NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6


Chapter 5 Ex.5.6 Question 1

If \(x\) and \(y\) are connected parametrically by the equations \(x = 2a{t^2},\;y = a{t^4}\), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = 2a{t^2},\; y = a{t^4}\)

Then,

\(\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {2a{t^2}} \right) = 2a.\frac{d}{{dt}}\left( {{t^2}} \right) = 2a.2t = 4at\)

\[\begin{align} & \frac{{dy}}{{dt}}= \frac{d}{{dt}}\left( {a{t^4}} \right)= a.\frac{d}{{dt}}\left( {{t^4}} \right) = a.4.{t^3} = 4a{t^3}\\&\therefore \frac{{dy}}{{dt}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{4a{t^3}}}{{4at}} = {t^2}\end{align}\]

Chapter 5 Ex.5.6 Question 2

If \(x\) and \(y\) are connected parametrically by the equations \(x = a\cos \theta,\;y = b\cos \theta \), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = a\cos \theta,\;y = b\cos \theta \)

Then,

\[\begin{align}&\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}\left( {a\cos \theta } \right) = a\left( { - \sin \theta } \right) = - a\sin \theta \\&\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left( {b\cos \theta } \right) = b\left( { - \sin \theta } \right) = - b\sin \theta \\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{ - b\sin \theta }}{{ - a\sin \theta }} = \frac{b}{a}\end{align}\]

Chapter 5 Ex.5.6 Question 3

If \(x\) and \(y\) are connected parametrically by the equations \(x = \sin t,\;y = \cos 2t\), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = \sin t,\;y = \cos 2t\)

Then, \(\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\sin t} \right) = \cos t\)

\[\begin{align}&\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\cos 2t} \right) = - \sin 2t.\frac{d}{{dt}}\left( {2t} \right) = - 2\sin 2t\\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{ - 2\sin 2t}}{{\cos t}} = \frac{{ - 2.2\sin t\cos t}}{{\cos t}} = - 4\sin t\end{align}\]

Chapter 5 Ex.5.6 Question 4

If \(x\) and \(y\) are connected parametrically by the equations \(x = 4t,\;y = \frac{4}{t}\), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = 4t,\;y = \frac{4}{t}\)

\[\begin{align}&\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {4t} \right) = 4\\&\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{4}{t}} \right) = 4.\frac{d}{{dt}}\left( {\frac{1}{t}} \right) = 4.\left( {\frac{{ - 1}}{{{t^2}}}} \right) = \frac{{ - 4}}{{{t^2}}}\\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\left( {\frac{{ - 4}}{{{t^2}}}} \right)}}{4} = \frac{{ - 1}}{{{t^2}}}\end{align}\]

Chapter 5 Ex.5.6 Question 5

If \(x\) and \(y\) are connected parametrically by the equations \(x = \cos \theta - \cos 2\theta,\;y = \sin \theta - \sin 2\theta \), without eliminating the parameter, find \(\)

 

Solution

 

Given, \(x = \cos \theta - \cos 2\theta ,\;y = \sin \theta - \sin 2\theta \)

Then,

\[\begin{align}\frac{{dx}}{{d\theta }} &= \frac{d}{{d\theta }}\left( {\cos \theta - \cos 2\theta } \right) = \frac{d}{{d\theta }}\left( {\cos \theta } \right) - \frac{d}{{d\theta }}\left( {\cos 2\theta } \right)\\[5pt]&= - \sin \theta - \left( { - 2\sin 2\theta } \right) \\[5pt]&= 2\sin 2\theta - \sin \theta \\[10pt] \frac{{dy}}{{d\theta }}& = \frac{d}{{d\theta }}\left( {\sin \theta - \sin 2\theta } \right) = \frac{d}{{d\theta }}\left( {\sin \theta } \right) - \frac{d}{{d\theta }}\left( {\sin 2\theta } \right)\\[5pt]&= \cos \theta - 2\cos 2\theta \\\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} &= \frac{{\cos \theta - 2\cos 2\theta }}{{2\sin 2\theta - \sin \theta }}\end{align}\]

Chapter 5 Ex.5.6 Question 6

If \(x\) and \(y\) are connected parametrically by the equations \(x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)\), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)\)

Then, \(\frac{{dx}}{{d\theta }} = a\left[ {\frac{d}{{d\theta }}\left( \theta \right) - \frac{d}{{d\theta }}\left( {\sin \theta } \right)} \right] = a\left( {1 - \cos \theta } \right)\)

\[\begin{align}\frac{{dy}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\left( 1 \right) + \frac{d}{{d\theta }}\left( {\cos \theta } \right)} \right] = a\left[ {0 + \left( { - \sin \theta } \right)} \right] = - a\sin \theta \\\therefore \frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{ - a\sin \theta }}{{a\left( {1 - \cos \theta } \right)}} = \frac{{ - 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\;\sin }^2}\frac{\theta }{2}}} = \frac{{ - \cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}} = - \cot \frac{\theta }{2}\end{align}\]

Chapter 5 Ex.5.6 Question 7

If \(x\) and \(y\) are connected parametrically by the equations \(x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},\;y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}\), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},\;y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}\)

Then,

\[\begin{align}\frac{{dx}}{{dt}} &= \frac{d}{{dt}}\left[ {\frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }}} \right]\\[5pt]&= \frac{{\sqrt {\cos 2t} .\frac{d}{{dt}}\left( {{{\sin }^3}t} \right) - {{\sin }^3}t.\frac{d}{{dt}}\sqrt {\cos 2t} }}{{\cos 2t}}\\[5pt]&= \frac{{\sqrt {\cos 2t} .3{{\sin }^2}t.\frac{d}{{dt}}\left( {\sin t} \right) - {{\sin }^3}t \times \frac{1}{{2\sqrt {\cos 2t} }}.\frac{d}{{dt}}\left( {\cos 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\sqrt {\cos 2t} .{{\sin }^2}t.\cos t - \frac{{{{\sin }^3}t}}{{2\sqrt {\cos 2t} }}.\left( { - 2\sin 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\cos 2t.{{\sin }^2}t\cos t + {{\sin }^3}t.\sin 2t}}{{\cos 2t\sqrt {\cos 2t} }}\\[5pt]\frac{{dy}}{{dt}}& = \frac{d}{{dt}}\left[ {\frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}} \right]\\&= \frac{{\sqrt {\cos 2t} .\frac{d}{{dt}}\left( {{{\cos }^3}t} \right) - {{\cos }^3}t.\frac{d}{{dt}}\left( {\sqrt {\cos 2t} } \right)}}{{\cos 2t}}\\[5pt]&= \frac{{\sqrt {\cos 2t} .3{{\cos }^2}t.\frac{d}{{dt}}\left( {\cos t} \right) - {{\cos }^3}t.\frac{1}{{2\sqrt {\cos 2t} }}.\frac{d}{{dt}}\left( {\cos 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\sqrt {\cos 2t} .{{\cos }^2}t\left( { - \sin t} \right) - {{\cos }^3}t.\frac{1}{{\sqrt {\cos 2t} }}.\left( { - 2\sin 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t.\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}\end{align}\]

\[\begin{align}\therefore \frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}}}{{\frac{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\sin 2t}}{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\sin 2t}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\left( {2\sin t\cos t} \right)}}{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\left( {2\sin t\cos t} \right)}}\\[5pt]&= \frac{{\sin t\cos t\left[ { - 3\cos 2t.\cos t + 2{{\cos }^3}t} \right]}}{{\sin t\cos t\left[ {3\cos 2t\sin t + 2{{\sin }^3}t} \right]}}\\[5pt]&= \frac{{\left[ { - 3\left( {2{{\cos }^2}t - 1} \right)\cos t + 2{{\cos }^3}t} \right]}}{{\left[ {3\left( {1 - 2{{\sin }^2}t} \right)\sin t + 2{{\sin }^3}t} \right]}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}\cos 2t = \left( {2{{\cos }^2}t - 1} \right)\\\cos 2t = \left( {1 - 2{{\sin }^2}t} \right)\end{array} \right]\\[5pt]&= \frac{{ - 4{{\cos }^3}t + 3\cos t}}{{3\sin t - 4{{\sin }^3}t}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}\cos 3t = 4{\cos ^3}t - 3\cos t\\\sin 3t = 3\sin t - 4{\sin ^2}t\end{array} \right]\\[5pt]&= \frac{{ - \cos 3t}}{{\sin 3t}} = - \cot 3t\end{align}\]

Chapter 5 Ex.5.6 Question 8

If \(x\) and \(y\) are connected parametrically by the equations \(x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),\;y = a\sin t\), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),\;y = a\sin t\)

Then,

\[\begin{align}\frac{{dx}}{{dt}} &= a.\left[ {\frac{d}{{dt}}\left( {\cos t} \right) + \frac{d}{{dt}}\left( {\log \tan \frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \frac{1}{{\tan \frac{t}{2}}}.\frac{d}{{dt}}\left( {\tan \frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \cot \frac{t}{2}.{{\sec }^2}\frac{t}{2}.\frac{d}{{dt}}\left( {\frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \frac{{\cos \frac{t}{2}}}{{\sin \frac{t}{2}}} \times \frac{1}{{{{\cos }^2}\frac{t}{2}}} \times \frac{1}{2}} \right]\\&= a\left[ { - \sin t + \frac{1}{{2\sin \frac{t}{2}\cos \frac{t}{2}}}} \right]\\&= a\left( { - \sin t + \frac{1}{{\sin t}}} \right)\\&= a\left( {\frac{{ - {{\sin }^2}t + 1}}{{\sin t}}} \right)\\&= a\left( {\frac{{{{\cos }^2}t}}{{\sin t}}} \right)\\\frac{{dy}}{{dt}} &= a\frac{d}{{dt}}\left( {\sin t} \right) = a\cos t\end{align}\]

Therefore,

\[\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{a\cos t}}{{\left( {a\frac{{{{\cos }^2}t}}{{\sin t}}} \right)}} = \frac{{\sin t}}{{\cos t}} = \tan t\]

Chapter 5 Ex.5.6 Question 9

If \(x\) and \(y\) are connected parametrically by the equations \(x = a\;\sec \theta,\;y = b\;\tan \theta \), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = a\;\sec \theta,\;y = b\;\tan \theta \)

Then,

\(\frac{{dx}}{{d\theta }} = a.\frac{d}{{d\theta }}\left( {\sec \theta } \right) = a\sec \theta \tan \theta \)

\(\frac{{dy}}{{d\theta }} = b.\frac{d}{{d\theta }}\left( {\tan \theta } \right) = b{\;\sec ^2}\theta \)

Therefore,

\[\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\\&= \frac{{b{{\;\sec }^2}\theta }}{{a\sec \theta \tan \theta }}\\&= \frac{b}{a}\sec \theta \cot \theta \\&= \frac{{b\cos \theta }}{{a\cos \theta \sin \theta }}\\&= \frac{b}{a} \times \frac{1}{{\sin \theta }}\\&= \frac{b}{a}{\mathop{\text cosec\;}\nolimits} \theta \end{align}\]

Chapter 5 Ex.5.6 Question 10

If \(x\) and \(y\) are connected parametrically by the equations \(x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right)\), without eliminating the parameter, find \(\frac{{dy}}{{dx}}\)

 

Solution

 

Given, \(x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right)\)

Then,

\[\begin{align}\frac{{dx}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\cos \theta + \frac{d}{{d\theta }}\left( {\theta \sin \theta } \right)} \right]\\&= a\left[ { - \sin \theta + \theta \frac{d}{{d\theta }}\left( {\sin \theta } \right) + \sin \theta \frac{d}{{d\theta }}\left( \theta \right)} \right]\\[5pt]&= a\left[ { - \sin \theta + \theta \cos \theta + \sin \theta } \right]\\[5pt]&= a\;\theta \cos \theta \end{align}\]

\[\begin{align}\frac{{dy}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\left( {\sin \theta } \right) - \frac{d}{{d\theta }}\left( {\theta \cos \theta } \right)} \right] \\&= a\left[ {\cos \theta - \left\{ {\theta \frac{d}{{d\theta }}\left( {\cos \theta } \right) + \cos \theta .\frac{d}{{d\theta }}\left( \theta \right)} \right\}} \right]\\[5pt]&= a\left[ {\cos \theta + \theta \sin \theta - \cos \theta } \right]\\[5pt]&= a\; \theta \sin \theta \end{align}\]

Therefore,

\[\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\\[5pt]&= \frac{{a\;\theta \sin \theta }}{{a\;\theta \cos \theta }}\\[5pt]&= \tan \theta \end{align}\]

Chapter 5 Ex.5.6 Question 11

If \(x = \sqrt {{a^{{{\sin }^{ - 1}}t}}},\;y = \sqrt {{a^{{{\cos }^{ - 1}}t}}} \), show that \(\frac{{dy}}{{dx}} = - \frac{y}{x}\)

 

Solution

 

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

Hence,

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

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,

\[\begin{align}&\Rightarrow \; \; \frac{1}{x}.\frac{{dx}}{{dt}} = \frac{1}{2}\log a.\frac{d}{{dt}}\left( {{{\sin }^{ - 1}}t} \right)\\&\Rightarrow \; \; \frac{{dx}}{{dt}} = \frac{x}{2}\log a.\frac{1}{{\sqrt {1 - {t^2}} }}\\&\Rightarrow \; \; \frac{{dx}}{{dt}} = \frac{{x\log a}}{{2\sqrt {1 - {t^2}} }}\end{align}\]

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,

\[\begin{align}&\Rightarrow \; \; \frac{1}{y}.\frac{{dy}}{{dt}} = \frac{1}{2}\log a.\frac{d}{{dt}}\left( {{{\cos }^{ - 1}}t} \right)\\&\Rightarrow \; \; \frac{{dy}}{{dt}} = \frac{y}{2}\log a.\frac{{ - 1}}{{\sqrt {1 - {t^2}} }}\\&\Rightarrow \; \; \frac{{dy}}{{dt}} = \frac{{ - y\log a}}{{2\sqrt {1 - {t^2}} }}\end{align}\]

Hence,

\[\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\left( {\frac{{ - y\log a}}{{2\sqrt {1 - {t^2}} }}} \right)}}{{\left( {\frac{{x\log a}}{{2\sqrt {1 - {t^2}} }}} \right)}} = - \frac{y}{x}\]

Instant doubt clearing with Cuemath Advanced Math Program