Class 12 ncert solution math exercise 5.6

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The  Class 12 NCERT solution math exercise 5.6 prepared by expert Mathematics teacher at gmath.in as per CBSE  guidelines. See our Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 Questions with Solutions to help you to revise complete Syllabus and Score More marks in your Board and School exams.

   EXERCISE 5.6

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find\frac{dy}{dx}.(Class 12 ncert solution math exercise 5.6)

Question 1: x=2 a t^2, y=a t^4,

Solution: Given, x=2 a t^2, y=a t^4
Then,

\frac{d x}{d t}=\frac{d}{d t}\left(2 a t^2\right)

=2 a \cdot \frac{d}{d t}\left(t^2\right)=2 a \cdot 2 t=4 a t

\frac{d y}{d t}=\frac{d}{d t}\left(a t^4\right)

=a \cdot \frac{d}{d t}\left(t^4\right)=a \cdot 4 \cdot t^3=4 a t^3

\therefore \frac{d y}{d t}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}

=\frac{4 a t^3}{4 a t}=t^2

Question 2: x=a \cos \theta, y=b \cos \theta

Solution: Given, x=a \cos \theta, y=b \cos \theta

Then,

\frac{d x}{d \theta}=\frac{d}{d \theta}(a \cos \theta)

=a(-\sin \theta)=-a \sin \theta

\frac{d y}{d \theta}=\frac{d}{d \theta}(b \cos \theta)

=b(-\sin \theta)=-b \sin \theta

\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}

=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}

Question 3: x=\sin t, y=\cos 2 t

Solution: Given, x=\sin t, y=\cos 2 t

Then, \frac{d x}{d t}=\frac{d}{d t}(\sin t)=\cos t

\frac{d y}{d t}=\frac{d}{d t}(\cos 2 t)

=-\sin 2 t \cdot \frac{d}{d t}(2 t)=-2 \sin 2 t

\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}

=\frac{-2 \sin 2 t}{\cos t}

=\frac{-2.2 \sin t \cos t}{\cos t}=-4 \sin t

Question 4: x=4 t, y=\frac{4}{t}

Solution: Given, x=4 t, y=\frac{4}{t}

\frac{d x}{d t}=\frac{d}{d t}(4 t)=4

\frac{d y}{d t}=\frac{d}{d t}\left(\frac{4}{t}\right)

=4 \cdot \frac{d}{d t}\left(\frac{1}{t}\right)

=4 \cdot\left(\frac{-1}{t^2}\right)=\frac{-4}{t^2}

\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}

=\frac{\left(\frac{-4}{t^2}\right)}{4}=\frac{-1}{t^2}

Question 5: x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta

Solution: Given, x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta

Then,

\frac{d x}{d \theta}=\frac{d}{d \theta}(\cos \theta-\cos 2 \theta)

=\frac{d}{d \theta}(\cos \theta)-\frac{d}{d \theta}(\cos 2 \theta)

=-\sin \theta-(-2 \sin 2 \theta)

=2 \sin 2 \theta-\sin \theta

\frac{d y}{d \theta}=\frac{d}{d \theta}(\sin \theta-\sin 2 \theta)

=\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\sin 2 \theta)

=\cos \theta-2 \cos 2 \theta

\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}

=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}

Question 6: x=a(\theta-\sin \theta), y=a(1+\cos \theta)

Solution: Given, x=a(\theta-\sin \theta), y=a(1+\cos \theta)

Then,

\frac{d x}{d \theta}=a\left[\frac{d}{d \theta}(\theta)-\frac{d}{d \theta}(\sin \theta)\right]

=a(1-\cos \theta)

\frac{d y}{d \theta}=a\left[\frac{d}{d \theta}(1)+\frac{d}{d \theta}(\cos \theta)\right]

=a[0+(-\sin \theta)]

=-a \sin \theta

\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}

=\frac{-a \sin \theta}{a(1-\cos \theta)}

=\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}

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

Solution: Given, x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}
Then,

\frac{d x}{d t}=\frac{d}{d t}\left[\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\right]

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

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

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

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

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

=\frac{\sin^2t\cos t(3\cos 2t+2\sin^2t)}{(\cos 2t)^{3/2}}

= \frac{\sin^2t\cos t[3(1-2\sin^2t)+2\sin^2t]}{(\cos 2t)^{3/2}}

= \frac{\sin^2t\cos t[3-6\sin^2 t+2\sin^2t]}{(\cos 2t)^{3/2}}

= \frac{\sin^2t\cos t[3-4\sin^2t]}{(\cos 2t)^{3/2}}

= \frac{\sin t\cos t(3\sin t-4\sin^3 t)}{(\cos 2t)^{3/2}}

= \frac{\sin t\cos t\sin 3t}{(\cos 2t)^{3/2}}

\frac{d y}{d t}=\frac{d}{d t}\left[\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\right]

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

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

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

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

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

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

=\frac{\cos^2t\sin t(-3\cos 2t+2\cos^2t)}{(\cos 2t)^{3/2}}

= \frac{\cos^2t\sin t[-3(2\cos^2t-1)+2\cos^2t]}{(\cos 2t)^{3/2}}

= \frac{-\cos^2t\sin t[6\cos^2 t-3-2\cos^2t]}{(\cos 2t)^{3/2}}

= \frac{-\cos^2t\sin t[4\cos^2t-3]}{(\cos 2t)^{3/2}}

= \frac{-\sin t\cos t(4\cos^3t-3\cos t)}{(\cos 2t)^{3/2}}

= \frac{-\sin t\cos t\cos 3t}{(\cos 2t)^{3/2}}

\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}

= \frac{\frac{-\sin t\cos t\cos 3t}{(\cos 2t)^{3/2}}}{\frac{\sin t\cos t\sin 3t}{(\cos 2t)^{3/2}}}

= \frac{-\cos 3t}{\sin 3t}

\Rightarrow \frac{dy}{dx} = -\cot 3t

Question 8: x=a(\cos t+\log\tan\frac{t}{2}),y=a\sin t

Solution: x=a(\cos t+\log\tan\frac{t}{2}),y=a\sin t

Then,

\frac{d x}{d t}=a \cdot\left[\frac{d}{d t}(\cos t)+\frac{d}{d t}\left(\log \tan \frac{t}{2}\right)\right]

=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{d t}\left(\tan \frac{t}{2}\right)\right]

=a\left[-\sin t+\cot \frac{t}{2} \cdot \sec ^2 \frac{t}{2} \cdot \frac{d}{d t}\left(\frac{t}{2}\right)\right]

=a\left[-\sin t+\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \times \frac{1}{\cos ^2 \frac{t}{2}} \times \frac{1}{2}\right]

=a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right]

=a\left(-\sin t+\frac{1}{\sin t}\right)

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

=a\left(\frac{\cos ^2 t}{\sin t}\right)

\frac{d y}{d t}=a \frac{d}{d t}(\sin t)=a \cos t

Therefore,

\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}

=\frac{a \cos t}{\left(a \frac{\cos ^2 t}{\sin t}\right)}

=\frac{\sin t}{\cos t}=\tan t

Question 9: x=a \sec \theta, y=b \tan \theta

Solution: Given, x=a \sec \theta, y=b \tan \theta
Then,

\frac{d x}{d \theta}=a \cdot \frac{d}{d \theta}(\sec \theta)=a \sec \theta \tan \theta

\frac{d y}{d \theta}=b \cdot \frac{d}{d \theta}(\tan \theta)=b \sec ^2 \theta

Therefore,

\frac{d y}{d x} =\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}

=\frac{b \sec ^2 \theta}{a \sec \theta \tan \theta}

=\frac{b}{a} \sec \theta \cot \theta

=\frac{b \cos \theta}{a \cos \theta \sin \theta}

=\frac{b}{a} \times \frac{1}{\sin \theta}

=\frac{b}{a} \operatorname{cosec} \theta

Question 10: x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)

Solution: Given, x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)

Then,

\frac{d x}{d \theta} =a\left[\frac{d}{d \theta} \cos \theta+\frac{d}{d \theta}(\theta \sin \theta)\right]

=a\left[-\sin \theta+\theta \frac{d}{d \theta}(\sin \theta)+\sin \theta \frac{d}{d \theta}(\theta)\right]

=a[-\sin \theta+\theta \cos \theta+\sin \theta]

=a \theta \cos \theta

\frac{d y}{d \theta} =a\left[\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\theta \cos \theta)\right]

=a\left[\cos \theta-\left\{\theta \frac{d}{d \theta}(\cos \theta)+\cos \theta \cdot \frac{d}{d \theta}(\theta)\right\}\right]

=a[\cos \theta+\theta \sin \theta-\cos \theta]

=a \theta \sin \theta

Therefore,

\frac{d y}{d x} =\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}

=\frac{a \theta \sin \theta}{a \theta \cos \theta}

=\tan \theta

Question 11: If x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}, show that \frac{d y}{d x}=-\frac{y}{x}

Solution: Given, x=\sqrt{a^{\sin ^{-1} t}} and y=\sqrt{a^{\cos ^{-1} t}}

Hence,

x=\sqrt{a^{\sin ^{-1} t}}

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

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

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

=\left(a^{\cos ^{-1} t}\right)^{\frac{1}{2}}

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

Consider x=a^{\frac{1}{2} \sin ^{-1} t}
Taking log on both sides, we get

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

Therefore,

\Rightarrow \frac{1}{x} \cdot \frac{d x}{d t}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\sin ^{-1} t\right)

\Rightarrow \frac{d x}{d t}=\frac{x}{2} \log a \cdot \frac{1}{\sqrt{1-t^2}}

\Rightarrow \frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^2}}

Now, y=a^{\frac{1}{2} \cos ^{-1} t}

Taking log on both sides, we get

\log x=\frac{1}{2} \cos ^{-1} t \log a

Therefore,

\Rightarrow \frac{1}{y} \cdot \frac{d y}{d t}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\cos ^{-1} t\right)

\Rightarrow \frac{d y}{d t}=\frac{y}{2} \log a \cdot \frac{-1}{\sqrt{1-t^2}}

\Rightarrow \frac{d y}{d t}=\frac{-y \log a}{2 \sqrt{1-t^2}}

Hence,

\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}

=\frac{\left(\frac{-y \log a}{2 \sqrt{1-t^2}}\right)}{\left(\frac{x \log a}{2 \sqrt{1-t^2}}\right)}

=-\frac{y}{x}

NCERT solution chapter 5 continuity and differentiability

1. Ncert solution  Exercise 5.1 continuity and differentiability

2. Ncert solution  Exercise 5.2 continuity and differentiability

3. Ncert solution  Exercise 5.3 continuity and differentiability

4. Ncert solution  Exercise 5.4 continuity and differentiability

5. Ncert solution  Exercise 5.5 continuity and differentiability

6. Ncert solution  Exercise 5.7 continuity and differentiability

7. Ncert solution  Chapter 5 Miscellaneous continuity and differentiability

 

continuity and differentiability multiple choice

Case study relation and function 3
An organization conducted bike race under 2 different categories- boys and girls.

 

 

 

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