NCERT Class 12 Maths Continuity and Differentiability Ex 5.6

Welcome to your comprehensive guide on continuity and differentiability ex 5.6 class 12 ncert. In previous exercises, you learned how to differentiate explicit functions where $y$ is directly written in terms of $x$, and implicit functions where $x$ and $y$ are mixed. However, in many physical and geometric scenarios, variables $x$ and $y$ are defined separately in terms of a third helper variable, known as a parameter. For instance, the coordinates of a moving projectile are often expressed as functions of time $t$. This exercise is dedicated to mastering the Chain Rule technique for differentiating these parametric equations. By understanding how to calculate $\frac{dy}{dx}$ when $x = f(t)$ and $y = g(t)$, you will build a solid foundation for finding rates of change and tangent slopes in engineering and physics. Let's dive deep into the concepts, step-by-step proofs, and essential board-level practice questions.

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: $\frac{\sqrt{2}b}{a}$

Frequently Asked Questions

What is a parameter in CBSE Class 12 mathematics?

A parameter is an auxiliary third variable (such as $t$ or $\theta$) that defines the relation between independent variable $x$ and dependent variable $y$ via separate equations. For example, in $x = a\cos t$ and $y = a\sin t$, $t$ is the parameter.

Can we find dy/dx without differentiating the parameter separately?

Yes, if we can easily eliminate the parameter to form an explicit or implicit equation in terms of $x$ and $y$, we can differentiate directly. However, in many cases, eliminating the parameter is algebraically tedious, making parametric differentiation much more practical.

Why must dx/dt not be equal to zero when finding dy/dx?

Since we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to obtain $\frac{dy}{dx}$, a value of $\frac{dx}{dt} = 0$ would lead to division by zero, making the derivative undefined at that point.