The largest right circular cone

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3 .

Solution

$ \displaystyle \text{Volume}\ \text{of}\ \text{right}\ \text{circular}\ \text{cone}\ =V=\frac{1}{3}\pi {{x}^{2}}(3+y)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{3}\pi {{x}^{2}}(\ 3+\sqrt{{9-{{x}^{2}}}}\ )$

$ \displaystyle \frac{{dV}}{{dx}}=\frac{1}{3}\pi \left[ {{{x}^{2}}.\frac{1}{2}{{{(9-{{x}^{2}})}}^{{-\frac{1}{2}}}}(-2x)+(\ 3+\sqrt{{9-{{x}^{2}}}}\ ).2x} \right]$

$ \displaystyle \ \ \ \ \ =\frac{1}{3}\pi \left[ {-\frac{{{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}+6x+2x\sqrt{{9-{{x}^{2}}}}} \right]$ 

$ \displaystyle \ \ \ \ \ =\frac{1}{3}\pi \left[ {6x+\frac{{-{{x}^{3}}+18x-2{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]$

$ \displaystyle \ \ \ \ \ =\frac{1}{3}\pi \left[ {6x+\frac{{18x-3{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]$

$ \displaystyle \ \ \ \ \ =\pi \left[ {2x+\frac{{6x-{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]$ 

$ \displaystyle \text{For}\ \text{stationary}\ \text{value}\ ,\ \frac{{dV}}{{dx}}=0$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \pi \left[ {2x+\frac{{6x-{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}} \right]=0$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{6x-{{x}^{3}}}}{{\sqrt{{9-{{x}^{2}}}}}}=-2x$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6-{{x}^{2}}=-2\sqrt{{9-{{x}^{2}}}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 36-12{{x}^{2}}+{{x}^{4}}=36-4{{x}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{4}}-8{{x}^{2}}=0$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{2}}=8$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\sqrt{8}$

$ \displaystyle \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=\pi \left[ {2+\frac{{\sqrt{{9-{{x}^{2}}}}(6-3{{x}^{2}})-(6x-{{x}^{3}})\frac{1}{2}{{{(9-{{x}^{2}})}}^{{-\frac{1}{2}}}}(-2x)}}{{9-{{x}^{2}}}}} \right]$

$ \displaystyle \ \ \ \ \ \ =\pi \left[ {2+\frac{{(9-{{x}^{2}})(6-3{{x}^{2}})+{{x}^{2}}(6-{{x}^{2}})}}{{(9-{{x}^{2}})\sqrt{{9-{{x}^{2}}}}}}} \right]$ 

$ \displaystyle \text{when}\ x=\sqrt{8}\ ,\ \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=\pi (2-18-16)=-32\pi <0\ $

$ \displaystyle V\ \text{is}\ \text{maximum}\ \text{value}\ \text{when}\ x=\sqrt{8}\ .$ 

$ \displaystyle \text{The}\ \text{volume}\ \text{of}\ \text{the}\ \text{largest}\ \text{right}\ \text{circular}\ \text{cone}\ =V=\frac{1}{3}\times \frac{{22}}{7}\times 32$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{704}}{{21}}\ \text{cube}\ \text{units}$

Calculus : Maximum and Minimum

Find the dimensions of the right circular cylinder of largest volume that can be inscribed in a sphere of radius  R .

Solution

$ \displaystyle \text{Let}\ \text{the}\ \text{volume}\ \text{of}\ \text{cylinder}\ =\ V=\pi {{r}^{2}}h$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ {{r}^{2}}+{{\left( {\frac{h}{2}} \right)}^{2}}={{R}^{2}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ {{r}^{2}}={{R}^{2}}-\frac{{{{h}^{2}}}}{4}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ V=\pi \left( {{{R}^{2}}-\frac{{{{h}^{2}}}}{4}} \right)h$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\pi \left( {{{R}^{2}}h-\frac{1}{4}{{h}^{3}}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \frac{{dV}}{{dh}}=\pi \left( {{{R}^{2}}-\frac{3}{4}{{h}^{2}}} \right)$ 

$ \displaystyle \text{For}\ \text{stationary}\ \text{value}\ ,\ \frac{{dV}}{{dh}}=0$ 

$ \displaystyle \ \ \ \ \ \ \ \pi \left( {{{R}^{2}}-\frac{3}{4}{{h}^{2}}} \right)=0$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ {{R}^{2}}-\frac{3}{4}{{h}^{2}}=0$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{h}^{2}}=\frac{{4{{R}^{2}}}}{3}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h=\frac{{2R}}{{\sqrt{3}}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \frac{{{{d}^{2}}V}}{{d{{h}^{2}}}}=-\frac{3}{2}\pi h$ 

$ \displaystyle \text{When}\ \ h=\frac{{2R}}{{\sqrt{3}}}\ \ ,\ \ \ \ \frac{{{{d}^{2}}V}}{{d{{h}^{2}}}}=-\sqrt{3}\ \pi R\ <\ 0$ 

$ \displaystyle V\ \text{is}\ \text{maximum}\ \text{value}\ \text{when}\ \ h=\frac{{2R}}{{\sqrt{3}}}\ $ 

$ \displaystyle \ \ \ \ \ \ \ {{r}^{2}}={{R}^{2}}-\frac{{{{R}^{2}}}}{3}=\frac{{2{{R}^{2}}}}{3}$ 

$ \displaystyle \therefore \ \ \ r\ =\sqrt{{\frac{2}{3}}}\ R$

The sum of the series

Find the value of sum of

$ \displaystyle \frac{1}{{11}}+\frac{1}{{11+22}}+\frac{1}{{11+22+33}}+...+\frac{1}{{11+22+33+...+2013}}$ .

Solution

$ \displaystyle \frac{1}{{11}}+\frac{1}{{11+22}}+\frac{1}{{11+22+33}}+...+\frac{1}{{11+22+33+...+2013}}$

$ \displaystyle =\frac{1}{{11}}\ \left[ {1+\frac{1}{{1+2}}+\frac{1}{{1+2+3}}+...+\frac{1}{{1+2+3+...+183}}} \right]$ 

$ \displaystyle =\frac{1}{{11}}\ \left[ {1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{{\frac{{183}}{2}(1+183)}}} \right]$ 

$ \displaystyle =\frac{1}{{11}}\left[ {1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{{183\times 184}}} \right]$ 

$ \displaystyle =\frac{2}{{11}}\left[ {\frac{1}{2}+\frac{1}{6}+\frac{1}{{12}}+...+\frac{1}{{183\times 184}}} \right]$

$ \displaystyle =\frac{2}{{11}}\left[ {\frac{1}{{1\times 2}}+\frac{1}{{2\times 3}}+\frac{1}{{3\times 4}}+...+\frac{1}{{183\times 184}}} \right]$ 

$ \displaystyle =\frac{2}{{11}}\ \left[ {\left( {\frac{1}{1}-\frac{1}{2}} \right)+\left( {\frac{1}{2}-\frac{1}{3}} \right)+\left( {\frac{1}{3}-\frac{1}{4}} \right)+...+\left( {\frac{1}{{183}}-\frac{1}{{184}}} \right)} \right]$ 

$ \displaystyle =\frac{2}{{11}}\left( {1-\frac{1}{{184}}} \right)$ 

$ \displaystyle =\frac{2}{{11}}\times \frac{{183}}{{184}}$ 

$ \displaystyle =\frac{{183}}{{1012}}$

Angle bisector theorem

Given : ABCD is a square , CF bisects $ \displaystyle \angle ACD$ , BPQ perpendicular CF .
Prove : DQ = 2PE .

$ \displaystyle \text{Proof}\ :\ \ \theta =\beta =\alpha =\gamma ={{22.5}^{\circ }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ In\ \Delta CQG\ \ and\ \ \Delta CPG\ ,$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \theta =\beta $ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \angle CGQ=\angle CGP={{90}^{\circ }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ CG=CG\ \ (\ \text{common}\ \text{side}\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \Delta CQG\cong \Delta CPG\ (\ A.S.A\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ CQ=CP$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{CQ}}{{DQ}}=\frac{{BC}}{{BD}}\ \ (\ \because \ \text{Angle}\ \text{bisector}\ \text{theorem}\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{CQ}}{{DQ}}=\frac{1}{{\sqrt{2}}}\ .........(1)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{PE}}{{CP}}=\frac{{BE}}{{BC}}\ \ (\ \because \ \text{Angle}\ \text{bisector}\ \text{theorem}\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{PE}}{{CP}}=\frac{1}{{\sqrt{2}}}\ .........(2)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (1)\ \times \ (2)\Rightarrow \ \ \frac{{CQ}}{{DQ}}\ \times \ \frac{{PE}}{{CP}}=\frac{1}{2}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{PE}}{{DQ}}=\frac{1}{2}\ (\ \because \ CQ=CP\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \ DQ\ =2PE$

Area: Trapezium

Parallel sides of a trapezium are equal to a and b . Find the length of the line segment which is parallel to them and divides the area of the trapezium into two equal parts.
( i.e Find x in terms of a and b .)

Solution

Draw AD and BC produced meet at G .

$ \displaystyle \text{Let}\ \alpha (\Delta GDC)=P\ ,\ \alpha (DEFG)=Q\ \ ,\ \alpha (ABFE)=R$ 

$ \displaystyle \Delta GDC\sim \Delta GEF\sim \Delta GAB\ \ (\because DC\parallel EF\parallel AB)$ 

$ \displaystyle \frac{P}{{P+Q}}=\frac{{{{b}^{2}}}}{{{{x}^{2}}}}$ 

$ \displaystyle \frac{P}{Q}=\frac{{{{b}^{2}}}}{{{{x}^{2}}-{{b}^{2}}}}\ ........(1)$

$ \displaystyle \frac{P}{{P+Q+R}}=\frac{{{{b}^{2}}}}{{{{a}^{2}}}}$ 

$ \displaystyle \frac{P}{{Q+R}}=\frac{{{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}$

$ \displaystyle \frac{P}{{2Q}}=\frac{{{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}\ \ (\because \ Q=R)$ 

$ \displaystyle \frac{P}{Q}=\frac{{2{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}\ ........(2)$

$ \displaystyle \text{From}\ (1)\ \text{and}\ (2)\ ,\ \frac{{{{b}^{2}}}}{{{{x}^{2}}-{{b}^{2}}}}\ =\frac{{2{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2{{x}^{2}}-2{{b}^{2}}={{a}^{2}}-{{b}^{2}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2{{x}^{2}}={{a}^{2}}+{{b}^{2}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \ x=\sqrt{{\frac{{{{a}^{2}}+{{b}^{2}}}}{2}}}$

Fermat point

$ \displaystyle \Delta ABC$ is a $ \displaystyle {{30}^{\circ }}-{{60}^{\circ }}$ right triangle with $ \displaystyle \angle A={{30}^{\circ }}\ ,\ \angle B={{60}^{\circ }}$. If P is a Fermat point of $ \displaystyle \Delta ABC$ , prove that AP , BP , CP are in a G.P.

Solution

$ \displaystyle \text{In}\ \Delta FCB\ \text{and}\ \Delta ACD\ ,$ 

$ \displaystyle \ \ \ \ \ FC\ =\ AC$ 

$ \displaystyle \ \ \ \ \ CB=CD$ 

$ \displaystyle \angle FCB=\angle ACD$ 

$ \displaystyle \Delta FCB\cong \Delta ACD\ \ (S.A.S)$ 

$\displaystyle \angle FBC=\angle ADC=\beta $ 

$ \displaystyle \therefore \ CDBP\ \text{is}\ \text{cyclic}\ .$

$ \displaystyle \angle PCB=\angle PDB=\theta $ 

$ \displaystyle AB\parallel \ CD\ \ (\because \ \angle BCD=\angle ABC)$

$ \displaystyle \therefore \ \angle ADC=\angle BAD\ \ (\because \text{alternate}\ \angle \text{s})$ 

$ \displaystyle \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{sines}\ ,$ 

$ \displaystyle \ \ \ \ \ \frac{{BP}}{{\sin \beta }}=\frac{{AP}}{{\sin \theta }}$ 

$ \displaystyle \ \ \ \ \ \ \ \frac{{BP}}{{AP}}=\frac{{\sin \beta }}{{\sin \theta }}\ \ \ \ \ \ \ ........(1)$ 

$ \displaystyle \ \ \ \ \ \ \frac{{CP}}{{\sin \beta }}=\frac{{BP}}{{\sin \theta }}$

$ \displaystyle \ \ \ \ \ \ \ \frac{{CP}}{{BP}}=\frac{{\sin \beta }}{{\sin \theta }}\ \ \ \ \ \ \ ........(2)$ 

$ \displaystyle \text{From}\,(1)\ \ \text{and}\ (2)\ ,\ \ \frac{{BP}}{{AP}}=\ \frac{{CP}}{{BP}}$ 

$ \displaystyle \therefore \ AP,BP,CP\ \text{are}\ \text{in}\ \text{a}\ G.P.\ $

Area of shaded region

In the figure , a square ABCD and a sector OAB of a circle centre O , radius r . Show that the area of square ABCD is $ \displaystyle 2{{r}^{2}}(1-\cos \theta )$ and then find the shaded area R , when $ \displaystyle \theta =\frac{\pi }{3}$ .

Solution

$ \displaystyle \text{Draw}\ \ OE\bot AB\ .$  

$ \displaystyle AE=r\sin \frac{\theta }{2}\ \ \ ,\ \ \ OE=r\cos \frac{\theta }{2}$ 

$ \displaystyle AB=2AE=2r\sin \frac{\theta }{2}\ $

$ \displaystyle \alpha (ABCD)={{(2r\sin \frac{\theta }{2}\ )}^{2}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{r}^{2}}{{\sin }^{2}}\frac{\theta }{2}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ .\ 2{{\sin }^{2}}\frac{\theta }{2}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ (1-\cos \theta )\ \ \ \ [\because \cos 2\theta =1-2{{\sin }^{2}}\theta ]$

$ \displaystyle \text{The}\ \text{shaded}\ \text{area}\ R\ =\alpha (ABCD)+\alpha (\Delta OAB)-\text{Area}\ \text{of}\ \text{sector}\ OAB$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ (1-\cos \theta )+\frac{1}{2}(2r\sin \frac{\theta }{2}\ )(r\cos \frac{\theta }{2})-\frac{1}{2}{{r}^{2}}\theta $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ (1-\cos \theta )+\frac{1}{2}{{r}^{2}}\sin \theta -\frac{1}{2}{{r}^{2}}\theta $

$ \displaystyle \text{When}\ \theta =\frac{\pi }{3}\ ,$

$ \displaystyle R=2{{r}^{2}}(1-\frac{1}{2})+\frac{1}{2}{{r}^{2}}(\frac{{\sqrt{3}}}{2})-\frac{1}{2}{{r}^{2}}(\frac{\pi }{3})$

$ \displaystyle \ \ \ =(1+\frac{{\sqrt{3}}}{4}-\frac{\pi }{6})\ {{r}^{2}}\ \text{sq}\ \text{units}$