2022-M2-07

$\begin{array}{rcl}y& =& \ln (x+2)\\ {\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{1}{x+2}}\\ {{\displaystyle \frac{dy}{dx}}|}_{x=h}& =& {\displaystyle \frac{1}{h+2}}\end{array}$

For $x=h$, $y=\ln (h+2)$. Therefore, the coordinates of the point of contact are $(h,\ln (h+2))$.

Hence, the equation of $L$ is

$\begin{array}{rcl}{\displaystyle \frac{y-\ln (h+2)}{x-h}}& =& {\displaystyle \frac{1}{h+2}}\\ y& =& {\displaystyle \frac{1}{h+2}}(x-h)+\ln (h+2)\\ y& =& {\displaystyle \frac{1}{h+2}}x-{\displaystyle \frac{h}{h+2}}+\ln (h+2)\end{array}$

Therefore, the required area $A$

$\begin{array}{cl}=& {\displaystyle {\int }_0^h({\displaystyle \frac{1}{h+2}}x-{\displaystyle \frac{h}{h+2}}+\ln (h+2)-\ln (x+2))dx}\\ =& {[{\displaystyle \frac{1}{h+2}}\times {\displaystyle \frac{1}{2}}{x}^2-{\displaystyle \frac{h}{h+2}}x+x\ln (h+2)]}_0^h-{\displaystyle {\int }_0^h\ln (x+2)dx}\\ =& {\displaystyle \frac{h^2}{2(h+2)}}+{\displaystyle \frac{h^2}{h+2}}+h\ln (h+2)-{[x\ln (x+2)]}_0^h+{\displaystyle {\int }_0^{h}xd\ln (x+2)}\\ =& {\displaystyle \frac{-h^2}{2(h+2)}}+h\ln (h+2)-h\ln (h+2)+{\displaystyle {\int }_0^h{\displaystyle \frac{x}{x+2}}dx}\\ =& {\displaystyle \frac{-h^2}{2h+4}}+{\displaystyle {\int }_0^h{\displaystyle \frac{x+2-2}{x+2}}dx}\\ =& {\displaystyle \frac{-h^2}{2h+4}}+{\displaystyle {\int }_0^h(1-{\displaystyle \frac{2}{x+2}})dx}\\ =& {\displaystyle \frac{-h^2}{2h+4}}+{[x-2\ln |x+2|]}_0^h\\ =& {\displaystyle \frac{-h^2}{2h+4}}+h-2\ln (h+2)+2\ln 2\\ =& {\displaystyle \frac{-h^2+2h^2+4h}{2h+4}}-2\ln (h+2)+2\ln 2\\ =& {\displaystyle \frac{h^2+4h}{2h+4}}-2\ln (h+2)+2\ln 2\end{array}$

$\begin{array}{rcl}A& =& {\displaystyle \frac{h^2+4h}{2h+4}}-2\ln (h+2)+2\ln 2\\ {\displaystyle \frac{dA}{dt}}& =& {\displaystyle \frac{(2h+4)(2h+4)-(h^2+4h)(2)}{(2h+4{)}^2}}\times {\displaystyle \frac{dh}{dt}}-{\displaystyle \frac{2}{h+2}}\times {\displaystyle \frac{dh}{dt}}\\ {\displaystyle \frac{dA}{dt}}& =& ({\displaystyle \frac{4h^2+16h+16-2h^2-8h}{4(h+2{)}^2}}-{\displaystyle \frac{2}{h+2}})\times {\displaystyle \frac{dh}{dt}}\\ {\displaystyle \frac{dA}{dt}}& =& ({\displaystyle \frac{2h^2+8h+16}{4(h+2{)}^2}}-{\displaystyle \frac{8(h+2)}{4(h+2{)}^2}})\times {\displaystyle \frac{dh}{dt}}\\ {\displaystyle \frac{dA}{dt}}& =& {\displaystyle \frac{2h^2+8h+16-8h-16}{4(h+2{)}^2}}\times {\displaystyle \frac{dh}{dt}}\\ {\displaystyle \frac{dA}{dt}}& =& {\displaystyle \frac{h^2}{2(h+2{)}^2}}\times {\displaystyle \frac{dh}{dt}}\dots \text{①}\end{array}$

Also,

$\begin{array}{rcl}h& =& 3^{-t}\\ {\displaystyle \frac{dh}{dt}}& =& (-\ln 3)3^{-t}\dots \text{②}\end{array}$

For $t=1$, $h=3^{-1}$.

Sub. $t=1$, $h=3^{-1}$ and $\text{②}$ into $\text{①}$, we have

$\begin{array}{rcl}{{\displaystyle \frac{dA}{dt}}|}_{t=1}& =& {\displaystyle \frac{(3^{-1}{)}^2}{2(3^{-1}+2{)}^2}}\times (-\ln 3)3^{-1}\\ {{\displaystyle \frac{dA}{dt}}|}_{t=1}& =& {\displaystyle \frac{-\ln 3}{2(3^3)(3^{-1}+2{)}^2}}\\ {{\displaystyle \frac{dA}{dt}}|}_{t=1}& =& {\displaystyle \frac{-\ln 3}{6(1+2(3){)}^2}}\\ {{\displaystyle \frac{dA}{dt}}|}_{t=1}& =& {\displaystyle \frac{-\ln 3}{294}}\end{array}$

Therefore, the rate of change of $A$ is ${\displaystyle \frac{-\ln 3}{294}}\text{sq. unit}/\text{s}$ when $t=1$.