Title: Default for 11.6 Parametric Differentiation
Course: HELM Workbooks 2023-2024

Parametric Differentiation

Sometimes the equation of a curve is not be given in Cartesian form \(y = f(x)\) but in parametric form: \(x = h(t)\), \(y = g(t)\). In this Section we see how to calculate the derivative \( \dfrac{dy}{dx}\) from a knowledge of the so-called parametric derivatives \( \dfrac{dx}{dt}\) and\( \dfrac{dy}{dt} \). We then extend this to the determination of the second derivative \( \dfrac{d^2y}{dx^2}\).

Parametric functions arise often in particle dynamics in which the parameter \(t\) represents the time and \((x(t), y(t))\) then represents the position of a particle as it varies with time.

Parametric differentiation

In this subsection we consider the parametric approach to describing a curve: \[ x = h(t), \hspace{0.6cm} y = g(t) \hspace{2cm}t_0 \leq t \leq t_1\] The functions \(x = h(t)\) and \(y = g(t)\) are known as the parametric equations.

The interval \(t_0 \leq t \leq t_1\) is known as the parametric range

As various values of \(t\) are chosen within the parameter range the corresponding values of \(x\), \(y\) are calculated from the parametric equations. When these points are plotted on an \(xy\) plane they trace out a curve. The Cartesian equation of this curve is obtained by eliminating the parameter \(t\) from the parametric equations. For example, consider the curve: \[ x = 2 \cos t \hspace{0.6cm} y = 2 \sin t \hspace{2cm} 0 \leq t \leq 2\pi .\] We can eliminate the \(t\) variable in an obvious way - square each parametric equation and then add: \[ x^2 + y^2 = 4 \cos^2 t + 4 \sin^2 t = 4 \hspace{1cm} \therefore \hspace{1cm} x^2 + y^2 = 4\] which we recognise as the standard equation of a circle with centre at \((0, 0)\) with radius \(2\). In a similar fashion the parametric equations \[ x = 2t \hspace{0.6cm} y = 4t^2 \hspace{2cm} - \infty < t < \infty \] describes a parabola. This follows since, eliminating the parameter \(t\): \[ t = \dfrac{x}{2} \hspace{0.6cm} \therefore \hspace{0.6cm} y = 4 \left( \dfrac{x^2}{4} \right) \hspace{0.6cm} \text{so } y = x^2 \] which we recognise as the standard equation of a parabola.

The question we wish to address in this Section is `how do we obtain the derivative \( \dfrac{dy}{dx} \) if a curve is given in parametric form?' To answer this we note the key result in this area:

Key Point

Parametric Differentiation

If \(x = h(t)\) and \(y = g(t)\) then \[ \dfrac{dy}{dx} = \left.\dfrac{dy}{dt} \middle/\dfrac{dx}{dt}\right.\]

We note that this result allows the determination of \( \dfrac{dy}{dx}\) without the need to find \(y\) as an explicit function of \(x\).

Example

Determine the equation of the tangent line to the semicircle with parametric equations \[ x = \cos t \hspace{0.6cm} y = \sin t \hspace{2cm} 0 \leq t \leq \pi \] at \(t = \pi/4\).

Solution

The semicircle is drawn in the Figure below. We have also drawn the tangent line at \(t = \pi/4\) (or, equivalently, at \(x = \cos \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}, y = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}\) )

[Image: semi circle and tangent]

Now \[ \dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt} = \dfrac{\cos t}{-\sin t} = -\cot t \]

Thus at \(t = \dfrac{\pi}{4}\) we have \( \dfrac{dy}{dx} = -\cot \left( \dfrac{\pi}{4} \right) = -1\).

The equation of the tangent line is \[ y = mx + c\] where \(m\) is the gradient of the line and \(c\) is a constant.

Clearly \(m = -1\) (since, at the point \(P\) the line and the circle have the same gradient).

To find \(c\) we note that the line passes through the point \(P\) with coordinates \( \left( \dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}} \right) \). Hence \[ \dfrac{1}{\sqrt{2}} = (-1) \dfrac{1}{\sqrt{2}} + c \hspace{1cm} \therefore c = \dfrac{2}{\sqrt{2}}\] Finally \[ y = -x + \dfrac{2}{\sqrt{2}}\] is the equation of the tangent line at the point in question.

We should note, before proceeding, that a derivative with respect to the parameter \(t\) is often denoted by a `dot'. Thus \[ \dfrac{dx}{dt} = \dot{x}, \hspace{1cm} \dfrac{dy}{dt} = \dot{y}, \hspace{1cm} \dfrac{d^2x}{dt^2} = \ddot{x}\]

Exercise

Here we are going to find  \( \dfrac{dy}{dx}\) if \[x = [value], \quad y =[value]. \]
1. First, specify the functions \( \dfrac{dx}{dt} \) and \( \dfrac{dy}{dt}\) for the composite functions
\( \dfrac{dx}{dt} = \)
\( \dfrac{dy}{dt} = \)
2. Now obtain \( \dfrac{dy}{dx}\). Remember to eliminate \(t\) from your answer.
\( \dfrac{dy}{dx} = \)
Feedback: We have \(x = [value]\) and \(y =[value].\) 1. So \( \dfrac{dx}{dt} = [value] \), and \( \dfrac{dy}{dt} = [value].\) 2. Thus \( \dfrac{dy}{dx} = \displaystyle{[value]}. \) Now eliminate \(t\) by the relation \( t = \dfrac{x}{[value]}\), so we have: \[ \dfrac{dy}{dx} = \displaystyle{[value]} = \displaystyle[value]. \]

Exercise

Here we are going find the value of \( \dfrac{dy}{dx}\) at \(t = [value]\) if \[ x = [value] , \quad y =[value]. \]
1. First, specify the functions \( \dfrac{dx}{dt} \) and \( \dfrac{dy}{dt}\) for the composite functions
\( \dfrac{dx}{dt} = \)
\( \dfrac{dy}{dt} = \)
2. Now obtain \( \dfrac{dy}{dx}\) at \(t = [value]\). Leave your answer in exact form.
\( \dfrac{dy}{dx} = \)
Feedback: We have \(x = [value]\) and \(y =[value]\). 1. First we find the derivatives with respect to \(t\): \(\displaystyle\frac{dx}{dt} = [value]\) and \(\displaystyle\frac{dy}{dt} = [value]\) 2. Thus \[ \dfrac{dy}{dx} = \displaystyle{[value]}, \] which when evaluated at \(t=[value]\) is \(\displaystyle[value] .\)

Higher derivatives

Having found the first derivative \( \dfrac{dy}{dx}\) using parametric differentiation we now ask how we might determine the second derivative \( \dfrac{d^2y}{dx^2}\).

By definition: \[ \dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \left( \dfrac{dy}{dx} \right) \] But \[ \dfrac{dy}{dx} = \dfrac{\dot{y}}{\dot{x}} \] and so \[ \dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \left( \dfrac{\dot{y}}{\dot{x}} \right) \] Now \( \dfrac{\dot{y}}{\dot{x}} \) is a function of \(t\) so we can change the derivative with respect to \(x\) into a derivative with respect to \(t\) since \[ \dfrac{d}{dx} \left( \dfrac{dy}{dx} \right) = \left\{ \dfrac{d}{dt} \left( \dfrac{dy}{dx} \right) \right\} \dfrac{dt}{dx} \] from the function of a function rule.

But, differentiating the quotient \( \dot{y}/\dot{x} \), we have \[ \dfrac{d}{dt} \left( \dfrac{\dot{y}}{\dot{x}} \right) = \dfrac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{\dot{x}^2} \] and \[ \dfrac{dt}{dx} = \dfrac{1}{ \left( \dfrac{dx}{dt} \right)} = \dfrac{1}{\dot{x}} \] So finally \[ \dfrac{d^2y}{dx^2} = \dfrac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{\dot{x}^3} \]

Key Point

If \(x = h(t)\), \(y = g(t)\) then the first and second derivatives of \(y\) with respect to \(x\) are: \[ \dfrac{dy}{dx} = \dfrac{ \dot{y} }{ \dot{x} }\] and \[ \dfrac{d^2y}{dx^2} = \dfrac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{\dot{x}^3} \]

Example

If the equations of a curve are \(x = 2t\), \(y = t^2 - 3\), determine \( \dfrac{dy}{dx} \) and \(\dfrac{d^2y}{dx^2}\).

Solution

Here \( \dot{x} = 2\), \(\dot{y} = 2t\), so \( \dfrac{dy}{dx} = \dfrac{\dot{y}}{\dot{x}} = \dfrac{2t}{2} = t\).

Also \( \ddot{x} = 0\), \(\ddot{y} = 2\), so \( \dfrac{d^2y}{dx^2} = \dfrac{2(2) - 2t(0)}{(2)^3} = \dfrac{1}{2} .\)

These results can easily be checked since \(t = \dfrac{x}{2}\) and \(y = t^2 -3\) which imply \(y = \dfrac{x^2}{4} -3\). Therefore the derivatives can be obtained directly:

\( \dfrac{dy}{dx} = \dfrac{2x}{4} = \dfrac{x}{2}\), and \( \dfrac{d^2y}{dx^2} = \dfrac{1}{2} \).

Exercise

Find \( \dfrac{dy}{dx}\) and \( \dfrac{d^2y}{dx^2}\) where \( x = [value]\) and \( y = [value] \).
1. \( \dfrac{dy}{dx} = \)
2. \( \dfrac{d^2y}{dx^2} = \)
Feedback: We have \(x = [value] \,\) and \(\, y =[value]\). This gives: \[ \dfrac{dx}{dt} = [value], \quad \dfrac{d^2x}{dt^2} = [value] \quad \text{and} \quad \dfrac{dy}{dt} = [value], \quad \dfrac{d^2y}{dt^2} = [value]\] Then: 1. \( \dfrac{dy}{dx} = \displaystyle{[value]} \) 2. \( \dfrac{d^2y}{dx^2} = \displaystyle{[value]} \)

Exercise

Find \( \dfrac{dy}{dx}\) and \( \dfrac{d^2y}{dx^2}\) where \( x = [value]\) and \( y = [value] \).
1. \( \dfrac{dy}{dx} = \)
2. \( \dfrac{d^2y}{dx^2} = \)
Feedback: We have \(x = [value] \quad \text{and} \quad y =[value]\). So \( \dfrac{dx}{dt} = [value], \quad \dfrac{d^2x}{dt^2} = [value] \quad \text{and} \quad \dfrac{dy}{dt} = [value], \quad \dfrac{d^2y}{dt^2} = [value]\). Thus: 1. \(\displaystyle \frac{dy}{dx} = [value]\) 2. \(\displaystyle \frac{d^2y}{dx^2} = [value]\)

Exercise

Find \( \dfrac{dy}{dx}\) and \( \dfrac{d^2y}{dx^2}\) where \( x = [value]\) and \( y = [value] \).
1. \( \dfrac{dy}{dx} = \)
2. \( \dfrac{d^2y}{dx^2} = \)
Feedback: We have \(x = [value] \quad \text{and} \quad y =[value]\). So \( \dfrac{dx}{dt} = [value], \quad \dfrac{d^2x}{dt^2} = [value] \quad \text{and} \quad \dfrac{dy}{dt} = [value] \quad \dfrac{d^2y}{dt^2} = [value]\). Thus: 1. \(\displaystyle \frac{dy}{dx} = [value] = [value]\) 2. \(\displaystyle \frac{d^2y}{dx^2} = [value] = [value]\)