , 0.89) \] The coordinates of point B corresponding to the solution \( \theta_2 \) are given by \[ (2 \cos \theta_2 , 2 \sin \theta_2) = (1.79 , 0.89) \]