Parametric Derivative

For an equation written in its parametric form, the first derivative is

$\frac{dy}{dx} = \frac{\hspace{2mm} \frac{dy}{dt}\hspace{2mm} }{\frac{dx}{dt}}$

The $x$ and $y$ time derivatives oscillate while the derivative (slope) of the function itself oscillates as well. ^[1]

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.

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 $x = 1?$

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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,

$\frac{dy}{dx} = - \frac14 e^{2t} = - \frac14 e^{-2 \ln \frac12} = -\frac14 e^{ \ln\big(\frac12\big)^{-2} } = -\frac14 e^{\ln 4} = -\frac14 (4) = -1.\ _\square$

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 $x?$

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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:

$\frac{d^2y}{dx^2} = \frac{ \dot{x}\ddot{y} - \dot{y} \ddot{x} } {\dot{x}^3} = \frac{ (8t)(60t^3) - (15t^4)(8) }{(8t)^3} = \frac{360t^4}{512t^3} = \frac{45}{64}t.\ _\square$

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 $\frac{dy}{dx}.$

1. Helsing, v. Animated involute of circle. Retrieved May 18, 2016, from https://commons.wikimedia.org/wiki/File:Animated_involute_of_circle.gif