Question 15 Marks
If $\text{x}=3\cot-2\cos^3\text{t},\text{y}=3\sin\text{t}-2\sin^3\text{t}$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
AnswerGiven,
$\text{x}=3\cot-2\cos^3\text{t},$
$\text{y}=3\sin\text{t}-2\sin^3\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}-6\cos^2\text{t}(-\sin\text{t})$
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}+6\cos^2\text{t}\sin\text{t}$
And $\text{y}=3\sin\text{t}-2\sin^2\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dy}}{\text{dt}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
Now,
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot-2\sin^2\text{t}\cos\text{t}}{-\sin\text{t}+2\cos^2\text{t}\sin\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot[1-2\sin^2\text{t]}}{\sin\text{t}[2\cos^2\text{t}-1]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cot\text{t}$
Differentiating both w.r.t. x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}(\cot\text{x})}{\text{dx}}=-\text{cosec}^2\text{x}$
View full question & answer→Question 25 Marks
If $\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},$ prove that $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0.$
AnswerWe have,
$\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},...(1)$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{nx}^{\text{n}-1}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}+\text{x}^\text{n}\\\Big\{-\text{a}\sin(\log\text{x})\times\frac{1}{\text{x}}+\text{b}\cos(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text {x})\}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}$
$\Rightarrow\frac{\text{n}}{\text{x}}\text{y}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\ [\text{from}(1)]$
$\Rightarrow\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}=\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}...(2)$
Differentiating $\frac{\text{dy}}{\text{dx}}$ with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\text{x}^{\text{n}-2}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\+\text{x}^{\text{n}-1}\Big\{-\text{a}\cos(\log\text{x})\times\frac{1}{\text{x}}-\text{b}\sin(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\frac{\text{dx}}{\text{dy}}-\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\frac{\text{x}^{\text{n}-1}}{\text{x}}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\-\frac{\text{x}^\text{n}}{\text{x}^2}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}$
$\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{ny}}{\text{x}^2}+\Big(\frac{\text{n}-1}{\text{x}}\Big)\Big(\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}\Big)-\frac{\text{y}}{\text{x}^2}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{n}+\text{n}-1}{\text{x}}\Big)-\frac{(\text{n}+\text{n}^2+\text{n}+1)\text{y}}{\text{x}^2}$
$=\Big(\frac{2\text{n}-1}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{(\text{n}^2+1)\text{y}}{\text{x}^2}$
$\Rightarrow\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}(2\text{n}-1)\frac{\text{dy}}{\text{dx}}+(\text{n}^2+1)\text{y}=0$
Hence, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\text{x}\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0$
View full question & answer→Question 35 Marks
If $\text{x}=\cos\text{t}+\log\tan\frac{\text{t}}{2},\text{y}=\sin\text{t},$ Then find the value of $\frac{\text{d}^2\text{y}}{\text{dt}^2}\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{4}.$
Answer$\text{x}=\cos\text{t}+\log\tan\frac{\text{t}}{2},\text{y}=\sin\text{t}$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)=-\sin\text{t}+\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}$
$=-\sin\text{t}+\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}=-\sin\text{t}+\frac{1}{\sin\text{t}}$
$=\frac{-\sin^2\text{t}+1}{\sin\text{t}}=\frac{-\sin^2\text{t}+1}{\sin\text{t}}$
$=\frac{\cos^2\text{t}}{\sin\text{t}}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\sin\text{t})=\cos\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dt}}\Big)=\frac{\text{d}}{\text{dt}}(\cos\text{t})=-\sin\text{t}$
$\Big(\frac{\text{d}^2\text{y}}{\text{dt}^2}\Big)_{(\text{t}=\frac{\pi}{4})}=-\sin\Big(\frac{\pi}{4}\Big)=-\frac{1}{\sqrt{2}}...(1)$
Also, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\cos\text{t}}{\frac{\cos^2\text{t}}{\sin\text{t}}}=\frac{\sin\text{t}}{\cos\text{t}}=\tan\text{t}$
Now, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{x})$
$\frac{\text{d}}{\text{dt}}(\tan\text{t})\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{\sin\text{t}}{\cos^2\text{t}}$
$=\frac{\sin\text{t}}{\cos^4\text{t}}$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{4}}=\frac{\sin\Big(\frac{\pi}{4}\Big)}{\cos^4\Big(\frac{\pi}{4}\Big)}=2\sqrt{2}...(2)$
Hence, at $\text{t}=\frac{\pi}{4},\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\frac{1}{\sqrt{2}}\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sqrt{2}$
View full question & answer→Question 45 Marks
If $\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{x}^2+\text{y}^2}{\text{y}^2}$
AnswerWe have$\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\text{t}-\text{b}\cos\text{t})=\text{a}\cos\text{t}+\text{b}\sin\text{t}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})=-\text{a}\sin\text{t}+\text{b}\cos\text{t}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos+\text{b}\sin\text{t}}$
Therefore
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos\text{t}+\text{b}\sin\text{t}}\Big)$
$=\frac{\text{d}}{\text{dt}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\sin\text{t}+\text{b}\sin\text{t}}\Big)\times\frac{\text{dt}}{\text{dx}}$
$=\frac{(\text{a}\cos\text{t}+\text{b}\sin\text{t})\frac{\text{d}}{\text{dt}}(-\text{a}\sin\text{t}+\text{b}\cos\text{t})-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2}$
$$$=\frac{(\text{a}\cos\text{t}+\text{b}\sin\text{t})(\text{a}\cos\text{t}+\text{b}\sin\text{t})(-\text{a}\sin\text{t}+\text{b}\cos\text{t})-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2}$
$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$
$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$
$=\frac{-\text{y}^2-\text{x}^2}{\text{y}^3}$
View full question & answer→Question 55 Marks
If $\text{y}=\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n},}$ prove that $(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}-\text{n}^2\text{y}=0.$
AnswerWe have
$\text{y}=\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n},}$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(1+\frac{1}{2\sqrt{\text{x}^2+1}\times2\text{x}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(1-\frac{1}{2\sqrt{\text{x}^2+1}\times2\text{x}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(1+\frac{\text{x}}{\sqrt{\text{x}^2+1}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(1-\frac{\text{x}}{\sqrt{\text{x}^2+1}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\sqrt{\text{x}^2+1}-\text{x}}{\sqrt{\text{x}^2+1}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\text{x}+\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big)+\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(\frac{\text{x}\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big)$
$=\Big\{\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\Big\}^{-\text{n}}\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)$
$=\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)\text{y}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\text{ny}$
Squaring both sides, we get
$(\text{x}^2+1)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{n}^2\text{y}^2...(2)$
Differentiating (2) with respect to x , we get
$(\text{x}^2+1)2\frac{\text{dy}}{\text{dx}}\times\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{n}^2\Big(2\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2(\text{y})$
$\Rightarrow(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2\text{y}=0$
Hence, $(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2\text{y}=0$
View full question & answer→Question 65 Marks
If $\text{x}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}),$evaluate $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{3}.$
AnswerWe have,
$\text{x}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}),$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\Big]=\text{a}\Bigg(-\sin\text{t}+\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}\Bigg)$
$=\text{a}\Bigg(-\sin\text{t}+\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}\Bigg)=\text{a}\Big(-\sin\text{t}+\frac{1}{\sin\text{t}}\Big)$
$=\text{a}\Big(\frac{-\sin^2\text{t}+1}{\sin\text{t}}\Big)=\text{a}\Big(\frac{\cos^2\text{t}}{\sin\text{t}}\Big)$
And
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\text{t})=\text{a}\cos\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dX}}{\text{dt}}}=\frac{\text{a}\cos\text{t}}{\text{a}\frac{\cos^2\text{t}}{\sin\text{t}}}=\tan\text{t}$
Therefore
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan(\text{t}))$
$=\frac{\text{d}}{\text{dt}}(\tan(\text{t}))\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{\sin\text{t}}{\text{a}\cos^2\text{t}}$
$=\Big(\frac{\sin\text{t}}{\text{a}\cos^4\text{t}}\Big)$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{3}}=\Bigg(\frac{\sin\Big(\frac{\pi}{3}\Big)}{\text{a}\cos^4\Big(\frac{\pi}{3}\Big)}\Bigg)=\frac{\frac{\sqrt{3}}{2}}{\text{a}\Big(\frac{1}{16}\Big)}=\frac{8\sqrt{3}}{\text{a}}$
Hence, $\text{at}\ \text{t}=\frac{\pi}{3},\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{8\sqrt{3}}{\text{a}}$
View full question & answer→Question 75 Marks
Find A and B so that $\text{y}=\text{A}\sin3\text{x}+\text{B}\cos3\text{x}$ satisfy the equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\frac{\text{dy}}{\text{dx}}+3\text{y}=10\cos3\text{x}.$
Answer$\text{y}=\text{Ae}^{-kt}\cos(\text{pt}+\text{c})$ Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\text{A}\Big\{\text{e}^{-\text{kt}}(-\sin(\text{pt}+\text{c})\times\text{p})+(\cos(\text{pt}+{c}))(-\text{re}^{-\text{kt}})\Big\}$
$\Rightarrow-\text{Ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{KAe}^{-\text{kt}}\cos(\text{pt}+\text{c})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=-\text{Ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{ky}$ Differentiating w.r.t.x, $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{Ap}\Big\{\text{e}^{-\text{kt}}(\cos(\text{pt}+\text{c})\times\text{p})+(\sin(\text{pt}+\text{c}))(\text{e}^{-\text{kt}}\times-\text{R})-\text{ky}^1\Big\}$
$=-\text{p}^2\text{y}+\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{ky}^1$
Adding & substracting $ky_1$ on RHS $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=+\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{y}-2\text{ky}^1+\text{ky}^1$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{y}-2\text{ky}^1-\text{k}\text{ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{k}^2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-(\text{p}+\text{k}^2)\text{y}-2\text{k}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+\text{n}^2\text{y}=0$
View full question & answer→Question 85 Marks
If $\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1+\cos\theta)$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
Answer$\text{x} =\text{a} (\theta -\sin\theta)\dots\text{ eq. } 1$
$\text{y}=\text{a}(1+\cos\theta)\dots\text{ eq. 2}$
To find: $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and differentiate it again.
$\frac{\text{dx}\text{}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(\theta-\sin\theta)=\text{a}(1-\cos\theta)\dots\ \text{eq. 3}$
Similarly,
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(1+\cos\theta)=-\text{a}\sin\theta\dots\text{ eq. }4$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}=\frac{-\sin\theta}{(1-\cos\theta)}\dots\ \text{eq. }5$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=-\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\theta}{1-\cos\theta}\Big)$
Using product rule and chain rule of differentiation together:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{-\frac{1}{1-\cos\theta}\frac{\text{d}}{\text{d}\theta}\sin\theta-\sin\theta\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}\Big\}\frac{\text{d}\theta}{\text{dx}}$
Apply chain rule to determine $\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}}=\Big\{\frac{-\cos\theta}{1-\cos\theta}+\frac{\sin^2\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}$ [using eq.3]
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta(1-\cos\theta)+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta+\cos^2\theta+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{\frac{1-\cos\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}[\because\cos^2\theta+\sin^2\theta=1]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a(1}-\cos\theta)^2}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a}\Big(2\sin^2\frac{\theta}{2}\Big)^2}\big[\because-\cos\theta=2\sin^2\frac{\theta}{2}\big]$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{4a}}\text{cosec}^4\frac{\theta}{2}$
View full question & answer→Question 95 Marks
If $\text{x}=\text{a}\sin\text{t}\ \text{and}\ \text{y}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2}),$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
Answer$\text{x}=\text{a}\sin\text{t}\ \text{and}\ \text{y}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2}),$
$\frac{\text{dx}}{\text{dt}}=\text{a}\cos\text{t}$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{a}\sin\text{t}$
$\frac{\text{dy}}{\text{dt}}=-\text{a}\sin\text{t}+\text{a}\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}$
$=-\text{a}\sin\text{t}+\text{a}\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}$
$=-\text{a}\sin\text{t}+\text{a}\ \text{cosec}\ \text{t}$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{a}\cos\text{t}-\text{a cosec t}\cot\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\frac{\text{dx}}{\text{dt}}\frac{\text{d}^2\text{y}}{\text{dt}^2}-\frac{\text{dy}}{\text{dt}}\frac{\text{d}^2\text{x}}{\text{dt}^2}}{\Big(\frac{\text{dx}}{\text{dt}}\Big)^3}$
$=\frac{\text{a}\cos\text{t}(-\text{a}\cos\text{t}-\text{a}\ \text{cosec t}\cot\text{t})-(-\text{a}\sin\text{t}+\text{a}\text{cosec t})(-\text{a}\sin\text{t})}{(\text{a}\cos\text{t})^3}$
$=\frac{-\text{a}^2\cos^2\text{t}-\text{a}^2\cot^2\text{t}-\text{a}^2\sin^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=\frac{-\text{a}^2\cos^2\text{t}-\text{a}^2\sin^2\text{t}-\text{a}^2\cot^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=\frac{-\text{a}^2(\cos^2\text{t}+\sin^2\text{t})-\text{a}^2\cot^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=-\frac{1}{\text{a}\sin^2\text{t}\cos\text{t}}$
View full question & answer→Question 105 Marks
Find the second order derivatives of the following functions:
$\text{e}^{6\text{x}} \cos \text{x}$
AnswerLet $\text{y}=\text{e}^{6\text{x}}\cos3\text{x}$
Then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(6\text{x})+\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(3\text{x})$
$=\cos3\text{x}.\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(6\text{x})+\text{e}^{6\text{x}}.(-\sin3\text{x}).\frac{\text{d}}{\text{dx}}(\cos3\text{x})$
$=6\text{e}^{6\text{x}}\cos3\text{x}-3\text{e}^{6\text{x}}\sin3\text{x}...(1)$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(6\text{e}^{6\text{x}}\cos3\text{x}-3\text{e}^{6\text{x}}\sin3\text{x})$
$=6.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}\cos3\text{x})-3.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}\sin3\text{x})$
$=6.[6\text{e}^{6\text{x}}\cos3\text{x}-3\text{e}^{6\text{x}}\sin3\text{x}]$
$-3.\Big[\sin3\text{x}.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}})+\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}\sin3\text{x})$
$=36\text{e}^{6\text{x}}\cos3\text{x}-18\text{e}^{6\text{x}}\sin3\text{x}-3[\sin3\text{x}.\text{e}^{6\text{x}}.\cos3\text{x}.3$
$=36\text{e}^{6\text{x}}\cos3\text{x}-18\text{e}^{6\text{x}}\sin3\text{x}-18\text{e}^{6\text{x}}\sin3\text{x}-9\text{e}^{6\text{x}}\cos3\text{x}$
$=27\text{e}^{6\text{x}}\cos3\text{x}-36\text{e}^{6\text{x}}\sin3\text{x}$
$=9\text{e}^{6\text{x}}(3\cos3\text{x}-4\sin3\text{x})$
View full question & answer→Question 115 Marks
Find $\text{y}=\text{Ae}^{-\text{kt}}\cos\text({pt}+\text{c})$prove that $\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+\text{n}^2\text{y}=0,$Where $\text{n}^2=\text{p}^2+\text{k}^2.$
AnswerWe have,
$\text{y}=\text{Ae}^{-\text{kt}}\cos\text({pt}+\text{c})...(1)$
Differentiating y with respect to t, we get
$\frac{\text{dy}}{\text{dt}}=-\text{KAe}^{-\text{kt}}\cos(\text{pt}+\text{c})-\text{PAe}^{-\text{kt}}\sin(\text{pt}+\text{c})$
$=-\text{ky}-\text{PAe}^{-\text{kt}}\sin(\text{pt}+\text{c})\ [\text{from}(1)]$
$\Rightarrow\text{pAe}^{-\text{kt}}\sin(\text{pt}+\text{c})=-\text{ky}-\frac{\text{dy}}{\text{dt}}...(2)$
Differentiating $\frac{\text{dy}}{\text{dt}}$ with respect to t, we get
$\frac{\text{d}^2\text{y}}{\text{dt}}=-\text{k}\frac{\text{dy}}{\text{dt}}+\text{pkAe}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{Ae}^{-\text{kt}}\cos(\text{pt}+\text{c})$
$=-\text{k}\frac{\text{dy}}{\text{dt}}+\text{k}\Big(-\text{ky}-\frac{\text{dy}}{\text{dt}}\Big)-\text{p}^2\text{y}\ [\text{from}(1)\ \text{and}\ (2)]$
$=-\text{k}\frac{\text{dy}}{\text{dt}}-\text{k}^2\text{y}-\text{k}\frac{\text{dy}}{\text{dt}}-\text{p}^2\text{y}$
$=-2\text{k}\frac{\text{dy}}{\text{dt}}-(\text{k}^2+\text{p}^2)\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+(\text{k}^2+\text{p}^2)\text{y}=0$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}} \text{n}^2\text{y}=0,$ where $\text{n}^2=\text{p}^2+\text{k}^2.$
Hence,
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}} \text{n}^2\text{y}=0,$ where $\text{n}^2=\text{p}^2+\text{k}^2.$
View full question & answer→Question 125 Marks
If $\text{x}=\text{a}(\cos\theta+\theta\sin\theta),\text{y}=\text{a}(\sin\theta-\theta\cos\theta)$ prove that $\frac{\text{d}^2\text{x}}{\text{d}\theta^2}=\text{a}(\cos\theta-\theta\sin\theta),\frac{\text{d}^2}{\text{d}\theta^2}$ $=\text{a}(\sin\theta-\theta\cos\theta)\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sec^3\theta}{\text{a}\theta}$
AnswerIt is given that, $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t}),\text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t})$
$\therefore\frac{\text{dx}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\cos\text{t}+\text{t}\sin\text{t})$
$=\text{a}\Big[-\sin\text{t}+\sin\text{t}.\frac{\text{d}}{\text{dt}.}(\text{t})+\frac{\text{d}}{\text{dt}}(\sin\text{t})\Big]$
$=\text{a}[-\sin\text{t}+\sin\text{t}+t\cos\text{t}]=\text{at}\cos\text{t}$
$\frac{\text{dy}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\sin\text{t}-\text{t}\cos\text{t})$
$=\text{a}\Big[\cos\text{t}-\Big\{\cos\text{t}\frac{\text{d}}{\text{dt}}(\text{t})+\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})\Big\}\Big]$
$=\text{a}[\cos\text{t}-\{\cos\text{t}-\text{t}\sin\text{t}\}]=\text{at}\sin\text{t}$
then, $\therefore\frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})=\sec^2\text{t}.\frac{\text{dt}}{\text{dx}}$
$=\sec^2\text{t}.\frac{1}{\text{at}\cos\text{t}}\ \Big[\frac{\text{dx}}{\text{dt}}=\text{at}\cos\text{t}\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\text{at}\cos\text{t}}\Big]$
$=\frac{\sec^3\text{t}}{\text{at}},0<\text{t}<\frac{\pi}{2}$
View full question & answer→Question 135 Marks
If $\text{y}\log(1+\cos\text{x}),$ prove that $\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^\text{y}}{\text{dx}^2}.\frac{\text{d}\text{y}}{\text{dx}}=0$
Answer$\text{y}\log(1+\cos\text{x}),$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\cos\text{x}}\times-\sin\text{x}=\frac{-\sin\text{x}}{1+\cos\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{(1+\cos\text{x})\cos\text{x}-\sin(-\sin\text{x})}{(1+\cos\text{x})^2}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]=-\Big[\frac{1+\cos\text{x}}{(1+\cos\text{x})^2}\Big]=\frac{-1}{1+\cos\text{x}}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^3}=-\Big(\frac{+1}{(1+\cos\text{x})^2}\times+\sin\text{x}\Big)\\=-\Big(\frac{-\sin\text{x}}{1+\cos\text{x}}\Big)\times\Big(\frac{-1}{1+\cos\text{x}}\Big)=-\frac{\text{dy}}{\text{dx}}.\frac{\text{d}^2\text{y}}{\text{dx}^2}$
$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}}{\text{dx}^2}.\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 145 Marks
If $\text{x}=\sin\text{t},\text{y}=\sin\text{pt}$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{y}=0$
AnswerHere,
$\text{x}=\sin\text{t},\text{y}=\sin\text{pt}$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{dt}}=\cos\text{t}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{o}\cos\text{pt}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{p}\cos\text{pt}}{\cos\text{t}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{o}^2\sin\text{pt}\cos\text{t}+\text{p}\cos\text{pt}\sin\text{t}}{\cos^3\text{t}}\times\frac{\text{dt}}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{o}^2\sin\text{pt}\cos\text{t}+\text{p}\cos\text{pt}\sin\text{t}}{\cos^3\text{t}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{p}\sin\text{pt}\cos\text{t}}{\cos^3\text{t}}+\frac{\text{p}\cos\text{pt}\sin\text{t}}{\cos^3\text{t}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{p}^2\text{y}}{\cos^2\text{t}}+\frac{\text{x}\frac{\text{dy}}{\text{dx}}}{\cos^3\text{t}}$
$\Rightarrow\cos^2\text{t}\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{p}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\sin^2\text{t})\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{p}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{y}=0$
View full question & answer→Question 155 Marks
If $\text{y}=\text{e}^{2\text{x}}(\text{ax}+\text{b}),$ show that $\text{y}_2-\text{4}\text{y}_1+4\text{y}=0$
Answer$\text{y}=\text{e}^{2\text{x}}(\text{ax}+\text{b}),$ Differentiating w.r.t.x, $\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{2\text{x}}(\text{a})+2(\text{ax}+b)(\text{e}^{2\text{x}})$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{ae}^{2\text{x}}+2\text{y} $ Differentiating w.r.t.x,$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{ae}^{2\text{x}}+2\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{dy}}{\text{dx}}+2\text{ae}^{2\text{x}}+4\text{y}-4\text{y}=2\frac{\text{dy}}{\text{dx}}+2\frac{\text{dy}}{\text{dx}}-4\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-4\frac{\text{dy}}{\text{dx}}+4\text{y}=0$
$\Rightarrow\text{y}_2-4\text{y}_1+4\text{y}=0$
Hence proved
View full question & answer→Question 165 Marks
Find the second order derivatives of the following functions:$\log(\sin\text{x})$
AnswerWe have,
$\text{y}=\text{e}^\text{e}\sin(5\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}\sin5\text{x}+\text{e}^\text{x}\cos5\text{x}\times5$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^\text{x}\sin5\text{x}+\text{e}^\text{x}\cos5\text{x}\times5+5\text{e}^\text{x}\cos5\text{x}$
$=-24\text{e}^\text{x}\sin5\text{x}+10\text{e}^\text{x}\cos5\text{x}$
$=2\text{e}^\text{x}(5\cos5\text{x}-12\sin5\text{x})$
View full question & answer→Question 175 Marks
If $\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
AnswerHere,
$\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sec\theta\tan\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=b\sec^2\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{b}\sec^2\theta}{\text{a}\sec\theta\tan\theta}=\frac{\text{b}\ \text{cosec}\theta}{\text{a}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\times-\text{cosec}\theta\cot\theta\times\frac{\text{d}\theta}{\text{dx}} $
$=-\frac{\text{b}}{\text{a}}\times\text{cosec}\theta\cot\theta\times\frac{1}{\text{a}\sec\theta\tan\theta}$
$=\frac{-\text{b}}{\text{a}^2}\times\frac{1}{\tan^3\theta}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Hence proved
View full question & answer→Question 185 Marks
If $\text{y}=500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}}$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=49\text{y}.$
AnswerIt is given that, $\text{y}=500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}}$
Then
$\frac{\text{dy}}{\text{dx}}=500.\frac{\text{d}}{\text{dx}}(\text{e}^{7\text{x}})+600.\frac{\text{d}}{\text{dx}}(\text{e}^{-7\text{x}})$
$=500.\text{e}^{7\text{x}}.\frac{\text{d}}{\text{dx}}(7\text{x})+600.\text{e}^{-7\text{x}}.\frac{\text{d}}{\text{dx}}(-7\text{x})$
$=3500\text{e}^{7\text{x}}-4200\text{e}^{-7\text{x}}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=3500.\frac{\text{d}}{\text{dx}}(\text{e}^{7\text{x}})-4200.\frac{\text{d}}{\text{dx}}({-7\text{x}})$
$=3500.e^{7\text{x}}.\frac{\text{d}}{\text{dx}}(7\text{x})-4200.\text{e}^{-7\text{x}}.\frac{\text{d}}{\text{dx}}(-7\text{x})$
$=7\times3500.\text{e}^{7\text{x}}+7\times4200.\text{e}^{-7\text{x}}$
$=49\times500\text{e}^{7\text{x}}+49\times600\text{e}^{-7\text{x}}$
$=49(500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}})$
$=49\text{y}$
Hence proved
View full question & answer→Question 195 Marks
If $\text{x}=2\cos\text{t}-\cos2\text{t},\text{y}=2\sin\text{t}-\sin2\text{t},$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{2}.$
AnswerGiven,
$\text{x}=2\cos\text{t}-\cos\text{2}\text{t}$
$\text{y}=2\sin\text{t}-\sin\text{2t}$
Differentiating w.r.t. t,
$\frac{\text{dy}}{\text{dx}}=2(-\sin\text{t})-2(-\sin\text{2t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=2\cot-2\cos\text{2t}$
Dividing both:
$\frac{\text{dy}}{\text{dx}}=\frac{2(\cos\text{t}-\cos\text{2t})}{2(\sin\text{2t}-\sin\text{t})}$
Differentiating w.r.t. t,
$\Rightarrow\frac{\text{d}\frac{\text{dy}}{\text{dx}}}{\text{dt}}=\frac{(\sin\text{2t}-\sin\text{t})(-\sin\text{t}+2\sin\text{2t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{(\sin\text{2t}-\sin\text{t})^2}$
Dividing:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{(\sin\text{2t}-\sin\text{t})(2\sin\text{t}-\sin\text{t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{2(\sin\text{2t}-\sin)^3}$
Putting: $\text{t}=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1+2}{-2}=-\frac{3}{2}$
View full question & answer→Question 205 Marks
Find the second order derivatives of the following functions:
$\text{y}=\tan^{-1}\text{x}$
AnswerWe have,
$\text{y}=\tan^{-1}\text{x}$
differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-(2\text{x})\times\frac{1}{(1+\text{x}^2)^2}$
$=\frac{-2\text{x}}{(1+\text{x}^2)^2}$
View full question & answer→Question 215 Marks
If $\text{x}=\text{a}(1-\cos^3\theta),\text{y}=\text{a}\sin^3\theta,$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{32}{27\text{a}}\text{ at}\ \theta=\frac{\pi}{6}$
AnswerHere,
$\text{x}=\text{a}(1-\cos^3\theta),\text{y}=\text{a}\sin^3\theta$
Differentiating w.r.t. x, we get
$\frac{\text{dx}}{\text{d}\theta}=3\text{a}\cos2\theta\sin\theta\text{ and }\frac{\text{dy}}{\text{d}\theta}=3\text{a}\sin2\theta\cos\theta$]
$\frac{\text{dy}}{\text{dx}}=\frac{3\text{a}\sin^2\theta\cos\theta}{3\text{a}\cos^2\theta\sin\theta}=\tan\theta$
Differentiating w.r.t. x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\sec^2\theta\frac{\text{d}\theta}{\text{dx}}$
$=\frac{\sec^2\theta}{3\text{a}\cos^2\theta\sin\theta}$
$=\frac{\sec^4\theta}{3\text{a}\sin\theta}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}}\text{ at }\theta=\frac{\pi}{6}$
$\frac{\text{d}^2\text{y}}{\text{dx}}=\frac{\Big(\sec\frac{\text{x}}{6}\Big)^4}{3\text{a}\sin\frac{\text{x}}{6}}=\frac{32}{27\text{a}}$
View full question & answer→Question 225 Marks
If $\text{x}=\text{a}(1+\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$ prove that
AnswerHere,
$\text{x}=\text{a}(1+\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=\text{a}+\text{a}\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{a}+\text{a}\cos\theta}{-\text{a}\sin\theta}=\frac{1+\cos\theta}{-\sin\theta}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{d}\theta}\Big\{\frac{\text{dy}}{\text{dx}}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big\{\frac{-\sin\theta\cos\theta-\cos^2\theta}{\sin^2\theta}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$=\frac{1+\cos\theta}{\sin^2\theta}\times\frac{-1}{\sin^2\theta}$
$\frac{-(1+\cos\theta)}{\sin^3\theta}$
At $\theta=\frac{\pi}{2}:\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-(1+\cos\frac{\pi}{2})}{\text{a}(\sin\frac{\pi}{2})^3}=\frac{-1}{\text{a}}$
View full question & answer→Question 235 Marks
If $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),$ find the value of $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{4}.$
AnswerWe have,
$\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{a}(\cos\text{t}+\text{t}\sin\text{t})]=-\text{a}\sin\text{t}+\text{a}\sin\text{t}+\text{at}\cos\text{t}=\text{at}\cos\text{at}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{a}(\sin\text{t}-\text{t}\cos\text{t})]=\text{a}\cos\text{t}-\text{a}\cos\text{t}+\text{at}\sin\text{t}=\text{at}\sin\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})$
$=\frac{\text{d}}{\text{dt}}(\tan\text{t})\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{1}{\text{at}\cos\text{t}}$
$=\frac{1}{\text{at}\cos^3\text{t}}$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{4}}=\frac{1}{\text{a}\Big(\frac{\pi}{4}\Big)\cos^3\Big(\frac{\pi}{4}\Big)}=\frac{8\sqrt{2}}{\text{a}\pi}$
Hence, $\text{at}\ \text{t}=\frac{\pi}{4},\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{8\sqrt{2}}{\text{a}\pi}$
View full question & answer→Question 245 Marks
If $\text{y}=\text{cosec}^{-1}\text{x},\text{x}>1$ prove that $\text{x}(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}=0.$
AnswerHere,
$\text{y}=\text{cosec}^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}\sqrt{{}\text{x}^2-1}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sqrt{\text{x}^2-1}+\frac{\text{x}^2}{\sqrt{\text{x}^2-1}}}{\text{x}^2(\text{x}^2-1)}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{x}^2-1+\text{x}^2}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}^2-1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{(\text{x}^2-1)\sqrt{\text{x}^2-1}}-\frac{1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{\sqrt{\text{x}^2-1}}-\frac{1}{\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=+(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}=0$
Hence proved
View full question & answer→Question 255 Marks
If $\text{a}(1-\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$ prove that, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{a}}$ at $\theta=\frac{\pi}{2}.$
Answer$\text{a}(1-\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}(0+\sin\theta);\ ...\text{Eq}\ 1$
$\frac{\text{dy}}{\text{d}\theta}\ \text{a}(1+\cos\theta)\ ...\text{Eq}\ 2$
Dividing (2) by (1)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{a}(1+\cos\theta)}{\text{a}\sin\theta}$
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{d}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\text{d}\theta}=\frac{\sin\theta(0-\sin\theta)-(1+\cos\theta)\cos\theta}{\sin^2\theta}...(3)$
$=-\frac{\sin^2\theta-\cos\theta-\cos^2\theta}{\sin^2\theta}$
$=-\frac{(1+\cos\theta)}{\sin^2\theta}...(4 )$
dividing (4) by (3)
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{(1+\cos\theta)}{\sin^2\theta\times\text{a}\sin\theta}$
putting $\theta=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{a}}$
Hence proved
View full question & answer→Question 265 Marks
If $\text{y}=\Big[\log\Big(\text{x}+\sqrt{\text{x}^2+1}\Big)\Big]^2$ show that $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}=2$
AnswerGiven:
$\text{y}=\Big[\log\Big(\text{x}+\sqrt{1+\text{x}}\Big)\Big]^2$
Differentiating w.r.t. x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}\big[\log\big(\text{x}+\sqrt{1+\text{x}^2}\big)\big]^2}{\text{dx}}$
Using formula (ii),
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\log\big(\text{x}+\sqrt{1+\text{x}}\big).\frac{1}{(\text{x}+\sqrt{1+\text{x}^2})}.\Big(1+\frac{2\text{x}}{2\sqrt{1+\text{x}^2}}\Big)$
Using formula (i),
$\Rightarrow\text{y}_1=\frac{2\log(\text{x}+\sqrt{1+\text{x}^2}}{\text{x}+\sqrt{1+\text{x}^2}}.\frac{\text{x}+\sqrt{1+\text{x}^2}}{\sqrt{1+\text{x}^2}}$
$\Rightarrow\text{y}_1=\frac{2\log(\text{x}\sqrt{1+\text{x}^2})}{\sqrt{1+\text{x}^2}}$
Squaring both sides:
$\text{(y}_1)^2=\frac{4}{1+\text{x}^2}[\log\big(\text{x}\sqrt{1+\text{x}^2}\big)$
Differentiating w.r.t. x,
$\Rightarrow(1+\text{x}^2)\text{y}_2\text{y}_1+2\text{x}(\text{y}_1)^2=4\text{y}_1$
Using formual (iii),
$\Rightarrow(1+\text{x}^2)\text{y}_2+\text{xy}_1=2$
Hence proved.
View full question & answer→Question 275 Marks
If $\text{y}=\sin(\sin\text{x})$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0$
AnswerGiven,
$\text{y} = \sin (\sin \text{x})\dots\text{ eq. } 1$
To prove: $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\sin\text{x})$
Using chain rule, we will differentiate the above expression:
Let $\text{t}=\sin\text{x}\Rightarrow\frac{\text{dt}}{\text{dx}}=\cos\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dy}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\cos\text{t}\cos\text{x}=\cos(\sin\text{x})\cos\text{x}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\cos\text{x}\frac{\text{d}}{\text{dx}}\cos(\sin\text{x})+\cos(\sin\text{x})\frac{\text{d}}{\text{dx}}\cos\text{x}$
Using chain rule again in the next step:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\cos\text{x}\cos\text{x}\sin(\sin\text{x})-\sin\text{x}\cos(\sin\text{x})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\cos^2\text{x}-\tan\text{x}\cos\text{x}\cos(\sin\text{x})$
$[$using eq. 1: $\text{y} = \sin (\sin \text{x})]$
And using eq. 2, we have:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\cos^2\text{x}-\tan\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}\cos^2\text{x}+\tan\text{x}\frac{\text{dy}}{\text{dx}}=0\dots\text{ proved.}$
View full question & answer→Question 285 Marks
If $\text{y}=\text{x}^3\log\text{x},$ Prove that $\frac{\text{d}^4\text{y}}{\text{dx}^4}=\frac{6}{\text{x}}$
Answerhere,
$\text{y}=\text{x}^3\log\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2\log{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log{\text{x}}+\text{x}^2$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\log\text{x}+6\text{x}\times\frac{1}{\text{x}}+5=6\log\text{x}+11$
Differentiating w.r.t.x, we get
View full question & answer→Question 295 Marks
If $\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x}),$ prove that $\text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0$
AnswerHere,
$\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x}),$
Differentiating w.r.t.x, we get
$\text{y}_1=-3\sin(\log\text{x})\times\frac{1}{\text{x}}+4\cos(\log\text{x})\times\frac{1}{\text{x}}$
$=\frac{-3\sin(\log\text{x})+4\cos(\log\text{x})}{\text{x}}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{\Big(\frac{-3\cos(\log\text{x})}{\text{x}}-\frac{4\sin(\log\text{x})}{\text{x}}\Big)\times \text{ x}-\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})-4\sin(\log\text{x})-\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})-4\sin(\log\text{x})}{\text{x}^2}-\frac{\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})+4\sin(\log\text{x})}{\text{x}^2}-\frac{\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-\text{y}}{\text{x}^2}-\frac{\text{y}_1}{\text{x}}$
$\Rightarrow\text{x}^2\text{y}_2=-\text{y}-\text{xy}_1$
$\Rightarrow\text{x}^2\text{y}_2+\text{y}+\text{xy}_1=0$
Hence proved
View full question & answer→Question 305 Marks
If $\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
AnswerGiven,
$\text{y}=(\sin^{-1}\text{x})^2\dots\text{ eq.1}$
To prove: $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})^2$
Using chain rule we will differentiate the above expression:
Let $\text{t}=\sin^{-1}\text{x}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\sqrt{(1-\text{x}^2)}}$ $[$using formula for derivative of $\sin^{-1}\text{x}]$
And $y = t^2$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=2\text{t}\frac{1}{\sqrt{(1-\text{x}^2)}}=2\sin^{-1}\text{x}\frac{1}{\sqrt{(1-\text{x}^2)}}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\frac{2}{\sqrt{(1-\text{x}^2)}}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2\sin^{-1}\text{x}}{2(1-\text{x}^2)\sqrt{1-\text{x}^2}}(-2\text{x})+\frac{2}{(1-\text{x}^2)}$ $\bigg[\text{using }\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n}-1}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}=\frac{1}{\sqrt{(1-\text{x}^2)}}\bigg]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}+\frac{2}{(1-\text{x}^2)}$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{2\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Using eq. 2:
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{\text{dy}}{\text{dx}}$
$\therefore(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0\dots\text{ proved.}$
View full question & answer→Question 315 Marks
If $\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{a}}{\text{y}^2}.$
AnswerHere
$\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}+\text{a}\cos\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\text{a}\sin\theta}{\text{a}+\text{a}\cos\theta}=\frac{-\sin\theta}{1+\cos\theta}$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big\{\frac{(1+\cos\theta)\cos\theta+\sin^2\theta}{(1+\cos\theta)^2}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$=\frac{-\cos\theta-\cos^2\theta-\sin^2\theta}{(1+\cos\theta)^2}\times\frac{1}{\text{a}+\text{a}\cos\theta}$
$=\frac{-(1+\cos\theta)}{\text{a}(1+\cos\theta)^3}$
$=\frac{-1}{\text{a}(1+\cos\theta)^2}$
$=\frac{-\text{a}}{\text{y}^2}\ [\because\text{y}=\text{a}(1+\cos\theta)]$
Hence proved
View full question & answer→Question 325 Marks
If $\text{y}=(\cot^{-1}\text{x})^2$ prove that $\text{y}^2(\text{x}^2+1)^2+2\text{x}(\text{x}^2+1)\text{y}_1=2.$
Answer$\text{y}=(\cot^{-1}\text{x})^2$Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}_1=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}$
$=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}\ (\text{chain rule})$
$\Rightarrow(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=-2\cot^{-1}\text{x}$
Differentiating w.r.t.x,
$\Rightarrow(1+\text{x}^2)\text{y}^2+2\text{xy}_1=+2\Big(\frac{+1}{1+\text{x}^2}\Big)$
(Multiplication rule on LHS)
$\Rightarrow(1+\text{x}^2)^2\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_1=2$
Hence proved
View full question & answer→Question 335 Marks
If $\text{x}=\cos\theta,\text{y}=\sin^3$ prove that $\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}^2}\Big)=3\sin^2\theta(5\cos^2\theta-1)$
AnswerHere$\text{x}=\cos\theta,\text{y}=\sin^3$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=-\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=3\sin^2\theta\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3\sin^2\theta\cos\theta}{-\sin\theta}=-3\sin\theta\cos\theta$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-3\cos^2\theta+3\sin^2\theta)\frac{\text{d}\theta}{\text{dx}}\frac{)-3\cos^2\theta+3\sin^2\theta)}{-\sin\theta}$
Now,
$\text{LHS}=\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$=\sin^3\theta\times\frac{(-3\cos^2\theta+3\sin^2\theta)}{\sin\theta}+(-3\sin\theta\cos\theta)^2$
$=3\sin^2\theta\cos^2\theta-3\sin^4\theta+9\sin^2\theta\cos^2\theta$
$=12\sin^2\theta\cos^2\theta-3\sin^4\theta$
$=3\sin^2\theta(4\cos^2\theta-\sin^2\theta)$
$=3\sin^2\theta(5\cos^2\theta-1)$
$=\text{RHS}$
View full question & answer→Question 345 Marks
If $\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta$ Show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
AnswerGiven,
$\text{x}=\text{a}\cos\theta\dots\text{ eq. }1$
$\text{y}=\text{b}=\sin\theta\dots\text{ eq. }2$
To prove: $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, let's first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and defferentiate it again.
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}\cos\theta=-\text{a}\sin\theta\dots\text{ eq. 3}$
Similarly, $\frac{\text{dy}}{\text{d}\theta}=\text{b}\cos\theta\dots\text{ eq. 4}$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x}\tan\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=-\frac{\text{b}\cos\theta}{\text{a}\sin\theta}=-\frac{\text{b}}{\text{a}}\cot\theta$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(-\frac{\text{b}}{\text{a}}\cot\theta\Big)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\text{cosec}^2\theta\frac{\text{d}\theta}{\text{dx}}\dots\text{ eq. 5}$
$\Big[$Using chain rule and $\frac{\text{d}}{\text{dx}}\cot\text{x}=-\text{cosec}^2\text{x}\Big]$
From equation 3:
$\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{d}\theta}{\text{dx}}=\frac{-1}{\text{a}\sin\theta}$
Putting the value in equation 5:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}}{a}\text{cosec}^2\theta\frac{1}{\text{a}\sin\theta}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\text{a}^2\sin^3\theta}$
From equation 1:
$\text{y}=\text{b}\sin\theta$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}\dots\text{proved.}$
View full question & answer→Question 355 Marks
If $\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$ show that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$
AnswerHere,
$\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$
$\Rightarrow\frac{1}{\text{a}}\log\text{y}=\sin^{-1}\text{x}$
$\Rightarrow\text{y}=\text{e}^\text{a}\sin^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\text{y}_1=\text{e}^{\text{a} \sin^{-1}\text{x}}\times\frac{\text{a}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\frac{\text{ay}}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{\text{ay}_1\sqrt{1-\text{x}^2}+\frac{\text{x}\text{ay}}{\sqrt{1-\text{x}^2}}}{(1-\text{x}^2)}$
$\Rightarrow\text{y}_2=\frac{\text{ay}_1(1-\text{x}^2)+\text{xay}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{\text{ay}_1}{\sqrt{1-\text{x}^2}}+\frac{\text{xay}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{\text{a}^2\text{y}}{1-\text{x}^2}+{\frac{\text{xy}_1}{(1-\text{x}^2)}}$
$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$
View full question & answer→Question 365 Marks
If $\text{y}=\text{e}^{\tan^{-1}\text{x}},$ prove that $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
AnswerGiven,
$\text{y}=\text{e}^{\tan^{-1}\text{x}}\dots\text{ eq. }1$
To prove: $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, let’s first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\text{e}^{\tan^{-1}\text{x}}$
Using chain rule we will differentiate the above expression:
Let $\text{t}=\tan^{-1}\text{x}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{1+\text{x}^2}\Big[\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}=\frac{1}{1+\text{x}^2}\Big]$
And $y = e^t$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{t}}\frac{1}{1+\text{x}^2}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^{\tan^{-1}\text{x}}\Big(\frac{1}{1+\text{x}}\Big)+\frac{1}{1+\text{x}^2}\frac{\text{d}}{\text{dx}}\text{e}^{\tan^{-1}\text{x}}$
Using chain rule we will differentiate the above expression:
$\frac{\text{d}^2\text{y}}{\text{dx}}=\Big(\frac{\text{e}^{\tan^{-1}\text{x}}}{(1+\text{x}^2)^2}\Big)\frac{2\text{xe}^{\tan^{-1}\text{x}}}{(1+\text{x}^2)^2}$ $\Big[\text{using eq. 2;}\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{x}-1}\text{ and }\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}=\frac{1}{1+\text{x}^2}\Big]$
$(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}-\frac{2\text{xe}^{\tan^{-1}\text{x}}}{1+\text{x}^2}$
$(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}(1-2\text{x})$
Using equation 2:
$(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}(1-\text{2x})$
$\therefore(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0\dots\text{ proved.}$
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