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

NCERT Solutions for Class 12 Maths Exercise 5.6 of Chapter 5 Continuity and Differentiability will clear your doubts on the concept of Continuity and Differentiability that help you in preparing your worksheet, doing your homework, and class assignments of maths.NCERT Solutions for Class 12 Maths Exercise 5.6 of Chapter 5 Continuity and Differentiability are created by Future Study Point for boosting your preparation of the CBSE Board Maths question paper 2021-22.NCERT Solutions of Maths class 12 are the best study materials for clearing your concept on Continuity and Differentiability.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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

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If x and y are connected parametrical by the equations given in the exercise 1 to 10 ,without eliminating the parameter, find dy/dx.

Q1. x = 2at², y = at⁴

Ans. The functions are x = 2at², y = at⁴

Taking x = 2at²

Differentiating the function with respect to t

$\frac{dx}{dt}=2a\frac{d}{dt}t^{2}$

$\frac{dx}{dt}=2a.2t$

$\frac{dx}{dt}=4at....\left ( i \right )$

Taking y = at⁴

Differentiating the function with respect to t

$\frac{dy}{dt}=a\frac{d}{dt}t^{4}$

$\frac{dy}{dt}=a.4t^{3}$

$\frac{dy}{dt}=4at^{3}....\left ( ii \right )$

Dividing equation (ii) by the equation (i)

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

$\frac{dy}{dx}=t^{2}$

Q2.x = acosθ, y = bcosθ

Ans. Taking x = acosθ

Differentiating the given function with respect to θ

$\frac{dx}{d\Theta }=a\frac{d}{d\Theta }cos\Theta$

$\frac{dx}{d\Theta }=a.\left ( -sin\Theta \right )$

$\frac{dx}{d\Theta }=-a.sin\Theta ...\left ( i \right )$

Taking y = bcosθ

Differentiating the given function with respect to θ

$\frac{dy}{d\Theta }=b.\frac{d}{d\Theta }cos\Theta$

$\frac{dy}{d\Theta }=b.\left ( -sin\Theta \right )$

$\frac{dy}{d\Theta }=-b.sin\Theta ....\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{-bsin\Theta }{-asin\Theta }$

$\frac{dy }{dx }=\frac{b}{a}$

Q3. x = sin t, y = cos 2t

Ans. Taking x = sin t

Differentiating the given function with respect to t

$\frac{dx }{dt }=\frac{d}{dt}sin\: t$

$\frac{dx }{dt }=cos\: t...\left ( i \right )$

Taking y = cos 2t

$\frac{dy }{dt }=\frac{d}{dt}cos\: 2t$

$\frac{dy }{dt }=-sin\: 2t\frac{d}{dt} 2t$

$\frac{dy }{dt }=-sin\: 2t.2$

$\frac{dy }{dt }=-2sin\: 2t.....\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/dt }{dx/dt }=\frac{-2sin2t}{cost}$

$\frac{dy}{dx }=\frac{-2sin2t}{cost}$

$\frac{dy}{dx }=\frac{-2\times 2sint.cost}{cost}$

$\frac{dy}{dx }=-4sin\: t$

$Q4.x= 4t,y=\frac{4}{t}$

Ans. Taking x = 4t

$\frac{dx}{dt}=4...\left ( i \right )$

Taking

$y=\frac{4}{t}$

$\frac{dy}{dt}= 4\frac{d}{dt}\frac{1}{t}$

$\frac{dy}{dt}= 4\frac{d}{dt}t^{-1}$

$\frac{dy}{dt}= -4t^{-2}$

$\frac{dy}{dt}= -\frac{4}{t^{2}}$ …….(ii)

Dividing equation (ii) by the equation (i)

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

$\frac{dy}{dx}=\frac{-4/t^{2}}{4}$

$\frac{dy}{dx}=\frac{-1}{t^{2}}$

Q5. x = cos θ – cos 2θ, y =sin θ – sin 2θ

Ans. The given functions are x = cos θ – cos 2θ, y =sin θ – sin 2θ

Taking x = cos θ – cos 2θ

$\frac{dx}{d\Theta }=\frac{d}{d\Theta }cos\Theta -\frac{d}{d\Theta }cos2\Theta$

$\frac{dx}{d\Theta }=-sin\Theta +sin2\Theta .\frac{d}{d\Theta }2\Theta$

$\frac{dx}{d\Theta }=-sin\Theta +sin2\Theta .2$

$\frac{dx}{d\Theta }=-sin\Theta +2sin2\Theta...\left ( i \right )$

Taking y =sin θ – sin 2θ

$\frac{dy}{d\Theta }=\frac{d}{d\Theta }sin\Theta -\frac{d}{d\Theta }sin2\Theta \frac{}{}$

$\frac{dy}{d\Theta }=cos\Theta -cos2\Theta .2$

$\frac{dy}{d\Theta }=cos\Theta -2cos2\Theta ...\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{cos\Theta -2cos2\Theta }{-sin\Theta +2sin2\Theta }$

$\frac{dy}{dx }=\frac{cos\Theta -2cos2\Theta }{2sin2\Theta -sin\Theta }$

Q6. x = a(θ – sin θ), y = a(1 + cos θ)

Ans.The given functions are x = a(θ – sin θ), y = a(1 + cos θ)

Taking x = a(θ – sin θ)

Differentiating the function with respect to x

$\frac{dx}{d\Theta }=\frac{d}{d\Theta }\left [a\left ( \Theta -sin\Theta \right ) \right ]$

$=a-a\frac{d}{d\Theta }sin\Theta$

$\frac{dx}{d\Theta }=a-a.cos\Theta ...\left ( i \right )$

Taking y = a(1 + cos θ)

Differentiating the function with respect to x

$\frac{dy}{d\Theta }=\frac{d}{d\Theta }\left [ a \left ( 1+cos\Theta \right ) \right ]$

$=0+a\frac{d}{d\Theta }cos\Theta$

$=a\left ( -sin\Theta \right )$

$\frac{dy}{d\Theta }=-a.sin\Theta ...\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{-asin\Theta }{a\left ( 1-cos\Theta \right )}$

$\frac{dy }{dx}=\frac{-asin\Theta }{a\left ( 1-cos\Theta \right )}$

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{-sin\Theta }{\left ( 1-cos\Theta \right )}$

$\frac{dy }{dx}=\frac{-2sin\frac{\Theta }{2}cos\frac{\Theta }{2} }{2sin^{2}\frac{\Theta }{2}}$

$\frac{dy }{dx}=-\frac{cos\frac{\Theta }{2}}{sin\frac{\Theta }{2}}=-cot\frac{\Theta }{2}$

$Q7.x=\frac{sin^{3}t}{\sqrt{cos2t}},y=\frac{cos^{3}t}{\sqrt{cos2t}}$

Ans. The given functions are

$x=\frac{sin^{3}t}{\sqrt{cos2t}},y=\frac{cos^{3}t}{\sqrt{cos2t}}$

Taking

$x=\frac{sin^{3}t}{\sqrt{cos2t}}$

Differentiating the function with respect to t

$\frac{dx}{dt}=\frac{d}{dt}\left (\frac{sin^{3}t}{\sqrt{cos2t}} \right )$

$=\frac{\sqrt{cos2t}\frac{d}{dt}sin^{3}t-sin^{3}t\frac{d}{dt}\sqrt{cos2t}}{\left ( \sqrt{cos2t} \right )^{2}}$

$=\frac{\sqrt{cos2t}.3sin^{2}t-sin^{3}t.\frac{1}{2}\left ( cos2t \right )^{-1/2}.\frac{d}{dt}cos2t}{cos2t}$

$=\frac{\sqrt{cos2t}.3sin^{2}t-sin^{3}t.\frac{1}{2}\left ( cos2t \right )^{-1/2}.\left (-sin2t \right )\frac{d}{dt}2t}{cos2t}$

$=\frac{\sqrt{cos2t}.3sin^{2}t.cost+\frac{sin^{3}t}{2\sqrt{cos2t}}.2sin2t}{cos2t}$

$=\frac{3sin^{2}t.cost. cos2t+sin^{3}t.sin2t}{\left ( cos2t \right )^{3/2}}$

$=\frac{3sin^{2}t.cost. cos2t+sin^{3}t.2sint.cost}{\left ( cos2t \right )^{3/2}}$

$\frac{dx}{dt}=\frac{sin^{2}t.cost\left ( 3cos2t+2sin^{2}t \right )}{\left ( cos2t \right )^{3/2}}....\left ( i \right )$

Taking

$y=\frac{cos^{3}t}{\sqrt{cos2t}}$

Differentiating the function with respect to t

$\frac{dy}{dt}=\frac{d}{dt}\left (\frac{cos^{3}t}{\sqrt{cos2t}} \right )$

$=\frac{\sqrt{cos2t}\frac{d}{dt}cos^{3}t-cos^{3}t\frac{d}{dt}.\sqrt{cos2t}}{\left ( \sqrt{cos2t} \right )^{2}}$

$=\frac{\sqrt{cos2t}.3cos^{2}t.\frac{d}{dt}cost-cos^{3}t\frac{1}{2}.(cos2t)^{-1/2}.\frac{d}{dt}cos2t}{\left ( \sqrt{cos2t} \right )^{2}}$

$=\frac{\sqrt{cos2t}.3cos^{2}t.\left ( -sint \right )-cos^{3}t\frac{1}{2\sqrt{cos2t}}.\left ( -sin2t \right )\frac{d}{dt}2t}{\left ( \sqrt{cos2t} \right )^{2}}$

$=\frac{-\sqrt{cos2t}.3cos^{2}t.sint+cos^{3}t\frac{1}{2\sqrt{cos2t}}.2sint.cost.2}{cos2t}$

Multiplying numerator and denominator by √(cos2t)

$=\frac{-cos2t.3cos^{2}t.sint+cos^{3}t.2sint.cost}{\left ( cos2t \right )^{3/2}}$

$\frac{dy}{dt}=\frac{cos^{2}t.sint\left ( 2cos^{2}t-3cos2t \right )}{\left ( cos2t \right )^{3/2}}....\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/dt}{dx/dt}=\frac{cost\left ( 2cos^{2}t-3cos2t \right )}{sint\left ( 3cos2t+2sin^{2}t \right )}$

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

$=\frac{cost}{sint}\left [ \frac{2cos^{2}t-6cos^{2}t+3}{3-6sin^{2}t+2sin^{2}t}\right ]$

$=\frac{cost}{sint}\left [ \frac{2cos^{2}t-6cos^{2}t+3}{3-6sin^{2}t+2sin^{2}t} \right ]$

$=\frac{cost}{sint}\left [ \frac{3-4cos^{2}t}{3-4sin^{2}t} \right ]$

$=\frac{cost}{sint}\left [ \frac{3-4cos^{2}t}{3-4sin^{2}t} \right ]=-\frac{4cos^{3}t-3cost}{3sint-4sin^{3}t}$

$\frac{dy}{dx}=-\frac{cos3t}{sin3t}=-cot3t$

Q8.x = a(cost + log tan t/2), y = a sin t

Ans. The given functions are x = a(cost + log tan t/2), y = a sin t

Taking x = a(cost + log tan t/2)

Differentiating the function with respect to x

$\frac{dx}{dt}=a\frac{d}{dt} cos t+a\frac{d}{dt}\left ( log\: tan\: \frac{t}{2} \right )$

$\frac{dx}{dt}=-asint+a.\frac{1}{tan\: t/2} .\frac{d}{dt}tan\: \frac{t}{2}$

$\frac{dx}{dt}=-asint+a.\frac{1}{tan\: t/2} .sec^{2}t/2.\frac{d}{dt}\frac{t}{2}$

$\frac{dx}{dt}=-asint+a.\frac{1}{2tan\: t/2} .sec^{2}t/2$

$=-asint+\frac{a}{2sin\: t/2\: cos\: t/2\: .cos^{2}\: t/2}$

$=-asint+\frac{a}{2sin\: t/2\: cos\: t/2\: }$

$=-asin\: t+\frac{a}{sin\: t}$

$\frac{dx}{dt}=a\left ( \frac{1-sin^{2}t}{sin\: t} \right )$

$\frac{dx}{dt}=\frac{acos^{2}t}{sint}....\left ( i \right )$

Taking y = a sin t

Differentiating the function with respect to t

$\frac{dy}{dt}=a\frac{d}{dt}sin\: t$

$\frac{dy}{dt}=acos\: t....\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/dt}{dx/dt}=\frac{acos\: t}{acos^{2}\: t/sin\: t}$

$\frac{dy}{dx}=\frac{sin\: t}{cos\: t}$

$\frac{dy}{dx}=tan\: t$

Q9. x = a sec θ, y = b tan θ

Ans. The given functions are x = a sec θ, y = b tan θ

Taking x = a sec θ

Differentiating the function with respect to θ

$\frac{dx}{d\Theta }=a\frac{d}{d\Theta }sec\Theta$

$\frac{dx}{d\Theta }=atan\Theta sec\Theta ...(i)$

Taking y = b tan θ

Differentiating the function with respect to θ

$\frac{dy}{d\Theta }=b\frac{d}{d\Theta }tan\Theta$

$\frac{dy}{d\Theta }=bsec^{2} \Theta ....\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{bsec^{2}\Theta }{atan\Theta sec\Theta }$

$\frac{dy}{dx }=\frac{bsec\Theta }{atan\Theta }$

$\frac{dy}{dx }=\frac{b/cos\Theta }{asin\Theta/cos\Theta }$

$\frac{dy}{dx }=\frac{b}{asin\Theta }$

$\frac{dy}{dx }=\frac{b}{a } cosec\Theta$

Q10. x = a(cos θ + θ sin θ), y = a (sinθ – θ cos θ)

Ans. Taking x = a(cos θ + θ sin θ)

Differentiating the function with respect to θ

$\frac{dx}{d\Theta }=a\left ( \frac{d}{d\Theta }cos\Theta +\Theta \frac{d}{d\Theta } sin\Theta +sin\Theta \frac{d}{d\Theta }\Theta \right )$

$=a\left ( -sin\Theta +\Theta cos\Theta +sin\Theta \right )$

$\frac{dx}{d\Theta }=a \Theta cos\Theta ...\left ( i \right )$

Taking y = a (sinθ – θ cos θ)

Differentiating the function with respect to θ

$\frac{dy}{d\Theta }=a\left ( \frac{d}{d\Theta }sin\Theta -\Theta \frac{d}{d\Theta } cos\Theta -cos\Theta \frac{d}{d\Theta }\Theta \right )$

$\frac{dy}{d\Theta }=a\left ( cos\Theta +\Theta sin\Theta -cos\Theta \right )$

$\frac{dy}{d\Theta }=a\Theta sin\Theta ....\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{a\Theta sin\Theta }{a\Theta cos\Theta }$

$\frac{dy }{dx }=\frac{sin\Theta }{ cos\Theta }$

$\frac{dy }{dx }=tan\Theta$

$Q11. x = \sqrt{a^{sin^{-1}t}},y =\sqrt{a^{cos^{-1}t}}$

Show that dy/dx = -y/x

Ans. Taking $x = \sqrt{a^{sin^{-1}t}}$

Differentiating the function with respect to t

$\frac{dx}{dt} = \frac{d}{dt}\sqrt{a^{sin^{-1}t}}$

$=\frac{1}{2\sqrt{a^{sin^{-1}t}}}.\frac{d}{dt}a^{sin^{-1}t}$

$\because \frac{d}{dx}a^{x}=a^{x}log a$

$=\frac{1}{2\sqrt{a^{sin^{-1}t}}}.a^{sin^{-1}t}loga.\frac{d}{dt}sin^{-1}t$

$=\frac{1}{2\sqrt{a^{sin^{-1}t}}}.a^{sin^{-1}t}loga.\frac{1}{\sqrt{1-t^{2}}}$

$=\frac{a^{sin^{-1}t}loga}{2\sqrt{a^{sin^{-1}t}}.\sqrt{1-t^{2}}}$

$\frac{dx}{dt}=\frac{\sqrt{a^{sin^{-1}t}}.loga}{2\sqrt{1-t^{2}}}...\left ( i \right )$

Taking

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

Differentiating the function similarly as above,we get

$\frac{dy}{dt}=-\frac{\sqrt{a^{cos^{-1}t}}.loga}{2\sqrt{1-t^{2}}}...\left ( ii \right )$

Dividing equation (ii) by the equation (i)

$\frac{dy/dt}{dx/dt}=-\frac{\sqrt{a^{cos^{-1}t}}.loga}{2\sqrt{1-t^{2}}}\div \frac{\sqrt{a^{sin^{-1}t}}.loga}{2\sqrt{1-t^{2}}}$

$\frac{dy}{dx}=-\frac{\sqrt{a^{cos^{-1}t}}}{\sqrt{a^{sin^{-1}t}}}$

Substituting the values $x = \sqrt{a^{sin^{-1}t}}$ , $y=\sqrt{a^{cos^{-1}t}}$

$\frac{dy}{dx}=-\frac{y}{x}$ , Hence proved

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

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability