12th NCERT Continuity And Differentiability Exercise 5.6 Questions 11

If x and y are connected parametrically by the equations given in Exercises 1 to 10. without eliminating the parameter, find dy/dx.

Question (1)

x = 2xt² , y = at⁴

Solution

\[x = 2a{t^2}\] \[diff.w.r.t.t\] \[\frac{{dx}}{{dt}} = 2a(2t) = 4at\] \[y = a{t^4}\] \[diff.w.r.t.t\] \[\frac{{dy}}{{dt}} = a(4{t^3}) = 4a{t^3}\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\] \[ = \frac{{4a{t^3}}}{{4at}} = {t^2}\]

Question (2)

x = a cos θ , y = b cos θ

Solution

\[x = a\cos \theta ,\] \[diff.w.r.t.\theta ,\] \[\frac{{dx}}{{d\theta }} = a( - \sin \theta ) = - a\sin \theta \] \[y = b\cos \theta \] \[diff.w.r.t.\theta \] \[\frac{{dy}}{{d\theta }} = b( - \sin \theta ) = - b\sin \theta \] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\] \[ = \frac{{ - b\sin \theta }}{{ - a\sin \theta }} = \frac{b}{a}\]

Question (3)

x= sin t, y = cos2t

Solution

\[x = \sin t\] \[diff.w.r.t.t\] \[\frac{{dx}}{{dt}} = \cos t\] \[y = \cos 2t\] \[diff.w.r.t.t \] \[\frac{{dy}}{{dt}} = ( - \sin 2t)2 = - 2\sin 2t\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{ - 2\sin 2t}}{{\cos t}}\] \[ = \frac{{ - 2(2\sin t\cos t)}}{{\cos t}} = - 4\sin t\]

Question (4)

x = 4t, y = 4/t

Solution

\[x = 4t\] \[diff.w.r.t.t,\] \[\frac{{dx}}{{dt}} = 4\] \[y = \frac{4}{t}\] \[diff.w.r.t.t,\] \[\frac{{dy}}{{dt}} = 4\left( { - \frac{1}{{{t^2}}}} \right) = - \frac{4}{{{t^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\] \[ = \frac{{\frac{{ - 4}}{{{t^2}}}}}{4} = \frac{{ - 1}}{{{t^2}}}\]

Question (5)

x = cos θ - cos 2θ , y = sin θ - sin 2θ

Solution

\[x = \cos \theta - \cos 2\theta \] \[diff.w.r.t.\theta ,\] \[\frac{{dx}}{{d\theta }} = ( - \sin \theta ) - ( - \sin 2\theta )2 = 2\sin 2\theta - \sin \theta \] \[y = \sin \theta - \sin 2\theta \] \[diff.w.r.t.\theta ,\] \[\frac{{dy}}{{d\theta }} = \cos \theta - \cos 2\theta (2) = \cos \theta - 2\cos 2\theta \] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\] \[ = \frac{{\cos \theta - 2\cos 2\theta }}{{2\sin 2\theta - \sin \theta }}\]

Question (6)

x = a(θ - sin θ), y = a( 1 + cos θ)

Solution

\[x = a(\theta - \sin \theta )\] \[diff.w.r.t.\theta ,\] \[\frac{{dx}}{{d\theta }} = a(1 - \cos \theta ) = a(2{\sin ^2}\frac{\theta }{2}) = 2a{\sin ^2}\frac{\theta }{2}\] \[y = a(1 + \cos \theta )\] \[diff.w.r.t.\theta ,\] \[\frac{{dy}}{{d\theta }} = a( - \sin \theta ) = - 2a\sin \frac{\theta }{2}\cos \frac{\theta }{2}\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\] \[ = \frac{{ - 2a\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2a{{\sin }^2}\frac{\theta }{2}}} = - \cot \frac{\theta }{2}\]

Question (7)

\[x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}\]

Solution

\[x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }}\] \[diff.w.r.t.t,\] \[\frac{{dx}}{{dt}} = \frac{{\sqrt {\cos 2t} \left( {3{{\sin }^2}t\cos t} \right) - {{\sin }^3}t\left( {\frac{1}{{2\sqrt {\cos 2t} }} \times ( - \sin 2t)2} \right)}}{{\cos 2t}}\] \[ = \frac{1}{{{{\cos }^{3/2}}2t}}\left[ {\cos 2t(3{{\sin }^2}t\cos t) + {{\sin }^3}t\sin 2t} \right]\] \[ = \frac{1}{{{{\cos }^{3/2}}2t}}\left[ {3{{\sin }^2}t\cos t\cos 2t + 2\sin t\cos t{{\sin }^3}t} \right]\] \[ = \frac{{{{\sin }^2}t\cos t}}{{{{\cos }^{3/2}}2t}}\left[ {3\cos 2t + 2{{\sin }^2}t} \right]\] \[ = \frac{{{{\sin }^2}t\cos t}}{{{{\cos }^{3/2}}2t}}\left[ {3(1 - 2{{\sin }^2}t) + 2{{\sin }^2}t} \right]\] \[ = \frac{{{{\sin }^2}t\cos t}}{{{{\cos }^{3/2}}2t}}\left[ {3 - 4{{\sin }^2}t} \right]\] \[y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}\] \[diff.w.r.t.t,\] \[\frac{{dy}}{{dt}} = = \frac{{\sqrt {\cos 2t} \left( {3{{\cos }^2}t( - \sin t)} \right) - {{\cos }^3}t\left( {\frac{1}{{2\sqrt {\cos 2t} }} \times ( - \sin 2t)2} \right)}}{{\cos 2t}}\] \[ = \frac{1}{{{{\cos }^{3/2}}2t}}\left[ {\cos 2t( - 3{{\cos }^2}t\sin t) + {{\cos }^3}t\sin 2t} \right]\] \[ = \frac{1}{{{{\cos }^{3/2}}2t}}\left[ { - 3\sin t{{\cos }^2}t\cos 2t + 2\sin t\cos t{{\cos }^3}t} \right]\] \[ = \frac{{\sin t{{\cos }^2}t}}{{{{\cos }^{3/2}}2t}}\left[ { - 3\cos 2t + 2{{\cos }^2}t} \right]\] \[ = \frac{{\sin t{{\cos }^2}t}}{{{{\cos }^{3/2}}2t}}\left[ { - 3(2{{\cos }^2}t - 1) + 2{{\cos }^2}t} \right]\] \[ = \frac{{\sin t{{\cos }^2}t}}{{{{\cos }^{3/2}}2t}}\left[ {3 - 4{{\cos }^2}t} \right]\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\] \[\frac{{dy}}{{dx}} = \frac{{\cos t(3 - 4{{\cos }^2}t)}}{{\sin t(3 - 4{{\sin }^2}t)}}\] \[ = \frac{{ - \left[ {4{{\cos }^3}t - 3\cos t} \right]}}{{3\sin t - 4{{\sin }^3}t}}\] \[ = \frac{{ - \cos 3t}}{{\sin 3t}}\] \[ = - \cot 3t\]

Question (8)

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

Solution

\[x = a\left( {\cos t + \log \tan \frac{t}{2}} \right)\] \[Diff.w.r.t.t\] \[\frac{{dx}}{{dt}} = a\left[ { - \sin t + \frac{1}{{\tan (t/2)}}{{\sec }^2}\left( {\frac{t}{2}} \right)\frac{1}{2}} \right]\] \[ = a\left[ { - \sin t + \frac{{\cos \frac{t}{2}}}{{\sin \frac{t}{2}}} \times \frac{1}{{2{{\cos }^2}t/2}}} \right]\] \[ = a\left[ { - \sin t + \frac{1}{{2\sin (t/2)\cos (t/2)}}} \right]\] \[ = a\left[ { - \sin t + \frac{1}{{\sin t}}} \right]\] \[ = a\left[ {\frac{{ - {{\sin }^2}t + 1}}{{\sin t}}} \right]\] \[ = a\left( {\frac{{{{\cos }^2}t}}{{\sin t}}} \right)\] \[y = a\sin t\] \[diff.w.r.t.t,\] \[\frac{{dy}}{{dt}} = a\cos t\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\] \[ = \frac{{a\cos t}}{{a\left( {\frac{{{{\cos }^2}t}}{{\sin t}}} \right)}}\] \[ = \frac{{\sin t}}{{\cos t}} = \tan t\]

Question (9)

x = a secθ , y = b tan θ

Solution

\[\begin{array}{l}x = a\sec \theta \\Diff.w.r.t.\theta \end{array}\] \[\frac{{dx}}{{d\theta }} = a\sec \theta \tan \theta \] \[\begin{array}{l}y = b\tan \theta \\Diff.w.r.t.\theta \end{array}\] \[\frac{{dy}}{{d\theta }} = b{\sec ^2}\theta \] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\] \[ = \frac{{b{{\sec }^2}\theta }}{{a\sec \theta \tan \theta }}\] \[ = \frac{{b\sec \theta }}{{a\tan \theta }}\] \[ = \frac{{\frac{b}{{\cos \theta }}}}{{\frac{{a\sin \theta }}{{\cos \theta }}}}\] \[ = \frac{b}{{a\sin \theta }} = \frac{b}{a}\cos ec\theta \]

Question (10)

x = a(cosθ + θ sin θ) , y = a( sin θ - θ cosθ)

Solution

\[x = a(\cos \theta + \theta \sin \theta )\] \[Diff.w.r.t.\theta \] \[\frac{{dx}}{{d\theta }} = a\left[ { - \sin \theta + \left( {\theta \frac{d}{{d\theta }}\sin \theta + \sin \theta \frac{d}{{d\theta }}\theta } \right)} \right]\] \[ = a\left[ { - \sin \theta + \theta \cos \theta + \sin \theta } \right]\] \[ = a\theta \cos \theta \] \[y = a(\sin \theta - \theta \cos \theta )\] \[Diff.w.r.t.\theta \] \[\frac{{dy}}{{d\theta }} = a\left[ {\cos \theta - \left( {\theta \frac{d}{{d\theta }}\cos \theta + \cos \theta \frac{d}{{d\theta }}\theta } \right)} \right]\] \[ = a\left[ {\cos \theta - \theta ( - \sin \theta ) - \cos \theta } \right]\] \[ = a\theta \sin \theta \] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\] \[ = \frac{{a\theta \sin \theta }}{{a\theta \cos \theta }} = \tan \theta \]

Question (11)

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

Solution

\[x = \sqrt {{a^{{{\sin }^{ - 1}}t}}} \] \[x = {\left( {{a^{{{\sin }^{ - 1}}t}}} \right)^{1/2}} = {a^{\frac{{{{\sin }^{ - 1}}t}}{2}}}\] \[Take\;\log \;on\;both\;sides\] \[\log x = \frac{{{{\sin }^{ - 1}}t}}{2}\log a\] \[Diff.w.r.t.t,\] \[\frac{1}{x}\frac{{dx}}{{dt}} = \frac{{\log a}}{2}\frac{d}{{dt}}{\sin ^{ - 1}}t\] \[\frac{{dx}}{{dt}} = \frac{{x\log a}}{2} \times \frac{1}{{\sqrt {1 - {t^2}} }}\] \[y = = \sqrt {{a^{{{\cos }^{ - 1}}t}}} \] \[y = {\left( {{a^{{{\cos }^{ - 1}}t}}} \right)^{1/2}} = {a^{\frac{{{{\cos }^{ - 1}}t}}{2}}}\] \[Take\;\log \;on\;both\;sides\] \[\log y = \frac{{{{\cos }^{ - 1}}t}}{2}\log a\] \[Diff.w.r.t.t,\] \[\frac{1}{y}\frac{{dy}}{{dt}} = \frac{{\log a}}{2}\frac{d}{{dt}}{\cos ^{ - 1}}t\] \[\frac{{dy}}{{dt}} = \frac{{y\log a}}{2} \times \frac{{ - 1}}{{\sqrt {1 - {t^2}} }}\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\] \[ = \frac{{\frac{{y\log a}}{2} \times \frac{{ - 1}}{{\sqrt {1 - {t^2}} }}}}{{\frac{{x\log a}}{2} \times \frac{1}{{\sqrt {1 - {t^2}} }}}}\] \[ = - \frac{y}{x}\]