Class 12 ncert solution math exercise 5.6

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   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}

 

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