Exercise 5.6 (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths

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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths

Ex 5.6 Question 1:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=2 a t^2, y=a t^4
$

Answer :
The given equations are $x=2 a t^2$ and $y=a t^4$
$
\begin{aligned}
& \text { 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 x}=\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
\end{aligned}
$

Ex 5.6 Question 2:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=a \cos \theta, y=b \cos \theta
$

Answer :
The given equations are $x=a \cos \theta$ and $y=b \cos \theta$
$
\begin{aligned}
& \text { 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}
\end{aligned}
$

Ex 5.6 Question 3 :

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=\sin t, y=\cos 2 t
$

Answer :
The given equations are $x=\sin t$ and $y=\cos 2 t$
$
\begin{aligned}
& \text { 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 \cdot 2 \sin t \cos t}{\cos t}=-4 \sin t
\end{aligned}
$

Ex 5.6 Question 4:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=4 t, y=\frac{4}{t}
$

Answer:
The given equations are $x=4 t$ and $y=\frac{4}{t}$
$
\begin{aligned}
& \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}
\end{aligned}
$

Ex 5.6 Question 5 :

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta
$

Answer :
The given equations are $x=\cos \theta-\cos 2 \theta$ and $y=\sin \theta-\sin 2 \theta$
$
\begin{aligned}
& \text { Then, } \begin{array}{l}
\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) \\
\quad=-\sin \theta-(-2 \sin 2 \theta)=2 \sin 2 \theta-\sin \theta
\end{array} \\
& \begin{aligned}
& \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
\end{aligned} \\
& \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}
\end{aligned}
$

Ex 5.6 Question 6:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=a(\theta-\sin \theta), y=a(1+\cos \theta)
$

Answer :
The given equations are $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$
$
\begin{aligned}
& \text { 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}
\end{aligned}
$

Ex 5.6 Question 7:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}
$

Answer:
The given equations are $x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}}$ and $y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}$
Then, $\frac{d x}{d t}=\frac{d}{d t}\left[\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\right]$
$
\begin{aligned}
& =\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 \cos t-\frac{\sin ^3 t}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t} \\
& =\frac{-3 \cos 2 t \sin ^2 t \cos t+\sin ^3 t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}
\end{aligned}
$

$\cos 2 t \cdot \sqrt{\cos 2 t}$
$
\begin{aligned}
\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)} & =\frac{-3 \cos 2 t \cdot \cos ^2 t \cdot \sin t+\cos ^3 t \sin 2 t}{3 \cos 2 t \sin ^2 t \cos t+\sin ^3 t \sin 2 t} \\
& =\frac{-3 \cos 2 t \cdot \cos ^2 t \cdot \sin t+\cos ^3 t(2 \sin t \cos t)}{3 \cos 2 t \sin ^2 t \cos t+\sin ^3 t(2 \sin t \cos t)} \\
& =\frac{\sin t \cos t\left[-3 \cos 2 t \cdot \cos t+2 \cos ^3 t\right]}{\sin t \cos t\left[3 \cos 2 t \sin t+2 \sin ^3 t\right]}
\end{aligned}
$

$\begin{array}{ll}
=\frac{\left[-3\left(2 \cos ^2 t-1\right) \cos t+2 \cos ^3 t\right]}{\left[3\left(1-2 \sin ^2 t\right) \sin t+2 \sin ^3 t\right]} & {\left[\begin{array}{c}
\cos 2 t=\left(2 \cos ^2 t-1\right), \\
\cos 2 t=\left(1-2 \sin ^2 t\right)
\end{array}\right]} \\
=\frac{-4 \cos ^3 t+3 \cos t}{3 \sin t-4 \sin ^3 t} & \\
=\frac{-\cos 3 t}{\sin 3 t} & {\left[\begin{array}{c}
\cos 3 t=4 \cos ^3 t-3 \cos t \\
\sin 3 t=3 \sin t-4 \sin ^3 t
\end{array}\right]} \\
=-\cot 3 t &
\end{array}$

Ex 5.6 Question 8:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t
$

Answer :
The given equations are $x=a\left(\cos t+\log \tan \frac{t}{2}\right)$ and $y=a \sin t$
Then, $\frac{d x}{d t}=a \cdot\left[\frac{d}{d t}(\cos t)+\frac{d}{d t}\left(\log \tan \frac{t}{2}\right)\right]$
$
\begin{aligned}
& =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]
\end{aligned}
$

$\begin{aligned}
& =a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right] \\
& =a\left(-\sin t+\frac{1}{\sin t}\right) \\
& =a\left(\frac{-\sin ^2 t+1}{\sin t}\right) \\
& =a \frac{\cos ^2 t}{\sin t} \\
& \frac{d y}{d t}=a \frac{d}{d t}(\sin t)=a \cos t \\
& \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \cos t}{\left(a \frac{\cos t}{\sin t}\right)}=\frac{\sin t}{\cos t}=\tan t
\end{aligned}$

Ex 5.6 Question 9 :

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=a \sec \theta, y=b \tan \theta
$

Answer :
The given equations are $x=a \sec \theta$ and $y=b \tan \theta$
$
\begin{aligned}
& \text { 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
\end{aligned}
$

Ex 5.6 Question 10:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$
x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)
$

Answer:
The given equations are $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$
$
\begin{aligned}
& \text { 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 \\
&
\end{aligned}
$
$
\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
$

Ex 5.6 Question 11:

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

Answer :
The given equations are $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$
$x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$
$\Rightarrow x=\left(a^{\sin ^{-1} t}\right)^{\frac{1}{2}}$ and $y=\left(a^{\cos ^{-1} t}\right)^{\frac{1}{2}}$
$\Rightarrow x=a^{\frac{1}{2} \sin ^{-1} t}$ and $y=a^{\frac{1}{2} \cos ^{-1},}$
Consider $x=a^{\frac{1}{2^{\sin ^{-1} t}}}$
Taking logarithm on both the sides, we obtain
$
\begin{aligned}
& \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}{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}}
\end{aligned}
$

Then, consider $y=a^{\frac{1}{2} \cos ^{-1} t}$
Taking logarithm on both the sides, we obtain
$
\log y=\frac{1}{2} \cos ^{-1} t \log a
$

$
\begin{aligned}
& \therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\cos ^{-1} t\right) \\
& \Rightarrow \frac{d y}{d t}=\frac{y \log a}{2} \cdot\left(\frac{-1}{\sqrt{1-t^2}}\right) \\
& \Rightarrow \frac{d y}{d t}=\frac{-y \log a}{2 \sqrt{1-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{-y \log a}{2 \sqrt{1-t^2}}\right)}{\left(\frac{x \log a}{\left.2 \sqrt{1-t^2}\right)}\right.}=-\frac{y}{x} .
\end{aligned}
$

Hence, proved.