2022-M2-07

$\begin{array}{rcl}
y & = & \ln (x+2) \\
\dfrac{dy}{dx} & = & \dfrac{1}{x+2} \\
\left. \dfrac{dy}{dx} \right|_{x=h} & = & \dfrac{1}{h+2}
\end{array}$

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

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

Therefore, the required area $A$

$\begin{array}{cl}
= & \dint_0^h \left(\dfrac{1}{h+2}x -\dfrac{h}{h+2}+\ln(h+2)-\ln(x+2)\right) dx \\
= &\left[ \dfrac{1}{h+2} \times \dfrac{1}{2}x^2-\dfrac{h}{h+2}x+x\ln(h+2)\right]_0^h-\dint_0^h\ln(x+2)dx \\
= & \dfrac{h^2}{2(h+2)} +\dfrac{h^2}{h+2}+h\ln (h+2)-\left[ x\ln(x+2)\right]_0^h +\dint_0^h xd\ln(x+2) \\
= & \dfrac{-h^2}{2(h+2)} +h\ln(h+2)-h\ln(h+2) +\dint_0^h\dfrac{x}{x+2} dx \\
= & \dfrac{-h^2}{2h+4} +\dint_0^h \dfrac{x+2-2}{x+2}dx \\
= & \dfrac{-h^2}{2h+4} +\dint_0^h\left( 1-\dfrac{2}{x+2}\right)dx \\
= & \dfrac{-h^2}{2h+4} +\left[ x-2\ln|x+2|\right]_0^h \\
= & \dfrac{-h^2}{2h+4} +h-2\ln(h+2)+2\ln 2\\
= & \dfrac{-h^2+2h^2+4h}{2h+4}-2\ln(h+2)+2\ln 2 \\
= & \dfrac{h^2+4h}{2h+4}-2\ln(h+2)+2\ln 2
\end{array}$

$\begin{array}{rcl}
A & = & \dfrac{h^2+4h}{2h+4}-2\ln(h+2)+2\ln 2 \\
\dfrac{dA}{dt} & = & \dfrac{(2h+4)(2h+4)-(h^2+4h)(2)}{(2h+4)^2}\times \dfrac{dh}{dt}-\dfrac{2}{h+2}\times \dfrac{dh}{dt} \\
\dfrac{dA}{dt} & = & \left(\dfrac{4h^2+16h+16-2h^2-8h}{4(h+2)^2}-\dfrac{2}{h+2}\right)\times \dfrac{dh}{dt} \\
\dfrac{dA}{dt} & = & \left(\dfrac{2h^2+8h+16}{4(h+2)^2}-\dfrac{8(h+2)}{4(h+2)^2}\right)\times \dfrac{dh}{dt} \\
\dfrac{dA}{dt} & = & \dfrac{2h^2+8h+16-8h-16}{4(h+2)^2}\times \dfrac{dh}{dt} \\
\dfrac{dA}{dt} & = & \dfrac{h^2}{2(h+2)^2}\times \dfrac{dh}{dt} \ \ldots \unicode{x2460}
\end{array}$

Also,

$\begin{array}{rcl}
h & = & 3^{-t} \\
\dfrac{dh}{dt} & = & (-\ln3) 3^{-t} \ \ldots \unicode{x2461}
\end{array}$

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

Sub. $t=1$, $h=3^{-1}$ and $\unicode{x2461}$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
\left. \dfrac{dA}{dt} \right|_{t=1} & = & \dfrac{(3^{-1})^2}{2(3^{-1}+2)^2}\times (-\ln3) 3^{-1} \\
\left. \dfrac{dA}{dt} \right|_{t=1} & = & \dfrac{-\ln 3}{2(3^3)(3^{-1}+2)^2} \\
\left. \dfrac{dA}{dt} \right|_{t=1} & = & \dfrac{-\ln 3}{6(1+2(3))^2} \\
\left. \dfrac{dA}{dt} \right|_{t=1} & = & \dfrac{-\ln 3}{294}
\end{array}$

Therefore, the rate of change of $A$ is $\dfrac{-\ln3}{294} \text{sq. unit}/\text{s}$ when $t=1$.