Application of Derivatives

$ \begin{array}{cc} \text { } \text { Given curve, } x^2-a^2=\frac{x^2 y^2}{a^2} \\ \Rightarrow 2 x=\frac{1}{a^2}\left[x^2(2 y) \frac{d y}{d x}+(2 x) y^2\right] \\ \Rightarrow 1=\frac{1}{a^2}\left[x y \frac{d y}{d x}+y^2\right] \end{array} $

$ \begin{aligned} & \Rightarrow \quad x y \frac{d y}{d x}+y^2=a^2 \Rightarrow \quad x y \frac{d y}{d x}=a^2-y^2 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{a^2-y^2}{x y} \ldots \\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)_{P(\alpha, y)} \quad=\frac{a^2-y^2}{\alpha y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &\text { Now, Since } P(\alpha, y) \text { lies on the curve. }\\ &\begin{aligned} & \therefore \alpha^2-a^2=\frac{\alpha^2 y^2}{a^2} \\ & \Rightarrow \quad y^2=\frac{a^2}{\alpha^2}\left(\alpha^2-a^2\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} \end{aligned} $

∴ Length of subnormal at $P(\alpha, y)$ is

$ y_1\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=y\left(\frac{a^2-y^2}{\alpha y}\right) $

(using Eq. (i))

$ \begin{aligned} & \quad=\frac{a^2-\left(\frac{a^2}{\alpha 2}\left(\alpha^2-a^2\right)\right)}{\alpha} \\ & \quad(\text { using Eq. }(\mathrm{ii})) \\ & \quad=\quad \frac{a^4}{\alpha^3}=\frac{k}{\alpha^3} \quad \text { (given) } \\ & \therefore \quad k=a^4 \end{aligned} $

Volume of rectangular parallelopiped is $27 \mathrm{~m}^3$

Let $A=$ Area of base and $h$ is height

$ \begin{aligned} & \therefore \quad V=A h \\ & \frac{d V}{d t}=A \frac{d h}{d t} \\ & \frac{d V}{d t}=A\left(\frac{3 d V}{d t}\right) \\ & \quad\left[\because \frac{d h}{d t}=\frac{3 d V}{d t}\right] \end{aligned} $

$ \begin{aligned} & A=1 / 3 \\ & \because \quad V=\frac{1}{3} h \\ & 27=\frac{h}{3} \Rightarrow h=81 \\ & \therefore \text { Height of tank }=81 \mathrm{~m} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} A.\,\,\, f(x) & =3 x^4-2 x^3-6 x^2+6 x+1 \\ f^{\prime}(x) & =12 x^3-6 x^2-12 x+6 \\ f^{\prime}(x) & =6\left(2 x^3-x^2-2 x+1\right) \\ f^{\prime}(x) & =6(x-1)(2 x-1)(x+1) \\ \text { Put, } \quad f^{\prime}(x) & =0 x=-1, \frac{1}{2}, 1 \end{aligned} \text { } \end{aligned} $

$f^{\prime}(x)$ is increasing in $\left(-1, \frac{1}{2}\right) \cup(1, \infty)$ decreasing value in $(\infty,-1) \cup(1 / 2,1)$ minimum value at $x=1$ or -1 maximum value of $x=\frac{1}{2}$

$ \begin{aligned} & \text { B. } \quad f(x)=x+\frac{1}{x} \\ & \quad f^{\prime}(x)=1-\frac{1}{x^2} \text { put } f^{\prime}(x)=0 \\ & \because \quad x^2=1 \Rightarrow x= \pm 1 \\ & \quad f^{\prime \prime}(x)=\frac{2}{x^3} \\ & \text { Maximum at } x=-1 \end{aligned} $

C. $f(x)=x^4(7-x)^3$

$ f^{\prime}(x)=4 x^3(7-x)^3-3 x^4(7-x)^2 $

Put

$ \begin{aligned} & f^{\prime}(x)=0 \Rightarrow x^3(7-x)^2(28-7 x)=0 \\ & \qquad x=4 \\ & \text { Maximum at } x=4 \end{aligned} $

D.

$ \begin{aligned} f(x) & =x^4+(8-x)^4 \\ f^{\prime}(x) & =4 x^3-4(8-x)^3, \text { put } \end{aligned} $

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

$x^3=(8-x)^3 \Rightarrow x=8-x$

$2 x=8 \Rightarrow x=4$

$f^{\prime \prime}(x)=12 x^2+12(8-x)^2$

$f^{\prime \prime}(4) \geq 0, \quad \therefore$ Minimum of $x=4$

Let $A$ and $P$ are area and perimeter of circle.

We have,

$ \begin{aligned} \frac{d A}{d t} & =\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d}{d t}\left(\pi r^2\right)=\frac{1}{\sqrt{\pi}} \\ \Rightarrow \quad 2 \pi r \frac{d r}{d t} & =\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d r}{d t}=\frac{1}{2 \pi r \sqrt{\pi}} \end{aligned} $

Now, $P=2 \pi r$

$ \begin{aligned} & \Rightarrow \frac{d P}{d r}=2 \pi \times \frac{d r}{d t}=2 \pi \times \frac{1}{2 \pi r \sqrt{\pi}}=\frac{1}{r \sqrt{\pi}} \\ & =\frac{1}{\frac{1}{2 \sqrt{\pi}} \times \sqrt{\pi}} \\ & {\left[\because P=\sqrt{\pi} \Rightarrow 2 \pi r=\sqrt{\pi} \Rightarrow r=\frac{1}{2 \sqrt{\pi}}\right]} \\ & =2 \text { unit } / \text { sec. } \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &y=\frac{x^n}{a^{n-1}} \quad \therefore \frac{d y}{d x}=\frac{n x^{n-1}}{a^{n-1}} \end{aligned} $

$ \begin{aligned} \therefore \text { Length of subnormal } & =y \frac{d y}{d x} \\ & =\frac{x^n}{a^{n-1}} \times n \frac{x^{n-1}}{a^{n-1}} \\ & =n \frac{x^{2 n-1}}{a^{2 n-2}} \end{aligned} $

∴ Length of subnormal at

$ (\alpha, \beta)=\frac{n \alpha}{a^{2 n-2}} $

Now, it is given that

Length of subnormal is proportional to $a^2$

$ \therefore \quad 2 n-1=2 \Rightarrow n=\frac{3}{2} $

Let $P(x)=a x^3+b x^2+c x+d$

$ \therefore \quad P^{\prime}(x)=3 a x^2+2 b x+c $

Since, $P(x)$ has extreme value at $x=1$

$ \therefore \quad P^{\prime}(1)=0 \quad \Rightarrow \quad 3 a+2 b+c=0 $

Now, we have

$ \begin{gathered} \lim _{x \rightarrow 0}\left(\frac{P(x)+4}{x^2}+2\right)=6 \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{a x^3+b x^2+c x+d+4}{x^2}=4 \\ \Rightarrow \quad \lim _{x \rightarrow 0} a x+b+\frac{c}{x}+\frac{d+4}{x^2}=4 \end{gathered} $

Since, value of limit is finite

$\therefore \quad c=0$ and $d+4=0 \Rightarrow c=0$,

$d=-4$

$ \therefore \quad \mathop {\lim }\limits_{x \to 0} a x+b=4 \Rightarrow b=4 $

$ \begin{aligned} &\text { Putting } b=4, c=0 \text { in Eq. (i), we get }\\ &\begin{gathered} 3 a+8=0 \\ \Rightarrow \quad a=\frac{-8}{3} \\ \therefore \quad P(x)=\frac{-8}{3} x^3+4 x^2-4 \\ \Rightarrow \quad \frac{d P(x)}{d x}=-8 x^2+8 x \\ \left.\therefore \frac{d P(x)}{d x}\right|_{x=1 / 2}=\frac{-8}{4}+\frac{8}{2}=2 \end{gathered} \end{aligned} $