Parametric Derivative
For an equation written in its parametric form, the first derivative is
\[\frac{dy}{dx} = \frac{\hspace{2mm} \frac{dy}{dt}\hspace{2mm} }{\frac{dx}{dt}}\]
and the second derivative is
\[\frac{d^2y}{dx^2} = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^3},\]
where \(x\) and \(y\) are the coordinates of the points described by the equation and \(t\) is the parameter.
First Derivative
A brute-force method for finding the first derivative \(\left(\frac{dy}{dx}\right)\) is to eliminate the parameter, but this makes problems involving the parameter more complex. More elegantly, consider the derivative of \(y\) with respect to \(t:\)
\[\frac{dy}{dt}.\]
By the chain rule,
\[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}.\]
The first term on the right side is the first derivative of the parametric function. In order to isolate it, divide both sides by \(\frac{dx}{dt}.\)
\[\frac{dy}{dx} = \frac{\hspace{2mm} \frac{dy}{dt}\hspace{1.5mm} }{\frac{dx}{dt}}\]
In practice, differentiate the parametric functions of \(x\) and \(y\) independently and plug them into the relation above.
If \(x = 2e^{-t}\) and \(y = \frac12 e^t,\) what is \(\frac{dy}{dx}\) when \(x = 1?\)
First, compute the derivatives of \(x\) and \(y\) in terms of \(t:\)
\[\frac{dx}{dt} = -2e^{-t},\quad \frac{dy}{dt} = \frac12 e^t.\]
Write down \(\frac{dy}{dx}:\)
\[\frac{dy}{dx} = \frac{\hspace{2mm} \frac{dy}{dt}\hspace{1mm} }{\frac{dx}{dt}} = \frac{\frac12 e^t}{-2e^{-t}} = - \frac14 e^{2t}.\]
Since \(x = 2e^{-t},\) when \(x = 1\)
\[t = - \ln \frac{x}{2} = - \ln \frac12.\]
Hence,
\[\frac{dy}{dx} = - \frac14 e^{2t} = - \frac14 e^{-2 \ln \frac12} = -\frac14 e^{ \ln\big(\frac12\big)^{-2} } = -\frac14 e^{\ln 4} = -\frac14 (4) = -1.\ _\square\]
Additional Derivatives
The second derivative is, by definition,
\[\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{dy}{dx}.\]
By the chain rule, \(\frac{d}{dx} = \frac{d}{dt} \frac{1}{\hspace{1mm} \frac{dx}{dt}\hspace{1mm} } = \frac{d}{dt} \frac{dt}{dx},\) so
\[\frac{d^2y}{dx^2} = \frac{d}{dt} \frac{dy}{dx} \frac{dt}{dx}.\]
From above, \(\frac{dy}{dx} = \frac{\hspace{1mm}\frac{dy}{dt}\hspace{1mm}}{\frac{dx}{dt}},\) so
\[\frac{d^2y}{dx^2} = \frac{d}{dt} \frac{\hspace{1mm} \frac{dy}{dt}\hspace{1mm}}{\frac{dx}{dt}} \frac{dt}{dx}.\]
Finally, evaluate \(\frac{d}{dt}\) on the right side by using the quotient rule:
\[\frac{d^2y}{dx^2} = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{\frac{dx}{dt}^2} \frac{dx}{dt} = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^3}.\]
This equation is less headache-inducing if written using Newton's dot notation, by which \(\dot{u}\) represents the first derivative of \(u\) with respect to \(t\) and \(\ddot{u}\) represents the second derivative of \(u\) with respect to \(t\).
\[\frac{d^2y}{dx^2} = \frac{ \dot{x}\ddot{y} - \dot{y} \ddot{x} } {\dot{x}^3}\]
If \(x = 4t^2\) and \(y = 3t^5,\) what is the second derivative of \(y\) in terms of \(x?\)
First, evaluate the first and second derivates needed:
\[\dot{x} = 8t,\quad \ddot{x} = 8,\quad \dot{y} = 15t^4,\quad \ddot{y} = 60t^3.\]
Next, evaluate the second derivative:
\[\frac{d^2y}{dx^2} = \frac{ \dot{x}\ddot{y} - \dot{y} \ddot{x} } {\dot{x}^3} = \frac{ (8t)(60t^3) - (15t^4)(8) }{(8t)^3} = \frac{360t^4}{512t^3} = \frac{45}{64}t.\ _\square\]
Equations of Tangent Lines
The tangent equation represents a straight linear line that creates a right angle at the point of tangency. The formula of a line is described in Algebra section as "point-slope formula":
\[y-y_1 = m(x-x_1).\]
In parametric equations, finding the tangent requires the same method, but with calculus:
\[y-y_1 = \frac{dy}{dx} (x-x_1).\]
Tangent of a line is always defined to be the derivative of the line. Thus, for slope, we use \(\frac{dy}{dx}.\)
References
- Helsing, v. Animated involute of circle. Retrieved May 18, 2016, from https://commons.wikimedia.org/wiki/File:Animated_involute_of_circle.gif