Question

Water is dropping out at a constant rate of 1 cubic cm/sec through a small hole at the vertex of the conical container, whose axis is vertical. If the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical angle of the conical vessel is π/6.

Answer 1

$\mapsto\bf{Given :-}$

Water is dripping out at a steady rate of 1cu cm /sec through a tiny hole at the vertex of the conical vessel whose axis is vertical. When the slant height of water in the vessel is 4cm find the rate of decrease of slant height where the semi vertical angle of the cone is π/6

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$\mapsto\bf{let's\:do\:it}$

rate of dripping of water from conical funnel

$\looparrowright$$\bold\pink{\fbox{\sf{=dv/dt=1cm\: cubic/sec}}}$

volume of water v$=1/3pi r^2h$

$\looparrowright$ also

$sin60=r/l$

$r=(root3)/2*l/$

$cos60=h/l$

$h=l/2$

$\looparrowright$ $\sf\color{red}$

v=$(1/3)*pi*3/4*l^2*l/2$

$=(1/8)*pi*l^3$

$dv/dt=(1/8)*pi*4(l^2)(dl/dt)$

$\boxed{\bf{atl=4, dv/dt=1}}$

$so dl/dt=(1*8)/64pi$

$\huge\textit{= (1/8)pi c.m./sec}$

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Thankyou :)

Answer 2

$\large{\mathbb{\colorbox{purple} {\boxed{\boxed{\colorbox{white} {-:Answer:-}}}}}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{peru}{Given::}}}}}}$

$\pink{➠}{ \sf{ \bf{ \frac {dv}{dt}} = \sf1 {cm}^{3}/s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\: \: \: \: \: \: \: ...(i) }}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{pink}{To \: find::}}}}}}$

$\pink{➠}{ \sf{ Rate \: of \: decrease \: of \: slant \: height.}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{pin}{Formula \: used::}}}}}}$

$\pink{➠}{ \sf{ Volume_{(cone)}= \frac{1}{3}\pi {r}^{2}h }}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{plum}{Construction::}}}}}}$

$\pink{➠}{ \sf{ Kindly \: refer \: the \: attachment\:also!!}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{blue}{Concept \: required::}}}}}}$

$\pmb{ \bf{From \: the \: above \: attachment..}}$

$\pink{➠}{ \sf{cos \: {\bf x} =\frac{h}{l} = \frac{ \sqrt{3} }{2} }}$

$\pink{➠}{ \sf{Radius_{(cone)},(r)= \frac{1}{2} }}$

$\pink{➠}{ \sf{Height_{(cone)},(h)= \frac{ \sqrt{3} }{2} l}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{cyan}{According \: to \: Question::}}}}}}$

$\pmb{ \bf{Let's \: start \: with \: help \: of \: formulas!!}}$

$\pink{➠}{ \sf{ Volume_{(cone)}= \frac{1}{3}\pi {r}^{2}h }}$

${: : \implies{ \sf{ Volume_{(cone)}= \frac{1}{3}\pi { \big( \frac{1}{2} \big)}^{2} \Big( \frac{ \sqrt{3} }{2}l \Big) }}}$

${: : \implies{ \sf{ Volume_{(cone)}= \frac{\pi}{8 \sqrt{3} } }}}$

$\pink{➠}{ \sf{ \bf{ \frac {dv}{dt}} = \sf1 {cm}^{3}/s \: \: \: \: \: \: \:\{ from \: eq. \: ..(i) \}}}$

$: : \implies{ \sf{ \bf{ \frac {d}{dt}}} \Big[\frac{\pi}{8 \sqrt{3}} {l}^{3} \Big ] = \sf1 {cm}^{3}/s }$

$: : \implies{ \sf \frac{\pi}{8 \sqrt{3}} { \bf{ \frac {d}{dt}}}{(l)}^{3} = \sf1 {cm}^{3}/s }$

$: : \implies{ \sf \frac{3 {l}^{2}\pi }{8 \sqrt{3}} { \bf{ \frac {d}{dt}}} = \sf1 {cm}^{3}/s }$

${: : \implies{ \sf \frac{ \sqrt{3} (16\pi )}{8 } { \bf{ \frac {dl}{dt}}} = \sf1 {cm}^{3}/s \: \: \: \: \: \: \: \: \{∵At \: l=4cm \}}}$

${: : \implies{ \sf2 \sqrt{3} \pi { \bf{ \frac {dl}{dt}}} = \sf1 {cm}^{3}/s}}$

${: : \implies{ \sf{ \bf{ \frac {dl}{dt}}} = \sf \: \frac{1}{2 \sqrt{3}\pi } {cm}^{3}/s}}$

$\pmb {\bf{Hence,}}$

$\purple᪣ {\boxed{ \sf{ Rate \: of \: decrease \: of \: slant \: height = \frac{1}{2 \sqrt{3} {\pi}}{cm}^{3} /s }}} ᪣$

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