Exercise 5.6 Class 12 Maths Continuity and Differentiability

NCERT Solutions Class 12 Maths Chapter 5 Ex 5.6 Detailed | LearnCBSEHub

💡 Golden Rule

If $x$ and $y$ are functions of a parameter $t$ (or $\theta$), then:

Formula: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Question 01

If $x = 2at^2, y = at^4$, find $\frac{dy}{dx}$.

Step 1: Differentiate w.r.t t

$$\frac{dx}{dt} = \frac{d}{dt}(2at^2) = 4at$$ $$\frac{dy}{dt} = \frac{d}{dt}(at^4) = 4at^3$$

Step 2: Divide

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at^3}{4at} = t^2$$

Answer: $t^2$

Question 02

If $x = a \cos\theta, y = b \cos\theta$, find $\frac{dy}{dx}$.

Step 1: Differentiate w.r.t $\theta$

$$\frac{dx}{d\theta} = -a \sin\theta$$ $$\frac{dy}{d\theta} = -b \sin\theta$$

Step 2: Divide

Answer: $\frac{-b \sin\theta}{-a \sin\theta} = \frac{b}{a}$

Question 03

If $x = \sin t, y = \cos 2t$, find $\frac{dy}{dx}$.

Step 1: Differentiate

$$\frac{dx}{dt} = \cos t$$ $$\frac{dy}{dt} = -\sin 2t \cdot 2 = -2 \sin 2t$$

Step 2: Simplify

$$\frac{dy}{dx} = \frac{-2 \sin 2t}{\cos t} = \frac{-2(2\sin t \cos t)}{\cos t} = -4 \sin t$$

Answer: $-4 \sin t$

Question 04

If $x = 4t, y = \frac{4}{t}$, find $\frac{dy}{dx}$.

Step 1: Differentiate

$$\frac{dx}{dt} = 4$$ $$\frac{dy}{dt} = 4(-1)t^{-2} = -\frac{4}{t^2}$$

Answer: $\frac{-4/t^2}{4} = -\frac{1}{t^2}$

Question 05

If $x = \cos\theta – \cos 2\theta, y = \sin\theta – \sin 2\theta$, find $\frac{dy}{dx}$.

Step 1: Differentiate w.r.t $\theta$

$$\frac{dx}{d\theta} = -\sin\theta – (-2\sin 2\theta) = 2\sin 2\theta – \sin\theta$$ $$\frac{dy}{d\theta} = \cos\theta – 2\cos 2\theta$$

Answer: $\frac{\cos\theta – 2\cos 2\theta}{2\sin 2\theta – \sin\theta}$

Question 06

If $x = a(\theta – \sin\theta), y = a(1 + \cos\theta)$, find $\frac{dy}{dx}$.

Step 1: Differentiate

$$\frac{dx}{d\theta} = a(1 – \cos\theta)$$ $$\frac{dy}{d\theta} = a(-\sin\theta) = -a \sin\theta$$

Step 2: Half Angle Formulas

$$\frac{dy}{dx} = \frac{-a \sin\theta}{a(1-\cos\theta)} = \frac{-2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)}$$

Answer: $-\cot(\theta/2)$

Question 07

If $x = \frac{3\sin t}{\cos^2 t}, y = \frac{3\cos t}{\cos^2 t}$, find $\frac{dy}{dx}$.

Step 1: Simplify Expressions

$x = 3 \tan t \sec t, \quad y = 3 \sec t$

Step 2: Differentiate

$$\frac{dx}{dt} = 3(\sec t \tan^2 t + \sec^3 t)$$ $$\frac{dy}{dt} = 3 \sec t \tan t$$

Step 3: Divide

$$\frac{dy}{dx} = \frac{3 \sec t \tan t}{3 \sec t (\tan^2 t + \sec^2 t)}$$

Answer: $\frac{\tan t}{\tan^2 t + \sec^2 t}$

Question 08

If $x = a(t + \log \tan \frac{t}{2}), y = a \sin t$, find $\frac{dy}{dx}$.

Step 1: Differentiate x

$$\frac{dx}{dt} = a\left(1 + \frac{1}{\tan(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}\right)$$ $$= a(1 + \csc t) = a\left(\frac{\sin t + 1}{\sin t}\right)$$

Step 2: Differentiate y

$\frac{dy}{dt} = a \cos t$

Answer: $\frac{a \cos t}{a(1+\csc t)} = \frac{\cos t}{1+\csc t}$

Question 09

If $x = a \sec\theta, y = b \tan\theta$, find $\frac{dy}{dx}$.

Step 1: Differentiate

$$\frac{dx}{d\theta} = a \sec\theta \tan\theta$$ $$\frac{dy}{d\theta} = b \sec^2\theta$$

Answer: $\frac{b \sec^2\theta}{a \sec\theta \tan\theta} = \frac{b}{a} \csc\theta$

Question 10

If $x = a(\cos\theta + \theta\sin\theta), y = a(\sin\theta – \theta\cos\theta)$, find $\frac{dy}{dx}$.

Step 1: Product Rule

$$\frac{dx}{d\theta} = a(-\sin\theta + \sin\theta + \theta\cos\theta) = a\theta\cos\theta$$ $$\frac{dy}{d\theta} = a(\cos\theta – (\cos\theta – \theta\sin\theta)) = a\theta\sin\theta$$

Answer: $\frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta$

Question 11

If $x = a \sin t, y = a \cos t$, find $\frac{dy}{dx}$.

Step 1: Differentiate

$$\frac{dx}{dt} = a \cos t$$ $$\frac{dy}{dt} = -a \sin t$$

Step 2: Substitute x and y

We know $\tan t = \frac{\sin t}{\cos t} = \frac{x/a}{y/a} = \frac{x}{y}$

Answer: $-\tan t = -\frac{x}{y}$