NCERT Class 12 Maths Continuity and Differentiability Ex 5.6

Understanding Derivatives of Functions in Parametric Form

When the relationship between two variables $x$ and $y$ is expressed through a third independent variable, say $t$ or $\theta$, this third variable is called a parameter. Equations of the form $x = f(t)$ and $y = g(t)$ are termed parametric equations.

To find the derivative $\frac{dy}{dx}$ without explicitly eliminating the parameter $t$, we employ the Chain Rule of differentiation. According to the Chain Rule:
$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

Rearranging this identity gives us the core formula for parametric differentiation:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \quad \text{provided} \quad \frac{dx}{dt} \neq 0$

Similarly, if the parameter is an angle $\theta$, the derivative is expressed as:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \quad \text{provided} \quad \frac{dx}{d\theta} \neq 0$

This elegant method allows us to find the instantaneous rate of change of $y$ with respect to $x$ directly from their individual rates of change with respect to the parameter $t$.

Step-by-Step Process to Solve Parametric Differentiation Problems

1. Identify the Parameter — Examine the given equations $x = f(t)$ and $y = g(t)$ to identify the parameter (commonly $t$ or $\theta$).
2. Differentiate Separately — Differentiate $x$ with respect to the parameter to find $\frac{dx}{dt}$. Next, differentiate $y$ with respect to the parameter to find $\frac{dy}{dt}$.
3. Apply the Division Formula — Divide the derivative of $y$ by the derivative of $x$: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
4. Simplify the Expression — Use algebraic manipulation and trigonometric identities to simplify the resulting fraction into its most compact form.

Fully Worked Out Master Examples

• Example 1: Find $\frac{dy}{dx}$ if $x = a(\theta - \sin\theta)$ and $y = a(1 - \cos\theta)$. Step 1: Differentiate $x$ with respect to $\theta$. $\frac{dx}{d\theta} = a(1 - \cos\theta)$ Step 2: Differentiate $y$ with respect to $\theta$. $\frac{dy}{d\theta} = a(0 - (-\sin\theta)) = a\sin\theta$ Step 3: Apply the parametric differentiation formula. $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\sin\theta}{a(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta}$ Step 4: Simplify using trigonometric identities. Recall that $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ and $1 - \cos\theta = 2\sin^2(\theta/2)$. $\frac{dy}{dx} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = \cot\left(\frac{\theta}{2}\right)$
• Example 2: Show that if $x = \sqrt{a^{\sin^{-1}t}}$ and $y = \sqrt{a^{\cos^{-1}t}}$, then $\frac{dy}{dx} = -\frac{y}{x}$. Step 1: Multiply the two equations to eliminate the parameter easily. $x \cdot y = \sqrt{a^{\sin^{-1}t}} \cdot \sqrt{a^{\cos^{-1}t}} = \sqrt{a^{\sin^{-1}t + \cos^{-1}t}}$ Step 2: Use the inverse trigonometric identity $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$. $xy = \sqrt{a^{\pi/2}}$ Since $a$ and $\pi/2$ are constants, the right-hand side is a constant, say $C$. $xy = C$ Step 3: Differentiate implicitly with respect to $x$. Using the product rule: $x\frac{dy}{dx} + y(1) = 0$ $x\frac{dy}{dx} = -y \implies \frac{dy}{dx} = -\frac{y}{x}$ (Note: This is a highly efficient board-exam trick instead of differentiating both parameters separately!)

Critical Exam Trap: Second-Order Derivatives of Parametric Functions

A very common error in CBSE board exams occurs when students are asked to calculate the second-order derivative $\frac{d^2y}{dx^2}$ for parametric equations.

The Trap: Students mistakenly calculate $\frac{d^2y}{dx^2}$ as $\frac{d^2y/dt^2}{d^2x/dt^2}$. This is mathematically incorrect!

The Correct Method:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$
Always remember to multiply by the factor $\frac{dt}{dx}$ (which is the reciprocal of $\frac{dx}{dt}$) at the end of your second derivative calculation.

Practice Questions with Solutions

• Q: Find $\frac{dy}{dx}$ if $x = 2at^2$ and $y = at^4$. A: Step 1: Differentiate $x$ with respect to $t$. $\frac{dx}{dt} = \frac{d}{dt}(2at^2) = 4at$ Step 2: Differentiate $y$ with respect to $t$. $\frac{dy}{dt} = \frac{d}{dt}(at^4) = 4at^3$ Step 3: Apply the formula for parametric differentiation. $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4at^3}{4at}$ Step 4: Simplify the expression. $\frac{dy}{dx} = t^2$ Final answer: $\frac{dy}{dx} = t^2$
• Q: If $x = a(\cos t + t\sin t)$ and $y = a(\sin t - t\cos t)$, find $\frac{dy}{dx}$. A: Step 1: Differentiate $x$ with respect to $t$ using the Product Rule for the term $t\sin t$. $\frac{dx}{dt} = a\left(-\sin t + \sin t + t\cos t\right) = at\cos t$ Step 2: Differentiate $y$ with respect to $t$ using the Product Rule for $t\cos t$. $\frac{dy}{dt} = a\left(\cos t - (\cos t - t\sin t)\right) = at\sin t$ Step 3: Calculate $\frac{dy}{dx}$. $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at\cos t} = \tan t$ Final answer: $\frac{dy}{dx} = \tan t$
• Q: Find $\frac{dy}{dx}$ if $x = a\cos^3 \theta$ and $y = a\sin^3 \theta$. A: Step 1: Differentiate $x$ with respect to $\theta$ using the Power and Chain Rule. $\frac{dx}{d\theta} = 3a\cos^2\theta \cdot (-\sin\theta) = -3a\cos^2\theta\sin\theta$ Step 2: Differentiate $y$ with respect to $\theta$. $\frac{dy}{d\theta} = 3a\sin^2\theta \cdot \cos\theta = 3a\sin^2\theta\cos\theta$ Step 3: Divide the derivatives. $\frac{dy}{dx} = \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}$ Step 4: Cancel common terms. $\frac{dy}{dx} = -\frac{\sin\theta}{\cos\theta} = -\tan\theta$ Final answer: $\frac{dy}{dx} = -\tan\theta$
• Q: Find the value of $\frac{dy}{dx}$ at $\theta = \frac{\pi}{4}$ if $x = a\sec\theta$ and $y = b\tan\theta$. A: Step 1: Find $\frac{dx}{d\theta}$. $\frac{dx}{d\theta} = a\sec\theta\tan\theta$ Step 2: Find $\frac{dy}{d\theta}$. $\frac{dy}{d\theta} = b\sec^2\theta$ Step 3: Apply the division formula. $\frac{dy}{dx} = \frac{b\sec^2\theta}{a\sec\theta\tan\theta} = \frac{b\sec\theta}{a\tan\theta} = \frac{b}{a} \cdot \frac{1/\cos\theta}{\sin\theta/\cos\theta} = \frac{b}{a\sin\theta} = \frac{b}{a}\csc\theta$ Step 4: Substitute $\theta = \frac{\pi}{4}$. $\left.\frac{dy}{dx}\right|_{\theta = \pi/4} = \frac{b}{a}\csc\left(\frac{\pi}{4}\right) = \frac{b}{a}\sqrt{2} = \frac{\sqrt{2}b}{a}$ Final answer: $\frac{\sqrt{2}b}{a}$

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