Application of Derivatives

Radius of ball = 10 cm

Given, $\frac{dV}{dt}=50$ cm$^3$/min

Let a be the thickness of ice.

Image

$\begin{aligned} & \text { Volume of ice part }=\frac{4}{3} \pi\left(R^3-r^3\right) \\ & V=\frac{4}{3} \pi\left[(10+a)^3-10^3\right] \\ & =\frac{4}{3} \pi\left[a^3+30 a^2+300 a\right] \\ & \frac{d V}{d t}=\frac{4}{3} \pi \frac{d}{d t}\left(a^3+30 a^2+300 a\right) \\ & 50=\frac{4}{3} \pi\left(3 a^2 \frac{d a}{d t}+60 a \frac{d a}{d t}+300 \frac{d a}{d t}\right) \\ & \Rightarrow \quad 50 \times \frac{3}{4 \pi}=\frac{d a}{d t}\left(3 a^2+60 a+300\right) \\ & \therefore \quad \frac{d a}{d t}=\frac{75}{2 \pi}\left(3 a^2+60 a+300\right) \\ \end{aligned}$

When, $\quad a=15$

$d a / d t=1 / 50 \pi$

Let $f(x, y)=2 x+3 y$,

Given,

$\begin{aligned} \text { Given, } \quad x y & =6 \\ \Rightarrow \quad y & =\frac{6}{x} \Rightarrow f(x)=2 x+3 \times \frac{6}{x} \\ & =\frac{2 x^2+18}{x}=2 x+\frac{18}{x} \end{aligned}$

Now, $f^{\prime}(x)=2-\frac{18}{x^2}$

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

$\begin{aligned} & \Rightarrow \quad 2-\frac{18}{x^2}=0 \Rightarrow \frac{18}{x^2}=2 \\ & \Rightarrow \quad x^2=9 \Rightarrow x= \pm 3 \\ & \Rightarrow \quad f^{\prime \prime}(x)=\frac{2 \times 18}{x^3}=\frac{36}{x^3} \Rightarrow f^{\prime \prime}(3) > 0 \\ \end{aligned}$

$\Rightarrow \quad x=3$ is minima.

$\therefore$ Minimum value $f(3)=2(3)+\frac{18}{3}=6+6=12$

Volume $=\frac{4}{3} \pi r^3$, given $\frac{d V}{d t}=30 \mathrm{~cm}^3 / \mathrm{min}$.

$\Rightarrow \quad \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^2 \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t} \Rightarrow 30=4 \pi r^2 \frac{d r}{d t}$/p>

or $\quad \frac{d r}{d t}=\frac{15}{2 \pi r^2}$ Surface Area $=4 \pi r^2$ ..... (i)

Then, $\frac{d S}{d t}=4 \pi \cdot 2 r \frac{d r}{d t}=8 \pi r \frac{d r}{d t}$

Use Eq. (i), $\frac{d S}{d t}=8 \pi r \cdot\left(\frac{15}{2 \pi r^2}\right)=\frac{60}{r}$

When, $r=6 \mathrm{~cm}$

$\left(\frac{d S}{d t}\right)_{r=6}=\frac{60}{6}=10 \mathrm{~cm}^2 / \mathrm{min} .$

$\begin{aligned} & g(x)=\frac{1}{6} f\left(3 x^2-1\right)+\frac{1}{2} f\left(1-x^2\right) \\ & g^{\prime}(x)=\frac{1}{6} f^{\prime}\left(3 x^2-1\right)(6 x)+\frac{1}{2} f^{\prime}(1-x)^2(-2 x) \\ & =x\left(f^{\prime}\left(3 x^2 \quad 1\right)+(-1) f^{\prime}\left(1-x^2\right)\right) \end{aligned}$

$g(x)$ is increasing, when $g^{\prime}(x) > 0$

Case 1 $f^{\prime}\left(3 x^2-1\right)-f^{\prime}\left(1-x^2\right) > 0$ and $x > 0$ ....... (a)

$\Rightarrow \quad f^{\prime}\left(3 x^2-1\right)>f^{\prime}\left(1-x^2\right)$ ....... (i)

Given that, $f^{\prime \prime}(x) > 0$

$\begin{array}{ll} \Rightarrow & f(x) \text { is increasing. } \\ \Rightarrow & f^{\prime}(x) > 0\end{array}$

From Eq. (i), $3 x^2-1 > 1-x^2$

$\begin{gathered} 4 x^2 > 2 \Rightarrow x^2>\frac{1}{2} \\ \Rightarrow \quad x < -\frac{1}{\sqrt{2}} \text { and } x > \frac{1}{\sqrt{2}} \\ \therefore \quad x < \frac{-1}{\sqrt{2}} \text { and } x > \frac{1}{\sqrt{2}} \\ \Rightarrow x \in\left(-\infty,-\frac{1}{\sqrt{2}}\right) \cup\left(\frac{1}{\sqrt{2}}, \infty\right) \quad \text{.......(b)} \end{gathered}$

From (a) and (b)

$x \in\left(\frac{1}{\sqrt{2}}, \infty\right) $

Case 2 When $x < 0$

$\begin{aligned} & \text { and } f^{\prime}\left(3 x^2-1\right) < f^{\prime}\left(1-x^2\right) \\ & \Rightarrow 3 x^2-1 < 1-x^2 \\ & \qquad x^2<\frac{1}{2} \Rightarrow x \in\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \\ & \because \quad x < 0 \Rightarrow x \in\left(-\frac{1}{\sqrt{2}}, 0\right)\end{aligned}$

Including case- 1 and case-2.

$x \in\left(-\frac{1}{\sqrt{2}}, 0\right) \cup\left(\frac{1}{\sqrt{2}}, \infty\right)$

Given,

$\begin{aligned} & f(x)=2 x^3-9 a x^2+12 a^2 x+1 \\ & f^{\prime}(x)=6 x^2-18 a x+12 a^2 \end{aligned}$

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

$\begin{aligned} & \Rightarrow \quad 6 x^2-18 a x+12 a^2=0 \\ & \text { or } \quad x^2-3 a x+2 a^2=0 \\ & \Rightarrow \quad(x-a)(x-2 a)=0 \\ & \Rightarrow \quad x=a, 2 a \\ & \text { Now, } \quad f^{\prime \prime}(x)=12 x-18 a \\ & \Rightarrow \quad f^{\prime \prime}(a)=12 a-18 a < 0 \\ & \Rightarrow \quad f^{\prime \prime}(2 a)=24-18 a > 0 \\ \end{aligned}$

$\therefore$ Minimum value attained at $x=2 a$

Maximum value attained at $x=a$

$\begin{aligned} & \therefore \quad p=a \text { and } q=2 a \Rightarrow p^2=q \text { gives, } a^2=2 a \\ & \Rightarrow \quad a(a-2)=0 \Rightarrow a=0 \text { and } a=2 \\ & \text { Since, } a \neq 0, a=2 .\end{aligned}$

$\begin{aligned} 2 y \frac{d y}{d x} & =4 a x^3 \\ (d y / d x)_{(3,6)} & =(2 a)\left(x^3 / y\right)_{3,6}=(2 a)(27 / 6)=9 a \quad {[\because \text { slope of } y=4 x-6 \text { is 4] }}\\ 9 a & =4 \\ \quad & \\ \Rightarrow \quad a & =\frac{4}{9} \Rightarrow 36=\frac{4}{9}(81)+b \Rightarrow b=0\end{aligned}$

$2\alpha+\beta=8$

$f'(x)=6x^2-18ax+12a^2$

Critical points

$f'(x)=0\Rightarrow 6x^2-18ax+12a^2=0$

$\Rightarrow (x-a)(x-2a)=0\Rightarrow x=a,x=2a$

Now, $f''(x)=12x-18a$

at $x=a$

$f''(a)=-6a < 0$,

as $a > 0$

$x=a$ is point of maxima

at $x=2a$

$f''(2a)=6a > 0$

$x=2a$ is point of minima

$\therefore \alpha=a,\beta=2a$

$\because 2\alpha+\beta=8$

$2a+2a=8$

$\Rightarrow a=2$

$A=4\pi r^2$

$dA=8\pi r\,dr=8\pi9 \,. (0.03) =2.16 \pi$ cm$^2$

$\begin{aligned} V & =\frac{\pi}{3}\left(r^2 h\right) \\ d V & =\frac{\pi}{3}\left(2 r h d r+r^2 d h\right) \quad[Q=10 \mathrm{~cm} \text { and } h=20 \mathrm{~cm}] \\ 0 & =200 \cdot 1+25 d h \Rightarrow d h=-8 \mathrm{~cm} / \mathrm{s}\end{aligned}$

Let $x=2, x+\Delta x=1.99$

$\Rightarrow \quad \Delta x=-0.01$

Now, $y=x^3-4 x$

$\begin{aligned} & \frac{d y}{d x}=3 x^2-4 \\ & \left(\frac{d y}{d x}\right)_{x=2}=3.2^2-4=8 \\ & \text { Now, } \Delta y=\left(\frac{d y}{d x}\right)_{x=2} \Delta x \\ & =8 \times(-0.01)=-0.08 \\ \end{aligned}$

Approximate change in $y=-0.08$

Given curves,

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1$ .... (i)

and ${y^3} = 16x$ .... (ii)

Differentiating Eq. (i) w.r.t. x

${{2x} \over {{a^2}}} + {y \over 2}{{dy} \over {dx}} = 0$

$ \Rightarrow {{dy} \over {dx}} = {{ - 4x} \over {{a^2}y}} = {m_1}$ (let)

Differentiating Eq. (ii) w.r.t. $x$

$\begin{aligned} 3 y^2 \frac{d y}{d x} & =16 \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{16}{3 y^2}=m_2 \end{aligned}$

$\because$ Eqs. (i) and (ii) meet each other at right angles

$\begin{array}{rlrl} & m_1 m_2 =-1 \\ \Rightarrow & \frac{16}{3 y^2} \cdot \frac{-4 x}{a^2 y} =-1 \\ \Rightarrow & 64 x =3 a^2 y^3 \Rightarrow a^2=\frac{64 x}{3 y^3} \\ \Rightarrow & \quad a^2 =\frac{64 x}{3 \times 16 x} \quad\left[\text { as } y^3=16 x\right] \\ \Rightarrow & a^2 =\frac{4}{3} \end{array}$

$y=x-x^2$

Area of first square $=A_x=x^2$

Area of second square

$\begin{aligned} & =A_y=y^2=\left(x-x^2\right)^2 \\ \text { Now, } \frac{d A_y}{d A_x} & =\frac{\frac{d A_y}{d x}}{\frac{d A_x}{d x}}=\frac{2\left(x-x^2\right)(1-2 x)}{2 x} \\ & =(1-x)(1-2 x)=1-3 x+2 x^2 \end{aligned}$

$\begin{aligned} & \text { Given, } g(x)=f\left(\tan ^2 x-2 \tan x+4\right), 0 < x < \frac{\pi}{2} \\ & g^{\prime}(x)=f^{\prime}\left(\tan ^2 x-2 \tan x+4\right) \times\left(2 \tan x \cdot \sec ^2 x-2 \sec ^2 x\right) \\ & \because g(x) \text { is increasing } \\&\Rightarrow \quad g^{\prime}(x) > 0 \\ & \Rightarrow \quad f^{\prime}\left(\tan ^2 x-2 \tan x+4\right) \cdot\left(2 \tan x \sec ^2 x-2 \sec ^2 x\right)>0 \\ & \Rightarrow \quad 2 \tan x \sec ^2 x>2 \sec ^2 x, \\ & \quad f^{\prime}\left(\tan ^2 x-2 \tan x+4\right)=f^{\prime}(3)=0, \\ & \Rightarrow \quad \tan x>1 \Rightarrow x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ & \because \text { At } x=\frac{\pi}{4} \Rightarrow x>\frac{\pi}{4}\end{aligned}$

Let the equation of line parallel to X-axis is $y=a$

Then, point of intersection of the line and the curve $y=\sqrt x$ is $(a^2,a)$.

The slope of the curve at any point

${{dy} \over {dx}} = {1 \over {2\sqrt x }}$

At ${({a^2},a)}$ curve and line makes an angle of 45$^\circ$

$ \Rightarrow {\left( {{{dy} \over {dx}}} \right)_{({a^2},a)}} = \tan 45\Upsilon = 1$

$ \Rightarrow {1 \over {2\sqrt {{a^2}} }} = 1 \Rightarrow a = {1 \over 2}$

$\therefore$ Line is $y = {1 \over 2}$

Given, $\frac{d r}{r}=0.05 \Rightarrow d r=0.05 r$

Area of circle

$\begin{gathered} A=\pi r^2 \Rightarrow \frac{d A}{d r}=2 \pi r \\ d A=2 \pi r d r \\ \therefore \quad \frac{d A}{A}=\frac{2 \pi r d r}{\pi r^2}=2 \frac{d r}{r}=2 \cdot(0.05)=0.1 \\ \therefore \text { Error in calculating area }=0.1 \% \end{gathered}$

$y=8 x^2-x^4-4$

Differentiating w.r.t. $x$

$\frac{d y}{d x}=16 x-4 x^3=4 x\left(4-x^2\right)=4 x(2-x)(2+x)$

For stationary points $\frac{d y}{d x}=0$

$\begin{array}{rlrl} \text { or } & & 4 x(2-x)(2+x) & =0 \\ \Rightarrow & & x=0, x=2, x=-2 \end{array}$

At $x=0, y=-4$

At $x=2, y=12$

At $x=-2, y=12$

Stationary points are $(0,-4),(2,12),(-2,12)$.

Given curve,

$y=e^{2 x}+x^2 \Rightarrow d y / d x=2 e^{2 x}+2 x $

At $x=0, y=e^0+0=1$

and $\frac{d y}{d x}=2 \cdot e^0+0=2$

Equation of normal at $(0,1)$ is

$y-1=\frac{-1}{\left(\frac{d y}{d x}\right)_{x=0}}(x-0) $

$\begin{array}{rlrl} y-1 & =\frac{-1}{2} \cdot x \\ \Rightarrow \quad 2 y-2 & =-x \\ \Rightarrow \quad x+2 y-2 & =0 \end{array}$

Distance between normal to the curve and origin is

$\frac{|1 \cdot 0+2 \cdot 0-2|}{\sqrt{1^2+2^2}}=\frac{2}{\sqrt{5}}$

We have, ${{dx} \over {dt}} = 2t + 3$ and ${{dy} \over {dt}} = 4t - 2$

$\therefore$ ${{dy} \over {dx}} = {{dy/dt} \over {dx/dt}} = {{4t - 2} \over {2t + 3}}$

Thus, slope of the tangent to the curve at the point t = 2 is

${\left[ {{{dy} \over {dx}}} \right]_{t = 2}} = {{4(2) - 2} \over {2(2) + 3}} = {6 \over 7}$