# 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

therefore the eq(1) can be rewritten as:
$V=\frac{1}{3}\mathrm{\pi }*{\left(\frac{Ι}{\sqrt{2}}\right)}^{2}*\frac{Ι}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3*2*\sqrt{2}}*{Ι}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{V}=\frac{\mathrm{\pi }}{6\sqrt{2}}{Ι}^{3}...........\left(3\right)$
differentiating wrt t :
$\frac{dV}{dt}=\frac{\mathrm{\pi }}{6\sqrt{2}}*3{l}^{2}*\frac{dl}{dt}\phantom{\rule{0ex}{0ex}}\frac{dV}{dt}=\frac{\mathrm{\pi }}{2\sqrt{2}}{l}^{2}*\frac{dl}{dt}\phantom{\rule{0ex}{0ex}}\frac{dl}{dt}=\frac{2\sqrt{2}}{{\mathrm{\pi l}}^{2}}*\frac{dV}{dt}.............\left(4\right)$
since it is given that rate of change (decrease) of volume of water wrt t is

therefore

hope this helps you

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