water is dripping out from a conical funnel of semi vertical angle 45 at uniform rate of 2 cm^2/s (is its sure areea) through a tiny hole at vertical of the bottom wht is the rate of decrese of slant height when the slant height of water is 4cm

let r be the radius , h be the height and V be the volume of the funnel at any time t.
$V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}..............\left(1\right)$
let l be the slant height of the funnel.
given: semi-vertical angle = 45 deg
in the triangle ADE;
$$\begin{gathered} \sin 45=\frac{DE}{AE}\Rightarrow \frac{1}{\sqrt{2}}=\frac{r}{l} \\ \cos 45=\frac{AD}{AE}\Rightarrow \frac{1}{\sqrt{2}}=\frac{h}{l} \\ r=\frac{l}{\sqrt{2}} and h=\frac{l}{\sqrt{2}}....\left(2\right) \end{gathered}$$
therefore the eq(1) can be rewritten as:
$$\begin{gathered} V=\frac{1}{3}\mathrm{\pi }*{\left(\frac{{\rm I}}{\sqrt{2}}\right)}^2*\frac{{\rm I}}{\sqrt{2}} \\ =\frac{\mathrm{\pi }}{3*2*\sqrt{2}}*{{\rm I}}^3 \\ \mathrm{V}=\frac{\mathrm{\pi }}{6\sqrt{2}}{{\rm I}}^3...........\left(3\right) \end{gathered}$$
differentiating wrt t :
$$\begin{gathered} \frac{dV}{dt}=\frac{\mathrm{\pi }}{6\sqrt{2}}*3l^2*\frac{dl}{dt} \\ \frac{dV}{dt}=\frac{\mathrm{\pi }}{2\sqrt{2}}{l}^2*\frac{dl}{dt} \\ \frac{dl}{dt}=\frac{2\sqrt{2}}{{\mathrm{\pi l}}^2}*\frac{dV}{dt}.............\left(4\right) \end{gathered}$$
since it is given that rate of change (decrease) of volume of water wrt t is
$\frac{dV}{dt}=-2 c{m}^3/sec$
therefore
$$\begin{gathered} \frac{dl}{dt}=\frac{2\sqrt{2}}{{\mathrm{\pi l}}^2}*(-2)=-\frac{4\sqrt{2}}{{\mathrm{\pi l}}^2} \\ {\overline{)\frac{dl}{dt}}}_{at l = 4}=-\frac{4\sqrt{2}}{\mathrm{\pi }*(4{)}^2} \\ =-\frac{\sqrt{2}}{4\mathrm{\pi }}cm/sec \end{gathered}$$

hope this helps you