NCERT Solutions for Exercise 5.6 Class 12 Maths Chapter 5 - Continuity and Differentiability

Question:2 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

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

Answer:

Given equations are
$x= a \cos \theta , y = b \cos \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\cos \theta)}{d\theta}= -a\sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\cos \theta)}{d\theta}= -b\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-b\sin \theta}{-a\sin \theta} = \frac{b}{a}$
Therefore, answer is $\frac{dy}{dx}= \frac{b}{a}$

Question:3 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx . $x = \sin t , y = \cos 2 t$

Answer:

Given equations are
$x = \sin t , y = \cos 2 t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t$
Similarly,
$\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t \ \ \ \ \ (\because \sin 2x = \sin x\cos x)$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t$
Therefore, the answer is $\frac{dy}{dx} = -4\sin t$

Question:4 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx

$x = 4t , y = 4/t$

Answer:

Given equations are
$x = 4t , y = 4/t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(4 t)}{dt}= 4$
Similarly,
$\frac{dy}{dt}=\frac{d(\frac{4}{t})}{dt}= \frac{-4}{t^2}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{ \frac{-4}{t^2} }{4} = \frac{-1}{t^2}$
Therefore, the answer is $\frac{dy}{dx} = \frac{-1}{t^2}$

Question:5 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta$

Answer:

Given equations are
$x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$
Therefore, answer is $\frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$

Question:7 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$

Answer:

Given equations are

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

Now, differentiate both w.r.t $t$:

$\frac{dx}{dt} = \frac{d\left( \frac{\sin^3 t}{\sqrt{\cos 2t}} \right)}{dt} = \frac{\sqrt{\cos 2t} \cdot \frac{d(\sin^3 t)}{dt} - \sin^3 t \cdot \frac{d(\sqrt{\cos 2t})}{dt}}{(\sqrt{\cos 2t})^2}$

$= \frac{3\sin^2 t \cos t \cdot \sqrt{\cos 2t} - \sin^3 t \cdot \frac{1}{2\sqrt{\cos 2t}} \cdot (-2\sin 2t)}{\cos 2t}$

$= \frac{3\sin^2 t \cos t \cdot \cos 2t + \sin^3 t \sin 2t}{\cos 2t \sqrt{\cos 2t}}$

$= \frac{\sin^3 t \sin 2t \left( 3 \cot t \cot 2t + 1 \right)}{\cos 2t \sqrt{\cos 2t}} \quad (\because \frac{\cos x}{\sin x} = \cot x)$

Similarly,

$\frac{dy}{dt} = \frac{d\left( \frac{\cos^3 t}{\sqrt{\cos 2t}} \right)}{dt} = \frac{\sqrt{\cos 2t} \cdot \frac{d(\cos^3 t)}{dt} - \cos^3 t \cdot \frac{d(\sqrt{\cos 2t})}{dt}}{(\sqrt{\cos 2t})^2}$

$= \frac{3\cos^2 t (-\sin t) \cdot \sqrt{\cos 2t} - \cos^3 t \cdot \frac{1}{2\sqrt{\cos 2t}} \cdot (-2\sin 2t)}{(\sqrt{\cos 2t})^2}$

$= \frac{-3\cos^2 t \sin t \cos 2t + \cos^3 t \sin 2t}{\cos 2t \sqrt{\cos 2t}}$

$= \frac{\sin 2t \cos^3 t \left( 1 - 3 \tan t \cot 2t \right)}{\cos 2t \sqrt{\cos 2t}}$

Now, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{\sin 2t \cos^3 t \left( 1 - 3 \tan t \cot 2t \right)}{\cos 2t \sqrt{\cos 2t}}}{\frac{\sin^3 t \sin 2t \left( 3 \cot t \cot 2t + 1 \right)}{\cos 2t \sqrt{\cos 2t}}}$

$= \frac{\cot^3 t \left( 1 - 3 \tan t \cot 2t \right)}{3 \cot t \cot 2t + 1}$

$= \frac{\cos^3 t \left( 1 - 3 \frac{\sin t}{\cos t} \frac{\cos 2t}{\sin 2t} \right)}{\sin^3 t \left( 3 \frac{\cos t}{\sin t} \frac{\cos 2t}{\sin 2t} + 1 \right)}$

$= \frac{\cos^2 t \left( \cos t \sin 2t - 3 \sin t \cos 2t \right)}{\sin^2 t \left( 3 \cos t \cos 2t + \sin t \sin 2t \right)}$

$= \frac{\cos^2 t \left( 2 \sin t \cos^2 t - 3 \sin t \cos^2 t + 3 \sin t \right)}{\sin^2 t \left( 3 \cos t - 6 \cos t \sin^2 t + 2 \sin^2 \cos t \right)}$

$= \frac{\sin t \cos t \left( -4 \cos^3 t + 3 \cos t \right)}{\sin t \cos t \left( 3 \sin t - 4 \sin^3 t \right)}$

$\frac{dy}{dx} = \frac{-4 \cos^3 t + 3 \cos t}{3 \sin t - 4 \sin^3 t} = \frac{-\cos 3t}{\sin 3t} = -\cot 3t \quad (\because \sin 3t = 3 \sin t - 4 \sin^3 t \text{ and } \cos 3t = 4 \cos^3 t - 3 \cos t)$
Therefore, the answer is $\frac{dy}{dx} = -\cot 3t$

Question:8 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = a ( \cos t + \log \tan t/2 ),y = a \sin t$

Answer:

Given equations are
$x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})$
$= a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})$
$=a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})$
Similarly,
$\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t$
Therefore, the answer is $\frac{dy}{dx} = \tan t$

Question:9 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = a \sec \theta , y = b \ tan \theta$

Answer:

Given equations are
$x = a \sec \theta , y = b \ tan \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}$
Therefore, the answer is $\frac{dy}{dx} = \frac{b cosec \theta}{a}$

Question:10 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )$

Answer:

Given equations are
$x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta$
Therefore, the answer is $\frac{dy}{dx}= \tan \theta$

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

Answer:

Given equations are
$x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$

$xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\ sin^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c$

differentiating with respect to x

$x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x}$