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.

Spammed Answers reported on the spot!​

Answer 1

${ \underline{ \huge{ \sf{Solution}}}}$

Given That,$\rm \frac{dv}{dt}=1c^{3}/s$where V is the Volume of Water in the conical Vessel.

From the Figure(Attached),

${ \implies \displaystyle\rm{l=4cm \: \: h = l\cos \frac{ \pi }{6} = \frac{ \sqrt{3} }{2} }}$

${ \implies \displaystyle \rm{l \: and \: r=l \: sin \frac{\pi}{6} = \frac{l}{2}.}}$

Therefore,

${ \implies \displaystyle \rm{v = \frac{1}{3} { \pi r}^{2} h = \frac{ \pi}{3} \frac{ {l}^{2} }{2} \frac{ \sqrt{3} }{2} l = \dfrac{ \sqrt{3}}{24} {l}^{3} }}$

${ \boxed{ \implies \displaystyle \rm{ \frac{dv}{dt} = \frac{ \sqrt{3 \pi} }{8} {l}^{2} \dfrac{dl}{dt} }}}$

Therefore Rate of decreases of Slant Height is 4cm.

${ \implies \displaystyle \rm{1 = \frac{ \sqrt{3 \pi} }{8} 16. \frac{dl}{dt} }}$

${ \implies \displaystyle \rm{ \frac{dl}{dt} = \frac{ 1}{2 \sqrt{3 \pi} } cm /s }}$

${ \boxed{ \color{hotpink} { \implies \displaystyle \sf{ \frac{dl}{dt} = \frac{ 1}{2}\sqrt{3\pi} } cm /s }}}$

Therefore,

The Rate of Slant Height =$\sf \frac{ 1}{2\sqrt{3 \pi} } cm /s .$

${ \rule{50000pt}{5pt }}$

☞︎︎︎Additional Information

Slant Height

The slant height is a very important property of a pyramid or a cone. Essentially, it measures the height of the apex, but along the slant of one of the lateral faces.

Unless the cone is a right circular cone or a right regular polygonal pyramid, the slant height will not be consistent from place to place, and thus it is not very useful. Thus, over here I will stick to right circular cones, right regular polygonal pyramids, and other right cones with simple bases.

The slant height is the shortest possible distance from the base to the apex along the surface of the solid. It is usually denoted either s or l.

For example, in a right circular cone, if we pick a point on the edge of the base and connect it to the apex with a straight segment, we will have our slant height. If we call the radius r and the height h, then the Pythagorean Theorem quickly tells us that 

• $\sf{s= \sqrt{r2+h2.}}$

The Pythagorean Theorem also helps us calculate the slant height for a right pyramid with a regular polygon base. The slant height is the altitude of one of the lateral faces. Let the apothem be a and the height be h. Then we have 

• $\sf{s= \sqrt{a2+h2.}}$

If we take a simple shape such as a rectangle, and then we have a right rectangular pyramid, then the same principle holds. However, the slant height will depend on the face. For example, if the rectangle has dimensions 18 and 32, and the pyramid has height 12, then two faces have slant height 15, and two have slant height 20.

The slant height finds its main use in calculating surface areas, and it can help to determine other information about the solid. These are generally the two main uses.

______________________

Answer 2

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

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

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

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

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

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

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

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

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

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

$\pmb{ \bf{From \: the \: above \: attachment..}}</p><p>\pink{➠}{ \sf{cos \: {\bf x} =\frac{h}{l} = \frac{ \sqrt{3} }{2} }}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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