Derivatives of Functions in Parametric Forms

Introduction to Parametric Forms

Sometimes the relation between two variables $x$ and $y$ is not explicit (like $y = f(x)$) nor implicit (like $f(x, y) = 0$). Instead, both $x$ and $y$ are expressed in terms of a third variable.

This third variable is called a parameter. The equations expressing $x$ and $y$ in terms of the parameter are called parametric equations. For example, $x = f(t)$ and $y = g(t)$, where $t$ is the parameter.

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Derivative of Parametric Functions

To find $\frac{dy}{dx}$ when $x$ and $y$ are given in parametric form, we use the Chain Rule.

Since $y$ is a function of $t$, and $x$ is also a function of $t$, we can write: $$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$

Rearranging this formula gives us: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ provided $\frac{dx}{dt} \ne 0$.

Working Rule

1. Differentiate $x$ with respect to the parameter $t$ to find $\frac{dx}{dt}$.
2. Differentiate $y$ with respect to the parameter $t$ to find $\frac{dy}{dt}$.
3. Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to obtain $\frac{dy}{dx}$.

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Solved Examples

Example 1: Basic Algebraic Parametric Equations

Find $\frac{dy}{dx}$, if $x = at^2$ and $y = 2at$.

Solution: Step 1: Differentiate $x$ with respect to $t$. $$\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$$

Step 2: Differentiate $y$ with respect to $t$. $$\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$$

Step 3: Apply the parametric differentiation formula. $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}$$

Answer: $\frac{1}{t}$

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Example 2: Trigonometric Parametric Equations (Circle)

Find $\frac{dy}{dx}$, if $x = a \cos\theta$ and $y = a \sin\theta$.

Solution: Step 1: Differentiate $x$ with respect to $\theta$. $$\frac{dx}{d\theta} = \frac{d}{d\theta}(a\cos\theta) = -a\sin\theta$$

Step 2: Differentiate $y$ with respect to $\theta$. $$\frac{dy}{d\theta} = \frac{d}{d\theta}(a\sin\theta) = a\cos\theta$$

Step 3: Use the parametric derivative formula. $$\frac{dy}{dx} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta$$

Answer: $-\cot\theta$

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Example 3: Finding Derivative at a Specific Point

Find $\frac{dy}{dx}$ at $\theta = \frac{\pi}{4}$ if $x = a \sec^3\theta$ and $y = a \tan^3\theta$.

Solution: Step 1: Differentiate $x$ with respect to $\theta$. $$\frac{dx}{d\theta} = a\cdot 3\sec^2\theta \cdot (\sec\theta\tan\theta) = 3a\sec^3\theta\tan\theta$$

Step 2: Differentiate $y$ with respect to $\theta$. $$\frac{dy}{d\theta} = a\cdot 3\tan^2\theta \cdot \sec^2\theta = 3a\tan^2\theta\sec^2\theta$$

Step 3: Find $\frac{dy}{dx}$. $$\frac{dy}{dx} = \frac{3a\tan^2\theta\sec^2\theta}{3a\sec^3\theta\tan\theta} = \frac{\tan\theta}{\sec\theta} = \sin\theta$$

Step 4: Evaluate at $\theta = \frac{\pi}{4}$. $$\left.\frac{dy}{dx}\right|_{\theta=\pi/4} = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$

Answer: $\frac{1}{\sqrt{2}}$

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Example 4: The Cycloid Curve

Find $\frac{dy}{dx}$, if $x = a(\theta - \sin\theta)$ and $y = a(1 - \cos\theta)$.

Solution: Step 1: Differentiate $x$ with respect to $\theta$. $$\frac{dx}{d\theta} = a(1 - \cos\theta)$$

Step 2: Differentiate $y$ with respect to $\theta$. $$\frac{dy}{d\theta} = a\sin\theta$$

Step 3: Divide to find $\frac{dy}{dx}$. $$\frac{dy}{dx} = \frac{a\sin\theta}{a(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta}$$

Step 4: Simplify using half-angle identities. $$\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, \qquad 1-\cos\theta = 2\sin^2\frac{\theta}{2}$$ So, $$\frac{dy}{dx} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = \cot\frac{\theta}{2}$$

Answer: $\cot\left(\frac{\theta}{2}\right)$

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Example 5: Product Rule in Parametric Forms

Find $\frac{dy}{dx}$, if $x = a(\cos t + t\sin t)$ and $y = a(\sin t - t\cos t)$.

Solution: Step 1: Differentiate $x$ with respect to $t$. $$\frac{dx}{dt} = a\left[-\sin t + \frac{d}{dt}(t\sin t)\right]$$ Using product rule, $$\frac{d}{dt}(t\sin t) = \sin t + t\cos t$$ Hence, $$\frac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = at\cos t$$

Step 2: Differentiate $y$ with respect to $t$. $$\frac{dy}{dt} = a\left[\cos t - \frac{d}{dt}(t\cos t)\right]$$ Using product rule, $$\frac{d}{dt}(t\cos t) = \cos t - t\sin t$$ Therefore, $$\frac{dy}{dt} = a[\cos t - (\cos t - t\sin t)] = at\sin t$$

Step 3: Find $\frac{dy}{dx}$. $$\frac{dy}{dx} = \frac{at\sin t}{at\cos t} = \tan t$$

Answer: $\tan t$

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Example 6: Complex Logarithmic/Trigonometric Parameter

Find $\frac{dy}{dx}$, if $x = a\left(\cos t + \log\tan\frac{t}{2}\right)$ and $y = a\sin t$.

Solution: Step 1: Differentiate $x$ with respect to $t$. $$\frac{dx}{dt} = a\left[-\sin t + \frac{d}{dt}\left(\log\tan\frac{t}{2}\right)\right]$$ Now, $$\frac{d}{dt}\left(\log\tan\frac{t}{2}\right) = \frac{1}{\tan(t/2)}\cdot \sec^2\frac{t}{2}\cdot \frac{1}{2}$$ This simplifies to $$\frac{1}{2\sin(t/2)\cos(t/2)} = \frac{1}{\sin t}$$ Hence, $$\frac{dx}{dt} = a\left(-\sin t + \frac{1}{\sin t}\right) = a\left(\frac{1-\sin^2 t}{\sin t}\right) = \frac{a\cos^2 t}{\sin t}$$

Step 2: Differentiate $y$ with respect to $t$. $$\frac{dy}{dt} = a\cos t$$

Step 3: Find $\frac{dy}{dx}$. $$\frac{dy}{dx} = \frac{a\cos t}{\frac{a\cos^2 t}{\sin t}} = \frac{\sin t}{\cos t} = \tan t$$

Answer: $\tan t$

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Example 7: The Substitution Trick

If $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$, show that $\frac{dy}{dx} = -\frac{y}{x}$.

Solution: Step 1: Square both equations. $$x^2 = a^{\sin^{-1} t}, \qquad y^2 = a^{\cos^{-1} t}$$

Step 2: Multiply them. $$x^2y^2 = a^{\sin^{-1} t} \cdot a^{\cos^{-1} t} = a^{\sin^{-1} t + \cos^{-1} t}$$

Step 3: Use the identity $$\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$$ So, $$x^2y^2 = a^{\pi/2}$$ which is a constant.

Step 4: Differentiate implicitly with respect to $x$. $$\frac{d}{dx}(x^2y^2) = 0$$ Using product rule, $$2xy^2 + x^2\cdot 2y\frac{dy}{dx} = 0$$

Step 5: Solve for $\frac{dy}{dx}$. $$2x^2y\frac{dy}{dx} = -2xy^2$$ $$\frac{dy}{dx} = -\frac{y}{x}$$

Answer: Proved.

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Example 8: Rational Parametric Forms

Find $\frac{dy}{dx}$, if $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$.

Solution: Step 1: Let $t = \tan\theta$. Then, $$x = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta, \qquad y = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$$

Step 2: Differentiate with respect to $\theta$. $$\frac{dx}{d\theta} = -2\sin 2\theta, \qquad \frac{dy}{d\theta} = 2\cos 2\theta$$

Step 3: Find $\frac{dy}{dx}$. $$\frac{dy}{dx} = \frac{2\cos 2\theta}{-2\sin 2\theta} = -\cot 2\theta$$ Since $x = \cos 2\theta$ and $y = \sin 2\theta$, $$-\cot 2\theta = -\frac{x}{y}$$

Answer: $-\frac{x}{y}$

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Example 9: Expressing $\frac{dy}{dx}$ in terms of parameter

Find $\frac{dy}{dx}$ if $x = 3\cos\theta - 2\cos^3\theta$ and $y = 3\sin\theta - 2\sin^3\theta$.

Solution: Step 1: Differentiate $x$ with respect to $\theta$. $$\frac{dx}{d\theta} = -3\sin\theta - 2\cdot 3\cos^2\theta(-\sin\theta) = -3\sin\theta + 6\sin\theta\cos^2\theta$$ $$\frac{dx}{d\theta} = 3\sin\theta(2\cos^2\theta - 1)$$ Using $2\cos^2\theta - 1 = \cos 2\theta$, $$\frac{dx}{d\theta} = 3\sin\theta\cos 2\theta$$

Step 2: Differentiate $y$ with respect to $\theta$. $$\frac{dy}{d\theta} = 3\cos\theta - 2\cdot 3\sin^2\theta\cos\theta = 3\cos\theta(1 - 2\sin^2\theta)$$ Using $1 - 2\sin^2\theta = \cos 2\theta$, $$\frac{dy}{d\theta} = 3\cos\theta\cos 2\theta$$

Step 3: Find $\frac{dy}{dx}$. $$\frac{dy}{dx} = \frac{3\cos\theta\cos 2\theta}{3\sin\theta\cos 2\theta} = \cot\theta$$

Answer: $\cot\theta$

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Example 10: Combining Parameter and Exponential

Find $\frac{dy}{dx}$ if $x = e^t\cos t$ and $y = e^t\sin t$.

Solution: Step 1: Differentiate $x$ with respect to $t$ using product rule. $$\frac{dx}{dt} = e^t\cos t + e^t(-\sin t) = e^t(\cos t - \sin t)$$

Step 2: Differentiate $y$ with respect to $t$ using product rule. $$\frac{dy}{dt} = e^t\sin t + e^t\cos t = e^t(\sin t + \cos t)$$

Step 3: Find $\frac{dy}{dx}$. $$\frac{dy}{dx} = \frac{e^t(\sin t + \cos t)}{e^t(\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t}$$

Step 4: Optional simplification: $$\frac{\sin t + \cos t}{\cos t - \sin t} = \frac{\tan t + 1}{1 - \tan t} = \tan\left(\frac{\pi}{4} + t\right)$$

Answer: $\frac{\sin t + \cos t}{\cos t - \sin t}$ or $\tan\left(\frac{\pi}{4} + t\right)$