Differentiation (Part-38) | S N De

Previously, we have stated about $\,29\,$ mathematical problems in  part-1, part-2 and part-3  and discussed about their solutions. This time, in this article , we will face few more mathematical problems. So, let's start.

30.Show that the lines $\,\, \frac{x}{\alpha}=\frac{y}{\beta}=\frac{z}{\gamma},\,\, \frac{x}{a\alpha}=\frac{y}{b \beta}=\frac{z}{c \gamma},\,\,\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\,\,$ will lie in the same plane if $\,\, \frac{l}{\alpha} (b-c)+ \frac{m}{\beta} (c-a)+\frac{n}{\gamma}(a-b)=0.$

Sol. Let three lines lie on the same plane $\,\,Ax+By+Cz+D=0.\\ \text{Then,}\,\, A \alpha +B \beta+C \gamma +D=0 \longrightarrow (1)\\ A (a\alpha) +B (b\beta)+C (c\gamma) +D=0 \longrightarrow (2)\\ Al+Bm+Cn=0 \longrightarrow (3)\\$

Now, eliminating A,B, C from (1),(2),(3) we get,  $\begin{vmatrix} \alpha & \beta & \gamma\\ a\alpha & b\beta & c\gamma\\ l& m &n \end{vmatrix}=0 \\ \Rightarrow \alpha(b\beta n-mc\gamma)+\beta(c\gamma l-na\alpha)\\+\gamma(am\alpha-bl\beta)=0 \\ \Rightarrow \alpha\beta n(a-b)+\beta \gamma l(b-c)+m\alpha \gamma(c-a)=0 \\ \Rightarrow \frac{l}{\alpha} (b-c)+ \frac{m}{\beta} (c-a)+\frac{n}{\gamma}(a-b)=0.$

31.If the lines $\,\,ax^2-2hxy+by^2=0\,\,$ form an equilateral triangle with the line $\,\, x \cos {\alpha}+y \sin{\alpha}=p,\,\,$ Show that $\,\, \frac{a}{1-2\cos{ 2 \alpha}}=\frac{b}{2 \sin{2 \alpha}}=\frac{c}{1+2 \cos{2 \alpha}}$

Sol. Let $\,\,AB: y=mx \longrightarrow (1)\\ BC: x \cos{\alpha}+y \sin{\alpha}=p \longrightarrow (2) \\ \Rightarrow y=-\frac{x \cos{\alpha}}{\sin{\alpha}} + \frac{p}{\sin{\alpha}} \\ \therefore m_1=-\cot{\alpha}$

For the triangle $\,\,\Delta ABC\,\,$ to be equilateral , $\,\,\angle{B}=\angle{C}=60^{\circ}\\ \therefore \tan{(\pm 60^{\circ})}=\frac{m+\cot{\alpha}}{1-m\cot{\alpha} } \\ \Rightarrow \pm \sqrt{3}=\frac{m+\cot{\alpha}}{1-m\cot{\alpha} } \\ \therefore 3= \left[ \frac{\frac{y}{x}+\frac{\cos{\alpha}}{\sin{\alpha}}}{1-\frac{y}{x}.\frac{\cos{\alpha}}{\sin{\alpha}} } \right]^2 \\ \Rightarrow 3(x\sin{\alpha}-y\cos{\alpha})^2=(y\sin{\alpha}+x\cos{\alpha} )^2 \\ \Rightarrow x^2(\cos^2{\alpha}-3\sin^2{\alpha})+8xy(\sin{\alpha}\cos{\alpha})\\+y^2(\sin^2{\alpha}-3\cos^2{\alpha})=0 \longrightarrow (3)$

Now, $\,\,\cos^2{\alpha}-3\sin^2{\alpha} \\ ~~~~~~=\frac{1}{2}(1+\cos{2\alpha}-\frac{3}{2}(1-\cos{2 \alpha})\\ ~~~~~~=\frac{1}{2}(1-3+\cos{2\alpha}+3\cos{2\alpha})\\~~~~~~=\frac{1}{2}(-2+4\cos{2\alpha})\\~~~~~~=2\cos{2\alpha}-1 \\ \sin^2{\alpha} -3\cos^2{\alpha}\\=\frac{1}{2}(1-\cos{2\alpha})-\frac{3}{2}(1+\cos{2\alpha})\\=\frac{1}{2}(1-\cos{2\alpha}-3-3\cos{2\alpha})\\=\frac{1}{2}(-2-4\cos{2\alpha})\\=-(1+2 \cos{\alpha})$

Hence from (3) we get , $\\ x^2(2 \cos{2 \alpha} -1)+4xy. \sin {2 \alpha} \\-y^2(1+2\cos{2\alpha})=0 \longrightarrow(4)$

Comparing (4), with $\,\,ax^2-2hxy+by^2=0\,\,$ we get,

$\frac{a}{2\cos{2\alpha}-1}=\frac{-2h}{4 \sin{2 \alpha}}=\frac{b}{-(1+2\cos{2\alpha})} \\  \Rightarrow \frac{a}{1-2\cos{2\alpha}}=\frac{h}{2 \sin{2 \alpha}}=\frac{b}{(1+2\cos{2\alpha})}$ 

32.Show that $\,\, 2x^2-6y^2-12z^2+18yz+2zx+xy=0\,\,$ represents a pair of planes and also find the angle between them. Sol. We know, $ax^2+by^2+cz^2+2fyz+ 2gzx +2hxy=0\longrightarrow (1) $ represents a pair of planes if $\\ \Delta= \begin{vmatrix} a & h& g\\ h& b&f\\ g & f& c \end{vmatrix} =0 \longrightarrow (2)$ Comparing (1) with the given equation , we get $\,\, a=2,\,\,b=-6,\,\,c=-12,\,\,f=9, \\g=1,\,\,h=\frac{1}{2}.\,\,$ 

 Now we see after putting this values in (2) $\,\,\Delta=0.\,\,$ If $\,\,\theta\,\,$ be the angle between the planes, then $\,\,\tan{\theta}=\pm \frac{2 \sqrt{f^2+g^2+h^2-ab-bc-ca}}{a+b+c} \\ = \pm \frac{2\sqrt{9^2+1^2+\frac{1}{4}+12-72+24}}{2-6-12}\\=\frac{\sqrt{185}}{16}\\ \Rightarrow \theta= \tan^{-1}(\frac{\sqrt{185}}{16})$

33. Show that the line $\,\,lx+my=n\,\,$ is a normal to the ellipse $\,\,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\,\,\text{if}\,\, \frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2)^2}{n^2}$

Sol. Equation of the normal to the ellipse at $\,\,(x_1,y_1)\,$ is : $\\ \frac{x-x_1}{\frac{x_1}{a^2}}=\frac{y-y_1}{\frac{y_1}{b^2}} \\ \Rightarrow xy_1.a^2-x_1y.b^2=x_1y_1(a^2-b^2) \longrightarrow (1)\\$

Now, comparing (1) with $\,\,lx+my=n,\,\, \frac{a^2y_1}{l}=\frac{-b^2x_1}{m}=\frac{x_1y_1(a^2-b^2)}{n} \\ \Rightarrow x_1=\frac{a^2n}{l(a^2-b^2)},\,\, y_1=\frac{-b^2n}{m(a^2-b^2)} \\ \therefore \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2} =\frac{n^2}{(a^2-b^2)^2} \left(\frac{a^2}{l^2}+\frac{b^2}{m^2} \right)\\ \Rightarrow 1=\frac{n^2}{(a^2-b^2)^2} \left(\frac{a^2}{l^2}+\frac{b^2}{m^2} \right) \\ \Rightarrow \left(\frac{a^2}{l^2}+\frac{b^2}{m^2} \right)=\frac{(a^2-b^2)^2}{n^2}$

 34. A line of constant length '2k' has its extremities on two fixed lines . Prove the locus of its mid-point is an ellipse whose axes are equally inclined to the fixed line.

Sol. Let the fixed lines be :

$\frac{x}{1}=\frac{y}{m}=\frac{z-c}{0}=r \longrightarrow (1) \,\,\text{and}\\ \frac{x}{1}=\frac{y}{-m}=\frac{z+c}{0}=r'\longrightarrow (2)\\$ 

The points on the line (1) and (2) are $\,\,(r,mr,c)\, \text{and}\,\, (r',-mr',-c).\,\,$ If these are the end points of the variable line at one position and $\,\,(\alpha,\beta,\gamma)\,\,$ be the coordinates of the mid-point of it, then $\,\,2\alpha=r+r',\,\,2 \beta=m(r-r'),\,\,2 \gamma=0.$

Since the length of the variable line is $\,\,2k,\,\, \\ (r-r')^2+m^2(r+r')^2+(c+c)^2=4k^2 \\ \Rightarrow (\frac{2\beta}{m})^2+m^2.(2 \alpha)^2+(2c)^2=4k^2 \\ \Rightarrow 4 \left[\frac{\beta^2}{m^2}+m^2\alpha^2+c^2 \right]=4k^2 \\ \Rightarrow m^2\alpha^2+\frac{\beta^2}{m^2}=k^2-c^2. $

Hence, the required locus is $\,\,m^2x^2+\frac{y^2}{m^2}=k^2-c^2, z=0.$

35. Prove that $\,\, (n+1)^{n-1} (n+2)^{n} > 3^n.(n!)^2$

Sol. Applying A.M. $>$ G.M. on $\{(1.2),(2.3),........,n(n+1)\}\\ \therefore \frac{(1.2)+(2.3)+.....+n(n+1)}{n}  \\ > \sqrt[n]{(1.2).(2.3).....n(n+1)} \longrightarrow (1)\\ \text{Let,}\,\, S_n=(1.2)+(2.3)+.....+n(n+1). \,\, \\t_n=n(n+1)=n^2+n \\ \therefore \sum{t_n}=\sum{n^2}+\sum{n} \\=\frac{n(n+1)(2n+1)}{6} +\frac{n(n+1)}{2}\\=\frac{1}{3}n(n+1)(n+2)\\ \text{Also,}\,\, (1.2).(2.3).....\{n(n+1)\}\\=1.2^2.3^2.....n^2(n+1)\\=(n!)^2.(n+1)\\ \text{Now, from(1),}\,\, \{\frac{1}{3}(n+1)(n+2)\}^n \\ >(n!)^2(n+1)\\ \Rightarrow (n+1)^{n-1}(n+2)^n >3^n.(n!)^2$

36. $\,\,\cos{(u+iv)}=x+iy ,\,\, \text{prove that}\\ \frac{x^2}{\cosh^2v} +\frac{y^2}{\sinh^2v}=1\,\,\text{and}\,\, \frac{x^2}{\cosh^2u} -\frac{y^2}{\sinh^2u}=1$

Sol. $\,\,x+iy=\cos{(u+iv)} \\  \Rightarrow x+iy=\cos{u}\cos{iv}-\sin{u}\sin{iv}\\ \Rightarrow x+iy=\cos{u} \cosh v -i\sin{u}\sinh v $

Using the fact, $\,\cos{iv}=\cosh v,\,\, \sin{iv}=i\sinh v \\ \therefore x= \cos{u}\cosh v,\,\,y=-\sin{u}\sinh v\\ \therefore \frac{x^2}{\cosh^2v} +\frac{y^2}{\sinh^2v}=\cos^2{u}+\sin^2{u}=1 \\ \frac{x^2}{\cos^2u} -\frac{y^2}{\sin^2u}=\cosh^2 v-\sinh^2 v=1$

37. If $\,\, x+iy=a \cos{(u+iv)}+ib\sin{(u+iv)}\\ \text{where x,y,u,v are real, show that}\\ \frac{x^2}{\cos^2u}-\frac{y^2}{\sin^2u}=a^2-b^2 $

Sol. $x+iy=a(\cos{u}.\cosh v-i\sin{u}\sinh v)\\+ib(\sin u \cosh v +i \cos u \sinh v)\\= (a \cos u \cosh v -b \cos u \sinh v)\\+i(-a \sin u \sinh v+b \sin u \cosh v)\\ \therefore x= \cos u(a \cosh v -b \sinh v), \\ y=\sin u(-a \sinh v+b \cosh v) \\ \therefore \frac{x^2}{\cos^2 u}-\frac{y^2}{\sin^2u}\\=(a \cosh v -b \sinh v)^2 \\-(-a \sinh v+b \cosh v)^2\\=a^2 \cosh^2v+b^2\sinh^2 v-2ab \sinh v \cosh v \\-(a^2\sinh^2v+b^2\cosh^2v-2ab\sinh v \cosh v)\\=a^2(\cosh^2 v-\sinh^2 v) \\-b^2(\cosh^2 v-\sinh^2 v)\\=a^2-b^2$

38.If $\,\,u+iv=\cot{(x+iy)},\,\,\text{prove that} \\ u=\frac{\sin{2x}}{\cosh{2y}-\cos{2x}}, \\ v=\frac{-\sinh{2y}}{\sinh{2y}-\cos{2x}}$

Sol.  $u+iv=\cot{(x+iy)},\,\, u-iv=\cot{(x-iy)} \\ \therefore 2u=\cot{(x+iy)}+\cot{(x-iy)}\\=\frac{\cos{(x+iy)}}{\sin{(x+iy)}}+\frac{\cos{(x-iy)}}{\sin{(x-iy)}}\\=\frac{\cos{(x+iy)}\sin{(x-iy)}+\cos{(x-iy)}\sin{(x+iy)}}{\sin{(x+iy)}\sin{(x-iy)}}\\=\frac{\sin\{(x-iy)+(x+iy)\}}{\sin^2x-\sin^2{(iy)}}\\=\frac{\sin{2x}}{\sin^2x-i^2\sinh^2y}\\=\frac{\sin{2x}}{\sin^2x+\sinh^2y}\\=\frac{\sin{2x}}{1-\cos^2x+\sinh^2y}\\=\frac{\sin{2x}}{\cosh^2y-\cos^2x}\\ \therefore 2u=\frac{2 \sin{2x}}{2\cosh^2y-2\cos^2x} \\=\frac{2 \sin{2x}}{1+\cosh{2y}-(1+\cos{2x})}\\ \Rightarrow u=\frac{\sin{2x}}{\cosh{2y}-\cos{2x}}\\ (u+iv)-(u-iv)=2iv \\ \Rightarrow 2iv= \frac{\cos{(x+iy)}}{\sin{(x+iy)}}-\frac{\cos{(x-iy)}}{\sin{(x-iy)}}\\=\frac{\cos{(x+iy)}\sin{(x-iy)}-\cos{(x-iy)}\sin{(x+iy)}}{\sin{(x+iy)}\sin{(x-iy)}}\\=\frac{\sin\{x-iy-x-iy\}}{\sin^2x-i^2\sinh^2y}\\ \Rightarrow 2iv=\frac{2\sin{(-2iy)}}{2(1-\cos^x+\sinh^2y)}\\ \Rightarrow 2iv= \frac{-2i \sinh{2y}}{2(\cosh^2-\cos^2x)}\\ \therefore v=\frac{-\sinh{2y}}{(1+\cosh{2y})-(1+\cos{2x})}=\frac{-\sinh{2y}}{\cosh{2y}-\cos{2x}}$

39.Solve $\,\,y^2\log{y}=xpy+p^2\,\,\text{where,}\,\,p=\frac{dy}{dx}$

Sol.  $\,\,y^2\log{y}=xpy+p^2 \\ \Rightarrow xpy=y^2\log{y}-p^2 \\ \Rightarrow x=\frac{y^2\log{y}-p^2}{py} \\ \Rightarrow \frac{dx}{dy}=\frac{1}{p}=\frac{1}{p}. \frac{d}{dy}(y\log{y})+y \log{y}(-\frac{1}{p^2})\frac{dp}{dy}-\{\frac{1}{y}\frac{dp}{dy}+p(-\frac{1}{y^2})\} \\ \Rightarrow \frac{1}{p}=\frac{1}{p}(1+\log{y})-\frac{y}{p^2}\log{y}\frac{dp}{dy}-\frac{1}{y}\frac{dp}{dy}+\frac{p}{y^2}\\ \Rightarrow 0=\frac{\log{y}}{p}-\frac{y\log{y}}{p^2}\frac{dp}{dy}-\frac{1}{y}\frac{dp}{dy}+\frac{p}{y^2} \\ \Rightarrow \frac{dp}{dy}\left(\frac{y \log{y}}{p^2}+\frac{1}{y} \right)=\frac{p}{y^2}+\frac{\log{y}}{p}\\ \Rightarrow \frac{y}{p} \frac{dp}{dy}\left(\frac{\log{y}}{p}+\frac{p}{y^2} \right)=\left(\frac{\log{y}}{p}+\frac{p}{y^2} \right)\\ \Rightarrow \frac{y}{p}\left( \frac{dp}{dy}-1 \right)\left(\frac{\log{y}}{p}+\frac{p}{y^2} \right)=0\\ \therefore \frac{y}{p}\frac{dp}{dy}-1=0 \\ \text{or,}\,\, p=cy.$

40. $\sqrt{p} + \sqrt{q}=1 \longrightarrow (1) \\ \Rightarrow \frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}=\frac{dz}{-(pf_p+qf_q)}=\frac{dx}{-f_p}=\frac{dy}{-f_q}\quad [\text{By Charpit's Method}] \\ \Rightarrow \frac{dp}{0}=\frac{dq}{0}=\frac{dz}{-\frac{p}{2\sqrt{p}}--\frac{q}{2\sqrt{q}}}=\frac{dx}{-\frac{1}{2\sqrt{p}}}=\frac{dy}{-\frac{1}{2\sqrt{q}}} \cdots \cdots (2) \\ \therefore \frac{dp}{0}=\frac{dx}{-\frac{1}{2\sqrt{p}}}  \Rightarrow dp=0 \Rightarrow p=a$

Also from (2) we get, $\frac{dq}{0}=\frac{dy}{-\frac{1}{2\sqrt{q}}} \Rightarrow dq=0 \Rightarrow q=b.$

So, by (1) we get, $\,\,\sqrt{a}+\sqrt{b}=1 \Rightarrow \sqrt{b} \\=1-\sqrt{a} \\  \Rightarrow b=(1-\sqrt{a})^2 \\ \therefore dz=pdx+qdy=adx+bdy \\ \therefore z=ax+by+c\\=ax+(1-\sqrt{a})^2y+c \quad \text{where  a,b are constants; Here by}\,\,\,\, f_x=\frac{\partial f}{\partial x},....$