Application of Derivatives

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

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