Parametric Derivative
For an equation written in its parametric form, the first derivative is
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]
and the second derivative is
\[\frac{d^2y}{dx^2} = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^3}\]
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 (\(\frac{dy}{dx}\)) 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 \(\frac{dx}{dt}.\)
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]
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 \(t.\)
\[\frac{dx}{dt} = -2e^{-t}\]
\[\frac{dy}{dt} = \frac12 e^t\]
Write down \(\frac{dy}{dx}.\)
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac12 e^t}{-2e^{-t}} = - \frac14 e^{2t}\]
Since \(x = 2e^{-t},\) when \(x = 1,\)
\[t = - \ln (\frac{x}{2}) = - \ln (\frac12).\]
Hence,
\[\frac{dy}{dx} = - \frac14 e^{2t} = - \frac14 e^{-2 \ln (\frac12)} = -\frac14 e^{ \ln(\frac12)^{-2} } = -\frac14 e^{\ln(4)} = -\frac14 (4) = -1.\]
Additional Derivatives
The second derivative is, by definition,
\[\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{dy}{dx}.\]
By the chain rule, \(\frac{d}{dx} = \frac{d}{dt} \frac{1}{\frac{dx}{dt}} = \frac{d}{dt} \frac{dt}{dx}.\)
\[\frac{d^2y}{dx^2} = \frac{d}{dt} \frac{dy}{dx} \frac{dt}{dx}\]
From above, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.\)
\[\frac{d^2y}{dx^2} = \frac{d}{dt} \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \frac{dt}{dx}\]
Finally, evaluate \(\frac{d}{dt}\) on the right side by using the quotient rule.
\[\frac{d^2y}{dx^2} = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{\frac{dx}{dt}^2} \frac{dx}{dt} = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^3}\]
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.
\[\dot{x} = 8t\]
\[\ddot{x} = 8\]
\[\dot{y} = 15t^4\]
\[\ddot{y} = 60t^3\]
Next, evaluate the second derivative.
\[\frac{d^2y}{dx^2} = \frac{ \dot{x}\ddot{y} - \dot{y} \ddot{x} } {\dot{x}^3} = \frac{ (8t)(60t^3) - (15t^4)(8) }{(8t)^3} = \frac{360t^4}{512t^3} = \frac{45}{64}t\]
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