Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Water is dripping out at the steady rate of

1 c u c m / s e c

through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel as

4 c m

, find the rate of decrease of slant height where the semi vertical angle of the cone is

\frac{π}{6}

.

Accepted Answer

Let $l$ be the slant height of conical vessel at any time $t$ .

and $V$ be the volume of cone at that instant.

Then $V = \frac{π}{3} r^{2} h$

Where $r =$ radius of cone ;

$h =$ height of cone

where $h = l \cos \frac{π}{6}$ and $r = l \sin \frac{π}{6}$

Now, $V = \frac{π}{3} {(l \sin \frac{π}{6})}^{2} (l \cos \frac{π}{6})$

$V = \frac{π}{3} l^{3} {(\frac{1}{2})}^{2} (\frac{\sqrt{3}}{2})$

$V = \frac{π l^{3} \sqrt{3}}{24}$

Differentiate both the sides w.r.t. $t$ , we get

$1 c u c m / s e c$

$\frac{d V}{d t} = \frac{π \sqrt{3}}{8} l^{2} \frac{d l}{d t} . . . . . . . . . . . . . . (1)$

Given $\frac{d V}{d t} = - 1 c u c m / s e c$ and $l = 4 c m$

$∴$ From $(1)$ ; we have

$- 1 = \frac{\sqrt{3} π}{8} \times 4^{2} \frac{d l}{d t}$

$\Rightarrow \frac{d l}{d t} = - \frac{1}{2 \sqrt{3} π} c m / s e c$

Hence, the slant height of cone be decreasing at the rate of $\frac{1}{2 \sqrt{3} π} c m / s e c$ .

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !

Water is dripping out at the steady rate of 1 cu 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 as 4 cm, find the rate of decrease of slant height where the semi vertical angle of the cone is π6.

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !