Chapter 6: Applications of Derivatives

Ans: $f(x) = {\cos ^2}x + \sin x$

${f^\prime }(x) = 2\cos x( - \sin x) + \cos x$

$=  - 2\sin x\cos x + \cos x$

${f^\prime }(x) = 0{\text{ }}$

$\Rightarrow 2\sin x\cos x = \cos x \Rightarrow \cos x(2\sin x - 1) = 0$

$\Rightarrow \sin x = \dfrac{1}{2}$ or $\cos x = 0$

$\Rightarrow x = \dfrac{\pi }{6}$, or $\dfrac{\pi }{2}$ as $x \in [0,\pi ]$

$f\left( {\dfrac{\pi }{6}} \right) = {\cos ^2}\dfrac{\pi }{6} + \sin \dfrac{\pi }{6} = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + \dfrac{1}{2} = \dfrac{5}{4}$

$f(0) = {\cos ^2}0 + \sin 0 = 1 + 0 = 1$

$f(\pi ) = {\cos ^2}\pi  + \sin \pi  = {( - 1)^2} + 0 = 1$

$f\left( {\dfrac{\pi }{2}} \right) = {\cos ^2}\dfrac{\pi }{2} + \sin \dfrac{\pi }{2} = 0 + 1 = 1$

absolute maximum value of $f$ is $\dfrac{5}{4}$ at $x = \dfrac{\pi }{6}$

The absolute minimum value of $f$ is 1 at $x = 0,x = \dfrac{\pi }{2}$, and $\pi $.

Ans: $V = \dfrac{1}{3}\pi {R^2}h$

$BC = \sqrt {{r^2} - {R^2}} $

$h = r + \sqrt {{r^2} - {R^2}} $

$\therefore V = \dfrac{1}{3}\pi {R^2}\left( {r + \sqrt {{r^2} - {R^2}} } \right) = \dfrac{1}{3}\pi {R^2}r + \dfrac{1}{3}\pi {R^2}\sqrt {{r^2} - {R^2}} $

$\therefore \dfrac{{dV}}{{dR}} = \dfrac{2}{3}\pi Rr + \dfrac{{2\pi }}{3}\pi R\sqrt {{r^2} - {R^2}}  + \dfrac{{{R^2}}}{3} \cdot \dfrac{{( - 2R)}}{{2\sqrt {{r^2} - {R^2}} }}$

$= \dfrac{2}{3}\pi Rr + \dfrac{{2\pi }}{3}\pi R\sqrt {{r^2} - {R^2}}  - \dfrac{{{R^3}}}{{3\sqrt {{r^2} - {R^2}} }}$

$= \dfrac{2}{3}\pi Rr + \dfrac{{2\pi Rr\left( {{r^2} - {R^2}} \right) - \pi {R^3}}}{{3\sqrt {{r^2} - {R^2}} }}$

$= \dfrac{2}{3}\pi Rr + \dfrac{{2\pi R{r^2} - 3\pi R{r^3}}}{{3\sqrt {{r^2} - {R^2}} }}$

$\dfrac{{dV}}{{d{R^2}}} = 0$

$\Rightarrow \dfrac{{2\pi rR}}{3} = \dfrac{{3\pi {R^2} - 2\pi R{r^2}}}{{3\sqrt {{r^2} - {R^2}} }}$

$\Rightarrow 2{\text{r}}\sqrt {{r^2} - {R^2}}  = 3{R^2} - 2{r^2}$

$\Rightarrow 4{r^2}\left( {{r^2} - {R^2}} \right) = {\left( {3{R^2} - 2{r^2}} \right)^2}$

$\Rightarrow 14{r^4} - 4{r^2}{R^2} = 9{R^4} + 4{r^4} - 12{R^2}{r^2}$

$\Rightarrow 9{R^4} - 8{r^2}{R^2} = 0$

$\Rightarrow 9{R^2} = 8{r^2}$

$\Rightarrow {R^2} = \dfrac{{8{r^2}}}{9}$

$\dfrac{{{d^2}V}}{{d{R^2}}} = \dfrac{{2\pi r}}{3} + \dfrac{{3\sqrt {{r^2} - {R^2}} \left( {2\pi {r^2} - 9\pi {R^2}} \right) - \left( {2\pi {R^3} - 3\pi {R^3}} \right)( - 6R)\dfrac{1}{{2{{\sqrt {{r^2} - R} }^2}}}}}{{9\left( {{r^2} - {R^2}} \right)}}$

$= \dfrac{{2\pi r}}{3} + \dfrac{{3\sqrt {{r^2} - {R^2}} \left( {2\pi {r^2} - 9\pi {R^2}} \right) - \left( {2\pi {R^3} - 3\pi {R^3}} \right)(3R)\dfrac{1}{{2{{\sqrt {{r^2} - R} }^2}}}}}{{9\left( {{r^2} - {R^2}} \right)}}$

when ${R^2} = \dfrac{{8{r^2}}}{9},\dfrac{{{d^2}V}}{{d{R^2}}} < 0$.

volume is the maximum when ${R^2} = \dfrac{{8{r^2}}}{9}$. ${R^2} = \dfrac{{8{r^2}}}{9}$, 

height of the cone $= r + \sqrt {{r^2} - \dfrac{{8{R^2}}}{9}}  = r + \sqrt {\dfrac{{{r^2}}}{9}}  = r + \dfrac{r}{3} = \dfrac{{4r}}{3}$

Ans: Consider I be the interval (a, b)

Given that ${{f}^{'}}(x)>0$for all x in an interval I. Consider ${{x}_{1}},{{x}_{1}},\in $I with ${{x}_{1}}<{{x}_{2}}$.

By Lagrange’s Mean Value Theorem, we have, 

$\frac{f({{x}_{2}})-f({{x}_{1}})}{{{x}_{2}}-{{x}_{1}}}={{f}^{'}}(c)\,$where ${{x}_{1}}<c<{{x}_{2}}$

$\Rightarrow f({{x}_{2}})-f({{x}_{1}})=({{x}_{2}}-{{x}_{1}}){{f}^{'}}(c)$where ${{x}_{1}}<c<{{x}_{2}}$

Now ${{x}_{1}}<{{x}_{2}}$

$\Rightarrow {{x}_{2}}-{{x}_{1}}>0$……….(1)

 Also, ${{f}^{'}}(x)>0$for all $x$ in an interval I

$\Rightarrow {{f}^{'}}(c)>0$

From equation (1), $f({{x}_{2}})-f({{x}_{1}})>0$

$\Rightarrow f({{x}_{1}})<f({{x}_{2}})$

Thus, for every pair of points ${{x}_{1}},{{x}_{1}},\in $I with ${{x}_{1}}<{{x}_{2}}$

$\Rightarrow f({{x}_{1}})<f({{x}_{2}})$

Therefore, f(x) is strictly increasing in I.

Ans: $h = 2\sqrt {{R^2} - {r^2}} $

$V = \pi {r^2}h = 2\pi {r^2}\sqrt {{R^2} - {r^2}} $

$\therefore \dfrac{{dV}}{{dr}} = 4\pi r\sqrt {{R^2} - {r^2}}  + \dfrac{{2\pi {r^2}( - 2r)}}{{2\sqrt {{R^2} - {r^2}} }}$

$= 4\pi r\sqrt {{R^2} - {r^2}}  - \dfrac{{2\pi {r^3}}}{{\sqrt {{R^2} - {r^2}} }}$

$= \dfrac{{4\pi r\left( {{R^2} - {r^2}} \right) - 2\pi {r^3}}}{{\sqrt {{R^2} - {r^2}} }}$

$= \dfrac{{4\pi r{R^2} - 6\pi {r^3}}}{{\sqrt {{R^2} - {r^2}} }}$

Now, $\dfrac{{dV}}{{dr}} = 0 \Rightarrow 4\pi r{R^2} - 6\pi {r^3} = 0$

$\Rightarrow {r^2} = \dfrac{{2{R^2}}}{3}$

$\dfrac{{{d^2}V}}{{d{r^2}}} = \dfrac{{\sqrt {{R^2} - {r^2}} \left( {4\pi {R^2} - 18\pi {r^2}} \right) - \left( {4\pi r{R^2} - 6\pi {r^3}} \right)\dfrac{{( - 2r)}}{{2\sqrt {{R^2} - {r^2}} }}}}{{\left( {{R^2} - {r^2}} \right)}}$

$= \dfrac{{\left( {{R^2} - {r^2}} \right)\left( {4\pi {R^2} - 18\pi {r^2}} \right) + r\left( {4\pi r{R^2} - 6\pi {r^3}} \right)}}{{{{\left( {{R^2} - {r^2}} \right)}^{\dfrac{3}{2}}}}}$

$= \dfrac{{4\pi {R^4} - 22\pi {r^2}{R^2} + 12\pi {r^4} + 4\pi {r^2}{R^2}}}{{{{\left( {{R^2} - {r^2}} \right)}^{\dfrac{3}{2}}}}}$

${r^2} = \dfrac{{2{R^2}}}{3},\dfrac{{{d^2}V}}{{d{r^2}}} < 0.$

volume is maximum when ${r^2} = \dfrac{{2{R^2}}}{3}$. ${r^2} = \dfrac{{2{R^2}}}{3}$.

height of the cylinder is $2\sqrt {{R^2} - \dfrac{{2{R^2}}}{3}}  = 2\sqrt {\dfrac{{{R^2}}}{3}}  = \dfrac{{2R}}{{\sqrt 3 }}$.

the volume of the cylinder is maximum when the height of the cylinder is $\dfrac{{2R}}{{\sqrt 3 }}$

Ans:

$r = h\tan a$

since $\Delta AOG$ is similar to $\Delta CEG$,

$\dfrac{{AO}}{{OG}} = \dfrac{{CE}}{{EG}}$

$\Rightarrow \dfrac{h}{r} = \dfrac{H}{{r - R}}$

$\Rightarrow H = \dfrac{h}{r}(r - R) = \dfrac{h}{{h\tan a}}(h\tan a - R) = \dfrac{1}{{\tan a}}(h\tan a - R)$

volume of the cylinder is $V = \pi {R^2}H = \dfrac{{\pi {R^2}}}{{\tan a}}(h\tan a - R)$

$= \pi {R^2}h - \dfrac{{\pi {R^3}}}{{\tan a}}$

$\therefore \dfrac{{dV}}{{dR}} = 2\pi Rh\dfrac{{3\pi {R^2}}}{{\tan a}}$

$\dfrac{{dV}}{{dR}} = 0$

$\Rightarrow 2\pi Rh = \dfrac{{3\pi {R^2}}}{{\tan a}}$

$\Rightarrow 2h\tan a = 3R$

$\Rightarrow R = \dfrac{{2h}}{3}\tan a$

$\dfrac{{{d^2}V}}{{d{R^2}}} = 2\pi Rh - \dfrac{{6\pi R}}{{\tan a}}$

And, for $R = \dfrac{{2h}}{3}\tan a$, we have:

$\dfrac{{{d^2}V}}{{d{R^2}}} = 2\pi h - \dfrac{{6\pi }}{{\tan a}}\left( {\dfrac{{2h}}{3}\tan a} \right) = 2\pi h - 4\pi h =  - 2\pi h < 0$

volume of the cylinder is greatest when $R = \dfrac{{2h}}{3}\tan a$. $R = \dfrac{{2h}}{3}\tan a,H = \dfrac{1}{{\tan a}}\left( {h\tan a - \dfrac{{2h}}{3}\tan a} \right) = \dfrac{1}{{\tan a}}\left( {\dfrac{{h\tan a}}{3}} \right) = \dfrac{h}{3}$.

the maximum volume of the cylinder can be obtained as

$\pi {\left( {\dfrac{{2h}}{3}\tan a} \right)^2}\left( {\dfrac{h}{3}} \right) = \pi \left( {\dfrac{{4{h^2}}}{9}{{\tan }^2}a} \right)\left( {\dfrac{h}{3}} \right) = \dfrac{4}{{27}}\pi {h^3}{\tan ^2}a$

Ans. $V = \pi {({\text{ radius }})^2}{\text{x}}$ height

$ = \pi {(10)^2}h\quad ($ radius $ = 10\;{\text{m}})$

$ = 100\pi h$

$\dfrac{{dV}}{{dt}} = 100\pi \dfrac{{dh}}{{dt}}$

Tank is being filled with wheat at rate of 314 cubic meters per hour.

 $\dfrac{{dV}}{{dt}} = 314\;{{\text{m}}^3}/{\text{h}}$

$314 = 100\pi \dfrac{{dh}}{{dt}}$

$ \Rightarrow \dfrac{{dh}}{{dt}} = \dfrac{{314}}{{100(3.14)}} = \dfrac{{314}}{{314}} = 1$

The depth of wheat is increasing at $1\;{\text{m}}/{\text{h}}$.

The correct answer is ${\mathbf{A}}$.