Ex 5.6 Class 12 Maths Ncert Solutions

It is given that
\(x=2 a t^{2}, y=a t^{4}\)
So, now
\(\frac{d x}{d t}=\frac{d\left(2 a t^{2}\right)}{d t}\)
\(=2 \mathrm{a} \frac{d\left(t^{2}\right)}{d t}\)
\(=2 \mathrm{a} \cdot 2 \mathrm{t}\)
\(=4 \mathrm{at} \ldots (1)\)
And
\( \frac{d y}{d t}=\frac{d\left(a t^{4}\right)}{d t} \)
\( =\mathrm{a} \frac{d\left(t^{4}\right)}{d t} \)
\( =\mathrm{a} \cdot 4 . \mathrm{t}^{3} \)
\( =4 a t^{3} \ldots(2)\)
Therefore, form equation (1) and (2). we get
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{4 a t^{3}}{4 a t}=t^{2}\)
Hence, the value of \( \frac{d y}{d x} \) is \( \mathrm{t}^{2} \)

It is given that
\(x=a \cos \theta, y=b \cos \theta\)
Then, we have
\( \frac{d x}{d \theta}=\frac{d(\operatorname{acos} \theta)}{d \theta} \)
\( =\mathrm{a}(-\sin \theta) \)
\(=-\mathrm{a} \sin \theta \ldots(1)\)
\(\frac{d y}{d \theta}=\frac{d(b \cos \theta)}{d \theta}\)
\(=\mathrm{b}(-\sin \theta)\)
\(=-\mathrm{b} \sin \theta \ldots (2)\)
From equation (1) and (2), we get
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}\)
Hence, the value of \( \frac{d y}{d x} \) is \( \frac{b}{a} \)

It is given that
\(x=\sin t, y=\cos 2 t\)
Then, we have
\(\frac{d x}{d t}=\frac{d(\sin t)}{d t}\)
\(=\cos t \ldots (1)\)
\(\frac{d y}{d x}=\frac{d(\cos 2 t)}{d t}=-\sin 2 t \frac{d(2 t)}{d t}\)
\(=-2 \sin 2 t \ldots (2)\)
So, equation (1) and (2), we get
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-2 \sin 2 t}{\cos t}\)
\(=\frac{-2.2 \sin t \cos t}{\cos t}, \text { Since } \sin 2 t=2 \sin t \cos t\)
\(=-4 \sin t\)
Hence, the value of \( \frac{d y}{d x} \) is \( -4 \sin t \)

It is given that
\(x=4 \mathrm{t}, y=\frac{4}{t}\)
Then, we have
\(\frac{d x}{d t}=\frac{d(4 t)}{d t}\)
\(=4 \ldots \ldots (1)\)
\(\frac{d y}{d t}=\frac{d\left(\frac{4}{t}\right)}{d t}=4 \frac{-1}{t^{2}}=\frac{-4}{t^{2}}\ldots(2)\)
Therefore, from equation (1) and (2), we get
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{-4}{t^{2}}}{4}=\frac{-1}{t^{2}}=-4 \sin \mathrm{t}\)
Hence, the value of \( \frac{d y}{d x} \) is \( \frac{-1}{t^{2}} \)

It is given that
\(x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta\)
Then, we have
\(\frac{d x}{d \theta}=\frac{d(\cos \theta-\cos 2 \theta)}{d \theta}\)
\(=\frac{d(\cos \theta)}{d \theta}-\frac{d(\cos 2 \theta)}{d \theta}\)
\(=-\sin \theta-(-2 \sin 2 \theta)\)
\(=2 \sin 2 \theta-\sin \theta \ldots(1)\)
\(\frac{d y}{d \theta}=\frac{d(\sin \theta-\sin \theta)}{d \theta}\)
\(=\frac{d(\sin \theta)}{d \theta}-\frac{d(\sin \theta)}{d \theta}\)
\(=\cos \theta-2 \cos 2 \theta\)
\(=-b \sin \theta \ldots (2)\)
From equation (1) and (2), we get,
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}\)
Hence, the value of \( \frac{d y}{d x} \) is \( \frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta} \)

It is given that
\(x=a(\theta-\sin \theta), y=a(1+\cos \theta)\)
Then, we have
\(\frac{d x}{d \theta}=a\left[\frac{d(\theta)}{d \theta}-\frac{d(\sin \theta)}{d \theta}\right]\)
\(=\mathrm{a}(1-\cos \theta) \ldots(1)\)
\(\frac{d y}{d \theta}=a\left[\frac{d(1)}{d \theta}-\frac{d(\cos \theta)}{d \theta}\right]\)
\(=\mathrm{a}[0+(-\sin \theta)]\)
\(=-\mathrm{a} \sin \theta \ldots (2)\)
From equation (1) and (2), we get
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-a \sin \theta}{a(1-\cos \theta)}\)
\(=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}\)
\(=\frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2}\)
Hence, the value of \( \frac{d y}{d x} \) is \( -\cot \frac{\theta}{2} \)

It is given that
\(x=a\left(\cos t+\log \tan \frac{t}{2}\right) y=a \sin t\)
Then, we have
\(\frac{d x}{d t}=a\left[\frac{d(\cos t)}{d t}+\frac{d\left(\log \tan \frac{t}{2}\right)}{d t}\right]\)
\(=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \frac{d\left(\tan \frac{t}{2}\right)}{d t}\right]\)
\(=a\left[-\sin t+\cot \frac{t}{2} \cdot \sec ^{2} \frac{t}{2} \frac{d\left(\frac{t}{2}\right)}{d t}\right]\)
\(=a\left[-\sin t+\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \times \frac{1}{\cos ^2 \frac{t}{2}} \times \frac{1}{2}\right]\)
\(=a\left[-\sin t+\frac{2}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right]\)
\(=a\left[-\sin t+\frac{1}{\sin t}\right]\)
\(=a\left[\frac{1-\sin ^{2} t}{\sin t}\right]\)
\(=a \frac{\cos ^{2} t}{\sin t} \ldots (1) \)
\(\frac{d y}{d t}=a \frac{d(\sin t)}{d t}\)
\(=\mathrm{a} \cos \mathrm{t} \ldots (2)\)
From equation (1) and (2), we get
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \cos t}{\left(a \frac{\cos ^{2} t}{\sin t}\right)}\)
\(=\frac{\sin t}{\cos t}\)\(=\tan t\)
Hence, the value of \( \frac{d y}{d x} \) is \( \tan \mathrm{t} \)

It is given that
\(x=\mathrm{a} \sec \theta, y=\mathrm{b} \tan \theta\)
Then, we have
\(\frac{d x}{d \theta}=a \frac{d(\sec \theta)}{d \theta}\)
\(=a \sec \theta \tan \theta \ldots(1)\)
\(\frac{d y}{d \theta}=b \frac{d(\tan \theta)}{d \theta}\)
\(=b \sec 2 \theta \ldots (2)\)
From equation (1) and (2), we get,
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{b \sec ^{2} \theta}{\operatorname{asec} \theta \tan \theta}\)
\(=\frac{b}{a} \sec \theta \cot \theta\)
\(=\frac{b \cos \theta}{a \cos \theta \sin \theta}\)
\(=\frac{b}{a} \times \frac{1}{\sin \theta}\)
\(=\frac{b}{a} \operatorname{cosec} \theta\)
Hence, the value of \( \frac{d y}{d x} \) is \( \operatorname{cosec} \theta \)

It is given that
\(x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)\)
Then, we have
\(\frac{d x}{d \theta}=a\left[\frac{d(\cos \theta)}{d \theta}+\frac{d(\theta \sin \theta)}{d \theta}\right]\)
\(=a\left[-\sin \theta+\frac{\theta d(\sin \theta)}{d \theta}+\sin \theta \frac{d(\theta)}{d \theta}\right]\)
\(=\mathrm{a}[-\sin \theta+\theta \cos \theta+\sin \theta]\)
\(=\mathrm{a} \theta \cos \theta \ldots(1)\)
\(\frac{d y}{d \theta}=a\left[\frac{d(\sin \theta)}{d \theta}-\frac{d(\theta \cos \theta)}{d \theta}\right]\)
\(=a\left[\cos \theta-\left\{\frac{\theta d(\cos \theta)}{d \theta}+\cos \theta \frac{d(\theta)}{d \theta}\right\}\right]\)
\(=\mathrm{a}[\cos \theta+\theta \sin \theta-\cos \theta]\)
\(=\mathrm{a} \theta \sin \theta \ldots(2)\)
From (1) and (2) we get,
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \theta \sin \theta}{a \theta \cos \theta}\)
\(=\tan \theta\)
Hence, the value of \( \frac{d y}{d x} \) is \( \tan \theta \)

It is given that
\(x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}\)
Now,
\(x=\sqrt{a^{\sin ^{-1} t}}=x=\left(a^{\sin ^{-1} t}\right)^{\frac{1}{2}}=x=a^{\frac{1}{2} \sin ^{-1} t}\)
Similarly, \( y=\sqrt{a^{\cos ^{-1} t}}=y\left(a^{\cos ^{-1} t}\right)^{\frac{1}{2}}=y=a^{\frac{1}{2} \cos ^{-1} t} \)
Let us 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, \( \frac{1}{x} \cdot \frac{d x}{d t}=\frac{1}{2} \log a \cdot \frac{d\left(\sin ^{-1} t\right)}{d t} \)
\(=\frac{d x}{d t}=\frac{x}{2} \log a \cdot \frac{1}{\sqrt{1-t^{2}}}\)
\(=\frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}} \ldots(1)\)
Now, Consider
\(y=a^{\frac{1}{2} \cos ^{-1} t}\)
Taking Log on both sides, we get
\(\log y=\frac{1}{2} \cos ^{-1} t \log a\)
Therefore, \( \frac{1}{y} \cdot \frac{d y}{d t}=\frac{1}{2} \log a \cdot \frac{d\left(\cos ^{-1} t\right)}{d t} \)
\(=\frac{d y}{d t}=\frac{y}{2} \log a \cdot \frac{-1}{\sqrt{1-t^{2}}}\)
\(=\frac{d y}{d t}=\frac{-y \log a}{2 \sqrt{1-t^{2}}} \ldots(2)\)
So, from equation (1) and (2), we get
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{\frac{-y \log a}{2 \sqrt{1-t^{2}}}}{\frac{x \log a}{2 \sqrt{1-t^{2}}}}=-\frac{y}{x}\)
Therefore, L.H.S. \( = \) R.H.S.
Hence Proved