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

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

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

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

Question (7)

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

Solution

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

Question (8)

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

Solution

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

Question (9)

x = a secθ , y = b tan θ

Solution

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

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

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