NCERT Solutions Class-12-Chapter-5-Differentiability and Continuity
Excercise-5.6
Note: When \( x \) and \( y \) are given as functions of a parameter \( t \) (i.e., \( x = f(t) \), \( y = g(t) \)), the derivative \( \frac{dy}{dx} \) is found using the formula: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \] provided \( \frac{dx}{dt} \neq 0 \).
Q1
If \( x = 2at^2, y = at^4 \), find \( \frac{dy}{dx} \).
Solution:
Given \( x = 2at^2 \) and \( y = at^4 \).
Differentiating w.r.t. \( t \):
\( \frac{dx}{dt} = 2a(2t) = 4at \)
\( \frac{dy}{dt} = a(4t^3) = 4at^3 \)
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at^3}{4at} = t^2 \]
Q2
If \( x = a \cos \theta, y = b \cos \theta \), find \( \frac{dy}{dx} \).
Solution:
Differentiating w.r.t. \( \theta \):
\( \frac{dx}{d\theta} = -a \sin \theta \)
\( \frac{dy}{d\theta} = -b \sin \theta \)
\[ \frac{dy}{dx} = \frac{-b \sin \theta}{-a \sin \theta} = \frac{b}{a} \]
Q3
If \( x = \sin t, y = \cos 2t \), find \( \frac{dy}{dx} \).
Solution:
\( \frac{dx}{dt} = \cos t \).
\( \frac{dy}{dt} = -\sin 2t \cdot 2 = -2 \sin 2t \).
\[ \frac{dy}{dx} = \frac{-2 \sin 2t}{\cos t} = \frac{-2 (2 \sin t \cos t)}{\cos t} = -4 \sin t \]
Q4
If \( x = 4t, y = \frac{4}{t} \), find \( \frac{dy}{dx} \).
Solution:
\( \frac{dx}{dt} = 4 \).
\( \frac{dy}{dt} = 4(-t^{-2}) = -\frac{4}{t^2} \).
\[ \frac{dy}{dx} = \frac{-4/t^2}{4} = -\frac{1}{t^2} \]
Q5
If \( x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2\theta \), find \( \frac{dy}{dx} \).
Solution:
\( \frac{dx}{d\theta} = -\sin \theta - (-\sin 2\theta \cdot 2) = 2\sin 2\theta - \sin \theta \).
\( \frac{dy}{d\theta} = \cos \theta - (\cos 2\theta \cdot 2) = \cos \theta - 2\cos 2\theta \).
\[ \frac{dy}{dx} = \frac{\cos \theta - 2\cos 2\theta}{2\sin 2\theta - \sin \theta} \]
Q6
If \( x = a(\theta - \sin \theta), y = a(1 + \cos \theta) \), find \( \frac{dy}{dx} \).
Solution:
\( \frac{dx}{d\theta} = a(1 - \cos \theta) \).
\( \frac{dy}{d\theta} = a(-\sin \theta) = -a \sin \theta \).
\[ \frac{dy}{dx} = \frac{-a \sin \theta}{a(1 - \cos \theta)} = \frac{-2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = -\cot(\theta/2) \]
Q7
If \( x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}} \), find \( \frac{dy}{dx} \).
Solution:
This involves differentiating quotient forms w.r.t \( t \).
Step 1: Find \( \frac{dx}{dt} \)
\[ \frac{dx}{dt} = \frac{\sqrt{\cos 2t} \cdot 3\sin^2 t \cos t - \sin^3 t \cdot \frac{1}{2\sqrt{\cos 2t}}(-2\sin 2t)}{\cos 2t} \]
\[ = \frac{3\sin^2 t \cos t \cos 2t + \sin^3 t \sin 2t}{(\cos 2t)^{3/2}} \]
Step 2: Find \( \frac{dy}{dt} \)
\[ \frac{dy}{dt} = \frac{\sqrt{\cos 2t} \cdot 3\cos^2 t (-\sin t) - \cos^3 t \cdot \frac{1}{2\sqrt{\cos 2t}}(-2\sin 2t)}{\cos 2t} \]
\[ = \frac{-3\cos^2 t \sin t \cos 2t + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}} \]
Step 3: Divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \)
\[ \frac{dy}{dx} = \frac{-3\cos^2 t \sin t \cos 2t + \cos^3 t \sin 2t}{3\sin^2 t \cos t \cos 2t + \sin^3 t \sin 2t} \]
Simplifying the numerator and denominator using trigonometric identities (like \( \sin 2t = 2\sin t \cos t \)):
\[ \frac{dy}{dx} = \frac{-\cos 3t}{\sin 3t} = -\cot 3t \]
Q8
If \( x = a(\cos t + \log \tan \frac{t}{2}), y = a \sin t \), find \( \frac{dy}{dx} \).
Solution:
\( \frac{dy}{dt} = a \cos t \).
For \( x \): \( \frac{dx}{dt} = a \left( -\sin t + \frac{1}{\tan(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2} \right) \)
\( \frac{dx}{dt} = a \left( -\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)} \right) = a \left( -\sin t + \frac{1}{\sin t} \right) = a \frac{\cos^2 t}{\sin t} \).
\[ \frac{dy}{dx} = \frac{a \cos t}{a \cos^2 t / \sin t} = \frac{\sin t}{\cos t} = \tan t \]
Q9
If \( x = a \sec \theta, y = b \tan \theta \), find \( \frac{dy}{dx} \).
Solution:
\( \frac{dx}{d\theta} = a \sec \theta \tan \theta \).
\( \frac{dy}{d\theta} = b \sec^2 \theta \).
\[ \frac{dy}{dx} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} = \frac{b \sec \theta}{a \tan \theta} = \frac{b}{a} \frac{1/\cos \theta}{\sin \theta / \cos \theta} = \frac{b}{a} \text{cosec } \theta \]
Q10
If \( x = a(\cos \theta + \theta \sin \theta), y = a(\sin \theta - \theta \cos \theta) \), find \( \frac{dy}{dx} \).
Solution:
\( \frac{dx}{d\theta} = a(-\sin \theta + \sin \theta + \theta \cos \theta) = a\theta \cos \theta \).
\( \frac{dy}{d\theta} = a(\cos \theta - (\cos \theta - \theta \sin \theta)) = a\theta \sin \theta \).
\[ \frac{dy}{dx} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta \]
Q11
If \( x = \sqrt{a^{\sin^{-1} t}}, y = \sqrt{a^{\cos^{-1} t}} \), show that \( \frac{dy}{dx} = -\frac{y}{x} \).
Solution:
Multiply x and y:
\[ xy = \sqrt{a^{\sin^{-1} t}} \cdot \sqrt{a^{\cos^{-1} t}} = \sqrt{a^{\sin^{-1} t + \cos^{-1} t}} \]
Since \( \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \), we have \( xy = \sqrt{a^{\pi/2}} \), which is a constant.
Differentiating \( xy = C \) w.r.t. \( x \):
\[ x \frac{dy}{dx} + y(1) = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \]