Differentiation (Part-38) | S N De

$~6(i)~~x=a\left(\cos t+\log\tan\frac t2\right),~~y=a\sin t.$

Sol. $~~x=a\left(\cos t+\log\tan\frac t2\right) \\ \therefore \frac{dx}{dt}=a\left[\frac{d}{dt}(\cos t)+\frac{d}{dt}(\log\tan\frac t2) \right] \\ \text{or,}~~  \frac{dx}{dt}\\=a\left[-\sin t+\frac{1}{\tan(t/2)}\frac{d}{dt}(\tan(t/2))\right]\\=a\left[-\sin t+\frac{1}{\tan(t/2)} \cdot \sec^2(t/2)~\frac{d}{dt}(t/2)\right]\\=a\left[-\sin t+\frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)}~\cdot\frac 12\right]\\=a\left[-\sin t +\frac{1}{2\sin(t/2)\cos(t/2)}\right]\\=a\left[-\sin t+\frac{1}{\sin t}\right]\\=a\left[\frac{1}{\sin t}-\sin t\right]\\=a\left[\frac{1-\sin^2t}{\sin t}\right]\\=a\left[\frac{\cos^2t}{\sin t}\right] \\ \therefore \frac{dt}{dx}=\frac{\sin t}{a\cos^2t}\rightarrow(1)\\~~\text{Again,}~~y=a\sin t \\ \therefore \frac{dy}{dt}=a\frac{d}{dt}(\sin t)=a\cos t\rightarrow(2)$

So, $~~~~\frac{dy}{dx}\\=\frac{dy}{dt}\cdot \frac{dt}{dx}\\=a\cos t \cdot \frac{\sin t}{a \cos^2 t}~~[\text{By (1),(2)}]\\=\frac{\sin t}{\cos t}\\=\tan t~~\text{(ans.)}$

$~(ii)~~x=a(t-\sin t),~y=a(1-\cos t)~~\text{at}~~t=\pi/2$

Sol. $~~x=a(t-\sin t)\\ \therefore \frac{dx}{dt}\\=a~\frac{d}{dt}(t-\sin t)\\=a\left[\frac{d}{dt}(t)-\frac{d}{dt}(\sin t)\right]\\=a~(1-\cos t) \\ \therefore \frac{dt}{dx}=\frac{1}{a(1-\cos t)}\rightarrow(1)$

Again, $~~y=a(1-\cos t) \\ \therefore \frac{dy}{dt}\\=a~\frac{d}{dt}(1-\cos t)\\=a~\left[\frac{d}{dt}(1)-\frac{d}{dt}(\cos t)\right]\\=a[0-(-\sin t)]\\=a\sin t\rightarrow(2)\\~\therefore \frac{dy}{dx}\\=\frac{dy}{dt} \cdot \frac{dt}{dx}\\=\frac{a\sin t}{a(1-\cos t)}~~[\text{By (1),(2)}]\\=\frac{\sin t}{1-\cos t}\\ \text{So,}~~\left[\frac{dy}{dx}\right]_{t=\pi/2}=\frac{\sin(\pi/2)}{1-\cos(\pi/2)}=\frac{1}{1-0}=1~~\text{(ans.)} $

$~(iii)~~\tan y=\frac{2t}{1-t^2},~~\sin x=\frac{2t}{1+t^2}$

Sol. $~~\tan y=\frac{2t}{1-t^2}\\ \therefore~y=\tan^{-1}\left(\frac{2t}{1-t^2}\right)\\ \text{or,}~~ y=\tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2{\theta}}\right) \\~~~[~~\text{let}~~t=\tan \theta~~] \\ \text{or,}~~  y=\tan^{-1}(\tan2\theta) \\ \text{or,}~~  y=2\theta \\ \text{or,}~~  \frac{dy}{d\theta}=2 \rightarrow(1)$

Again, $~~\sin x=\frac{2t}{1+t^2} \\ \therefore x=\sin^{-1}\left(\frac{2t}{1+t^2}\right) \\ \text{or,}~~ x=\sin^{-1}\left(\frac{2\tan \theta}{1+\tan^2\theta}\right) \\ \text{or,}~~ x=\sin^{-1}(\sin 2\theta) \\ \text{or,}~~ x=2\theta \\ \therefore \frac{dx}{d\theta}=2 \\ \text{or,}~~  \frac{d\theta}{dx}=\frac 12 \rightarrow(2)$

So, $~~\frac{dy}{dx}\\=\frac{dy}{d\theta}\cdot \frac{d\theta}{dx}\\=2 \cdot \frac 12~~~[\text{By (1),(2)}]\\=1~~(\text{ans})$

$~(iv)~~x=a(2t+\sin2t),~~y=a(1-\cos2t)$

Sol. $~~x=a(2t+\sin2t)\\ \therefore \frac{dx}{dt}=a~\frac{d}{dt}(2t+\sin2t)\\ \text{or,}~~\frac{dx}{dt}\\=a~\left[\frac{d}{dt}(2t)+\frac{d}{dt}(\sin2t)\right]\\=a\left[2+\cos2t \cdot~\frac{d}{dt}(2t)\right]\\=a(2+2\cos2t)\\=2a(1+\cos2t)\\ \therefore \frac{dt}{dx}=\frac{1}{2a(1+\cos2t)}\rightarrow(1)$

Again, $~~y=a(1-\cos 2t) \\ \therefore \frac{dy}{dt}=a~\frac{d}{dt}(1-\cos2t)\\~\text{or,}~~\frac{dy}{dt}\\=a~\left[\frac{d}{dt}(1)-\frac{d}{dt}(\cos2t)\right]\\=a[0-(-\sin2t)~\frac{d}{dt}(2t)]\\=2a\sin2t \rightarrow(2)$

So, $~~\frac{dy}{dx}\\=\frac{dy}{dt} \cdot \frac{dt}{dx}\\=\frac{2a\sin2t}{2a(1+\cos2t)}~~[\text{By (1),(2)}]\\=\frac{\sin2t}{1+\cos2t}\\=\frac{2\sin t\cos t}{2\cos^2t}\\=\frac{\sin t}{\cos t}\\=\tan t~~\text{(ans.)}$

$~(v)~~x=\sec^{-1}\frac{1+t^2}{1-t^2},~y=\tan^{-1}\frac{3t-t^3}{1-3t^2}$

Sol. $~~x=\sec^{-1}\frac{1+t^2}{1-t^2}\rightarrow(1)\\ y=\tan^{-1}\frac{3t-t^3}{1-3t^2}\rightarrow(2)$

Let us put $~t=\tan \theta~$ in $\,(1),\,(2)~$ and thus get,

$~~~~~~~x=\sec^{-1}\left(\frac{1+\tan^2\theta}{1-\tan^2\theta}\right)\\ \text{or,}~~x=\sec^{-1}\left[\frac{1}{\frac{1-\tan^2\theta}{1+\tan^2\theta}}\right] \\ \text{or,}~~x=\sec^{-1}\left(\frac{1}{\cos2\theta}\right)\\ \text{or,}~~x=\sec^{-1}(\sec2\theta) \\  \text{or,}~~x=2\theta \\ \therefore \frac{dx}{d\theta}=2\rightarrow(3)$

Similarly, $~~y=\tan^{-1}\left(\frac{3\tan\theta-\tan^2\theta}{1-3\tan^2\theta}\right) \\ \therefore y=\tan^{-1}(\tan3\theta) \\ \text{or,}~~y=3\theta \\ \text{or,}~~\frac{dy}{d\theta}=3 \rightarrow(4)$

Hence, $~~\frac{dy}{dx}\\=\frac{dy}{d\theta} \cdot\frac{d\theta}{dx}\\=3 \cdot\frac 12 ~~~[~\text{By (3),(4)}~]\\=\frac 32~~\text{(ans.)}$

$~(vi)~~x=\cos^{-1}(8t^4-8t^2+1),\\ ~~~~~~~~~y=\sin^{-1}(3t-4t^3),~~[~0<t<\frac 12~].$

Sol.  $~~~x=\cos^{-1}(8t^4-8t^2+1)\\ \text{or,}~~  x=\cos^{-1} [2(4t^4-4t^2+1)-1] \\ \text{or,}~~  x=\cos^{-1}[2(2t^2-1)^2-1] \\ \text{or,}~~  x=\cos^{-1}[2(2\cos^2\theta-1)^2-1]\\~~~[~~\text{let}~t=\cos\theta] \\ \text{or,}~~  x=\cos^{-1}[2\cos^22\theta-1] \\ \text{or,}~~  x=\cos^{-1}(\cos4\theta) \\ \text{or,}~~  x=4\theta \\ \therefore ~~\frac x4=\cos^{-1}t \rightarrow(1) $

$~~~~y=\sin^{-1}(3t-4t^3) \\ \text{or,}~~  y=\sin^{-1}(3\sin\phi-4\sin^3\phi)\\~~~[~\text{where}~~t=\sin\phi] \\ \text{or,}~~ y=\sin^{-1}(\sin3\phi) \\ \text{or,}~~  y=3\phi \\ \text{or,}~~  \frac y3=\phi=\sin^{-1}t \rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~~~~\frac x4+\frac y3=\cos^{-1}t+\sin^{-1}t \\ \text{or,}~~  \frac x4+\frac y3=\frac{\pi}{2}\\~~\therefore~~\text{Diff. w.r.t.}~~x,\\~~~\frac 14+\frac 13~\frac{dy}{dx}=0 \\ \text{or,}~~  \frac 13~\frac{dy}{dx}=-\frac 14 \\ \text{or,}~~  \frac{dy}{dx}=-\frac 34~~\text{(ans.)}$