Comprehensive solutions for Class 12 Maths (NCERT) Exercise 5.6. Learn how to find the derivative $\frac{dy}{dx}$ for functions defined parametrically, mastering the formula $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ and techniques for simplifying complex trigonometric results.

image 117 Rbse Solutions Class 12 Maths Chapter 5 Exercise 5.6: Parametric Differentiation

For functions defined parametrically by $x = f(t)$ and $y = g(t)$, the derivative $\frac{dy}{dx}$ is found using the formula:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

1. $x = 2at^2$, $y = at^4$

Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$$\frac{dx}{dt} = \frac{d}{dt}(2at^2) = 4at$$$$\frac{dy}{dt} = \frac{d}{dt}(at^4) = 4at^3$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{4at^3}{4at} = t^2$$$$\mathbf{\frac{dy}{dx} = t^2}$$

2. $x = a \cos \theta$, $y = b \cos \theta$

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:$$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta) = -a \sin \theta$$$$\frac{dy}{d\theta} = \frac{d}{d\theta}(b \cos \theta) = -b \sin \theta$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{-b \sin \theta}{-a \sin \theta} = \frac{b}{a}$$$$\mathbf{\frac{dy}{dx} = \frac{b}{a}}$$

3. $x = \sin t$, $y = \cos 2t$

Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$$\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t$$$$\frac{dy}{dt} = \frac{d}{dt}(\cos 2t) = -\sin(2t) \cdot 2 = -2 \sin 2t$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{-2 \sin 2t}{\cos t}$$Using the identity $\sin 2t = 2 \sin t \cos t$:$$\frac{dy}{dx} = \frac{-2 (2 \sin t \cos t)}{\cos t} = -4 \sin t$$$$\mathbf{\frac{dy}{dx} = -4 \sin t}$$

4. $x = 4t$, $y = \frac{4}{t}$

Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$$\frac{dx}{dt} = \frac{d}{dt}(4t) = 4$$$$\frac{dy}{dt} = \frac{d}{dt}(4t^{-1}) = 4(-1)t^{-2} = -\frac{4}{t^2}$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{-\frac{4}{t^2}}{4} = -\frac{1}{t^2}$$$$\mathbf{\frac{dy}{dx} = – \frac{1}{t^2}}$$

5. $x = \cos \theta – \cos 2\theta$, $y = \sin \theta – \sin 2\theta$

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:$$\frac{dx}{d\theta} = -\sin \theta – (-\sin 2\theta) \cdot 2 = 2 \sin 2\theta – \sin \theta$$$$\frac{dy}{d\theta} = \cos \theta – (\cos 2\theta) \cdot 2 = \cos \theta – 2 \cos 2\theta$$
Find $\frac{dy}{dx}$:$$\mathbf{\frac{dy}{dx} = \frac{\cos \theta – 2 \cos 2\theta}{2 \sin 2\theta – \sin \theta}}$$

6. $x = a (\theta – \sin \theta)$, $y = a (1 + \cos \theta)$

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:$$\frac{dx}{d\theta} = a (1 – \cos \theta)$$$$\frac{dy}{d\theta} = a (0 – \sin \theta) = -a \sin \theta$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{-a \sin \theta}{a (1 – \cos \theta)} = – \frac{\sin \theta}{1 – \cos \theta}$$Using half-angle identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 – \cos \theta = 2 \sin^2 \frac{\theta}{2}$:$$\frac{dy}{dx} = – \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}} = – \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = – \cot \frac{\theta}{2}$$$$\mathbf{\frac{dy}{dx} = – \cot \frac{\theta}{2}}$$

7. $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$, $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$

Find $\frac{dy}{dx}$:Notice that the denominator is the same for both $x$ and $y$.$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$Since $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v u’ – u v’}{v^2}$, if $v = \sqrt{\cos 2t}$, the $v^2$ term in the denominator of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ will cancel out.$$\frac{dy}{dt} = \frac{\sqrt{\cos 2t} \frac{d}{dt}(\cos^3 t) – \cos^3 t \frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t}$$$$\frac{dx}{dt} = \frac{\sqrt{\cos 2t} \frac{d}{dt}(\sin^3 t) – \sin^3 t \frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t}$$$$\frac{dy}{dx} = \frac{\sqrt{\cos 2t} \frac{d}{dt}(\cos^3 t) – \cos^3 t \frac{d}{dt}(\sqrt{\cos 2t})}{\sqrt{\cos 2t} \frac{d}{dt}(\sin^3 t) – \sin^3 t \frac{d}{dt}(\sqrt{\cos 2t})}$$
Calculate the derivatives of the parts:
- $\frac{d}{dt}(\cos^3 t) = 3 \cos^2 t (-\sin t)$
- $\frac{d}{dt}(\sin^3 t) = 3 \sin^2 t (\cos t)$
- $\frac{d}{dt}(\sqrt{\cos 2t}) = \frac{1}{2\sqrt{\cos 2t}} (-\sin 2t)(2) = -\frac{\sin 2t}{\sqrt{\cos 2t}}$
Substitute into $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{\sqrt{\cos 2t} (3 \cos^2 t (-\sin t)) – \cos^3 t (-\frac{\sin 2t}{\sqrt{\cos 2t}})}{\sqrt{\cos 2t} (3 \sin^2 t (\cos t)) – \sin^3 t (-\frac{\sin 2t}{\sqrt{\cos 2t}})}$$Multiply the numerator and the denominator by $\sqrt{\cos 2t}$ to clear the internal fractions:$$\frac{dy}{dx} = \frac{(\cos 2t) (3 \cos^2 t (-\sin t)) – \cos^3 t (-\sin 2t)}{(\cos 2t) (3 \sin^2 t (\cos t)) – \sin^3 t (-\sin 2t)}$$Factor out $\sin t \cos t$ from both the numerator and denominator (since $\sin 2t = 2 \sin t \cos t$):$$\frac{dy}{dx} = \frac{-3 \cos 2t \cos t + 2 \cos^3 t}{3 \cos 2t \sin t + 2 \sin^3 t} \cdot \frac{\sin t \cos t}{\sin t \cos t}$$$$\frac{dy}{dx} = \frac{\cos t (-3 \cos 2t + 2 \cos^2 t)}{\sin t (3 \cos 2t + 2 \sin^2 t)}$$Using $\cos 2t = 2 \cos^2 t – 1$ (Numerator):$$-3(2 \cos^2 t – 1) + 2 \cos^2 t = -6 \cos^2 t + 3 + 2 \cos^2 t = 3 – 4 \cos^2 t$$Using $\cos 2t = 1 – 2 \sin^2 t$ (Denominator):$$3(1 – 2 \sin^2 t) + 2 \sin^2 t = 3 – 6 \sin^2 t + 2 \sin^2 t = 3 – 4 \sin^2 t$$$$\frac{dy}{dx} = \frac{\cos t (3 – 4 \cos^2 t)}{\sin t (3 – 4 \sin^2 t)} = \frac{\cos t (-\cos 3t / \cos t)}{\sin t (\sin 3t / \sin t)}$$Wait, this trigonometric identity manipulation is getting complicated and error-prone. Let’s use the simplest identity for $\cos 3t$ and $\sin 3t$:
- $\cos 3t = 4 \cos^3 t – 3 \cos t \implies 3 \cos t – 4 \cos^3 t = -\cos 3t$
- $\sin 3t = 3 \sin t – 4 \sin^3 t$
Let’s go back to the expression before factoring $\sin t \cos t$:$$\frac{dy}{dx} = \frac{\cos t [-3 \cos 2t + 2 \cos^2 t]}{\sin t [3 \cos 2t + 2 \sin^2 t]}$$Let’s use $\cos 2t = \cos^2 t – \sin^2 t$:$$\text{Numerator part: } \cos t [-3(\cos^2 t – \sin^2 t) + 2 \cos^2 t]$$$$= \cos t [-3 \cos^2 t + 3 \sin^2 t + 2 \cos^2 t] = \cos t [3 \sin^2 t – \cos^2 t]$$$$\text{Denominator part: } \sin t [3(\cos^2 t – \sin^2 t) + 2 \sin^2 t]$$$$= \sin t [3 \cos^2 t – 3 \sin^2 t + 2 \sin^2 t] = \sin t [3 \cos^2 t – \sin^2 t]$$$$\frac{dy}{dx} = \frac{\cos t [3 \sin^2 t – \cos^2 t]}{\sin t [3 \cos^2 t – \sin^2 t]}$$$$\frac{dy}{dx} = \frac{\cos t (3 (1 – \cos^2 t) – \cos^2 t)}{\sin t (3 \cos^2 t – (1 – \cos^2 t))} = \frac{\cos t (3 – 4 \cos^2 t)}{\sin t (4 \cos^2 t – 1)}$$Now, use the triple angle formulas:$$3 – 4 \cos^2 t = -(4 \cos^2 t – 3) = -\frac{\cos 3t}{\cos t}$$$$4 \cos^2 t – 1 = 4(1 – \sin^2 t) – 1 = 3 – 4 \sin^2 t = \frac{\sin 3t}{\sin t}$$$$\frac{dy}{dx} = \frac{\cos t \cdot (-\frac{\cos 3t}{\cos t})}{\sin t \cdot (\frac{\sin 3t}{\sin t})} = \frac{-\cos 3t}{\sin 3t} = -\cot 3t$$$$\mathbf{\frac{dy}{dx} = – \cot 3t}$$

8. $x = a \left(\cos t + \log \tan \frac{t}{2}\right)$, $y = a \sin t$

Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$$\frac{dx}{dt} = a \left[ -\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2} \right]$$Simplify the $\log$ part:$$\frac{1}{\tan \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2} = \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \cdot \frac{1}{\cos^2 \frac{t}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}} = \frac{1}{\sin t}$$So, $\frac{dx}{dt} = a \left[ -\sin t + \frac{1}{\sin t} \right] = a \left[ \frac{1 – \sin^2 t}{\sin t} \right] = a \frac{\cos^2 t}{\sin t}$$$\frac{dy}{dt} = a \cos t$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} = \frac{\cos t \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = \tan t$$$$\mathbf{\frac{dy}{dx} = \tan t}$$

9. $x = a \sec \theta$, $y = b \tan \theta$

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:$$\frac{dx}{d\theta} = a (\sec \theta \tan \theta)$$$$\frac{dy}{d\theta} = b (\sec^2 \theta)$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} = \frac{b}{a} \frac{\sec \theta}{\tan \theta}$$Using $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:$$\frac{\sec \theta}{\tan \theta} = \frac{1/\cos \theta}{\sin \theta / \cos \theta} = \frac{1}{\sin \theta} = \csc \theta$$$$\mathbf{\frac{dy}{dx} = \frac{b}{a} \csc \theta}$$

10. $x = a (\cos \theta + \theta \sin \theta)$, $y = a (\sin \theta – \theta \cos \theta)$

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:Use the Product Rule for $\theta \sin \theta$ and $\theta \cos \theta$.$$\frac{dx}{d\theta} = a \left[ -\sin \theta + (1 \cdot \sin \theta + \theta \cdot \cos \theta) \right]$$$$\frac{dx}{d\theta} = a \left[ -\sin \theta + \sin \theta + \theta \cos \theta \right] = a \theta \cos \theta$$$$\frac{dy}{d\theta} = a \left[ \cos \theta – (1 \cdot \cos \theta + \theta \cdot (-\sin \theta)) \right]$$$$\frac{dy}{d\theta} = a \left[ \cos \theta – \cos \theta + \theta \sin \theta \right] = a \theta \sin \theta$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$$$$\mathbf{\frac{dy}{dx} = \tan \theta}$$

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

Given: $x = (a^{\sin^{-1} t})^{1/2}$ and $y = (a^{\cos^{-1} t})^{1/2}$.

Method 1: Direct Parametric Differentiation (Longer)

Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ (using Chain Rule and Logarithmic Differentiation for $a^u$):Recall $\frac{d}{dt}(a^u) = a^u \log a \frac{du}{dt}$.$$\frac{dx}{dt} = \frac{1}{2\sqrt{a^{\sin^{-1} t}}} \cdot \frac{d}{dt}(a^{\sin^{-1} t})$$$$\frac{dx}{dt} = \frac{1}{2x} \cdot a^{\sin^{-1} t} \cdot \log a \cdot \frac{1}{\sqrt{1 – t^2}}$$Since $a^{\sin^{-1} t} = x^2$:$$\frac{dx}{dt} = \frac{1}{2x} \cdot x^2 \cdot \log a \cdot \frac{1}{\sqrt{1 – t^2}} = \frac{x \log a}{2 \sqrt{1 – t^2}}$$Similarly for $y$:$$\frac{dy}{dt} = \frac{1}{2y} \cdot a^{\cos^{-1} t} \cdot \log a \cdot \left(-\frac{1}{\sqrt{1 – t^2}}\right) = -\frac{y \log a}{2 \sqrt{1 – t^2}}$$
Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{-\frac{y \log a}{2 \sqrt{1 – t^2}}}{\frac{x \log a}{2 \sqrt{1 – t^2}}} = -\frac{y}{x}$$$$\mathbf{\frac{dy}{dx} = -\frac{y}{x}}$$

Method 2: Elimination (Simpler)

Square both equations:

$$x^2 = a^{\sin^{-1} t}$$

$$y^2 = a^{\cos^{-1} t}$$

Multiply the equations:

$$x^2 y^2 = a^{\sin^{-1} t} \cdot a^{\cos^{-1} t} = a^{\sin^{-1} t + \cos^{-1} t}$$

Using the identity $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$:

$$x^2 y^2 = a^{\pi/2}$$

Since $a^{\pi/2}$ is a constant, let $C = a^{\pi/2}$.

$$x^2 y^2 = C$$

Differentiate implicitly w.r.t. $x$:

$$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(C)$$