Differentiation (Part-38) | S N De

 

$~18(i)~~x=a\cos\theta,~y=b\sin\theta$

Sol. $~~~~~x=a\cos\theta \\ \therefore~\frac{dx}{d\theta}=a~\frac{d}{d\theta}(\cos\theta)=-a\sin\theta\rightarrow(1)$

Again, $~~y=b\sin\theta \\ \therefore \frac{dy}{d\theta}=b~\frac{d}{d\theta}(\sin\theta)=b\cos\theta\rightarrow(2)$

So, $~\frac{dy}{dx}\\=\frac{dy}{d\theta} ~\cdot\frac{d\theta}{dx}\\=b\cos\theta~\cdot \frac{1}{(-a\sin\theta)}~~[\text{By (1),(2)}]\\=-\frac ba~\cot\theta~~\text{(ans.)}$

$~(ii)~~x=a\cos(\theta/2),~y=b\sin(\theta/2)$

Sol. $~~x=a\cos(\theta/2) \\ \therefore \frac{dx}{d\theta}\\=a~\frac{d}{d\theta}[\cos(\theta/2)]\\=a[-\sin(\theta/2)]~\frac{d}{d\theta}(\theta/2)\\=-a\sin(\theta/2)~\cdot \frac 12\\=-\frac a2\sin(\theta/2)\rightarrow(1)$

Again, $~~y=b\sin(\theta/2) \\ \therefore~\frac{dy}{d\theta}\\=b~\frac{d}{d\theta}[\sin(\theta/2)]\\=b~\cos(\theta/2)~\frac{d}{d\theta}(\theta/2)\\=b\cos(\theta/2)~\cdot \frac 12\\=\frac b2 \cos(\theta/2)\rightarrow(2)$

So, $~\frac{dy}{dx}\\=\frac{dy}{d\theta} \cdot \frac{d\theta}{dx}\\=\frac b2\cos(\theta/2) \cdot \frac{1}{-\frac a2 \sin(\theta/2)}~~[\text{By (1),(2)}]\\=-\frac ba~\cot(\theta/2)~~\text{(ans.)}$

$~(iii)~~x=at^2,~y=2at~~~~~~\text{[NCERT]}$

Sol. $~~x=at^2 \\ \therefore \frac{dx}{dt}\\=a~\frac{d}{dt}(t^2)\\=a \cdot(2t)\\=2at\rightarrow(1)$

Again, $~~y=2at \\ \therefore~\frac{dy}{dt}\\=2a~\frac{d}{dt}(t)\\=2a \cdot 1\\=2a \rightarrow(2)$

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

$~(iv)~~x=a\cos^3 \theta,~~y=a\sin^3\theta$

Sol. $~~x=a\cos^3\theta \\ \therefore~\frac{dx}{d\theta}\\=a\frac{d}{d\theta}(\cos^3\theta)\\=a~(3\cos^2\theta)~\frac{d}{d\theta}(\cos\theta)\\=3a\cos^2\theta~(-\sin\theta)\\=-3a\sin\theta\cos^2\theta\rightarrow(1)$

Again, $~y=a\sin^3\theta \\ \therefore~\frac{dy}{d\theta}\\=a~\frac{d}{d\theta}(\sin^3\theta)\\=a~(3\sin^2\theta)~\frac{d}{d\theta}(\sin\theta)\\=3a\sin^2\theta~\cos\theta \rightarrow(2)$

So, $~\frac{dy}{dx}\\=\frac{dy}{d\theta}~\cdot \frac{d\theta}{dx}\\=(3a\sin^2\theta~\cos\theta)~\cdot \frac{1}{-3a\sin\theta\cos^2\theta}\\~~~~~[\text{By (1),(2)}]\\=-\frac{\sin\theta}{\cos\theta}\\=-\tan\theta~~\text{(ans.)}$

$~(v)~~x=t~\log t,~~y=\frac{\log t}{t}~~\text{when}~~t=1.$

Sol. $~~x=t~\log t \\ \therefore~\frac{dx}{dt}\\=\frac{d}{dt}(t~\log t)\\=\frac{d}{dt}(t)~\cdot \log t+\frac{d}{dt}(\log t)~\cdot t\\=1 \cdot \log t+\frac 1t~\cdot t\\=\log t+1 \rightarrow(1)$

Again, $~y=\frac{\log t}{t} \\ \therefore~\frac{dy}{dt}\\=\frac{d}{dt}\left(\frac{\log t}{t}\right)\\=\frac{t\cdot \frac{d}{dt}(\log t)-\log t~\cdot \frac{d}{dt}(t)}{t^2}\\=\frac{t \cdot \frac 1t-\log t\cdot 1}{t^2}\\=\frac{1-\log t}{t^2}\rightarrow(2)$

So, $~\frac{dy}{dx}\\=\frac{dy}{dt}~\cdot \frac{dt}{dx}\\=\frac{1-\log t}{t^2}~\cdot \frac{1}{\log t+1 }~~~[\text{By (1),(2)}]\\=\frac{1-\log t}{t^2(1+\log t)}\rightarrow(3)$

Hence, $~\left[\frac{dy}{dx}\right]_{t=1}\\=\frac{1-\log 1}{1^2(1+\log 1)}~~[\text{By (3)}]\\=\frac{1-0}{1(1+0)}\\=1~~\text{(ans.)}$