An inverted cone has a depth of 40 cm and a base of radius 5cm water is poured into - Maths - Application of Derivatives

Question

An inverted cone has a depth of 40 cm and a base of radius 5cm
water is poured into it at a rate of 3/2 cubic centimetres per
minute find the rate at which the level of water in the cone is
rising when the depth is 4 cm

Duke Biswass · Accepted Answer

Let V be the volume in the inverted cone

$v o l u m e = \frac{1}{3} π r^{2} h$

By similar triangles $\frac{5}{40} = \frac{r}{h}$

⇒ $r = \frac{1}{8} h$

Hence volume becomes $V = \frac{1}{192} π h^{3}$

Water is getting filled in funnel at the rate 3/2 cc/sec

Hence $\frac{d v}{d t} = \frac{3}{2}$

$\frac{d}{d t} (\frac{1}{192} π h^{3}) = \frac{1}{64} π h^{2} \frac{d h}{d t} = \frac{3}{2}$

⇒ $\frac{d h}{d t} = \frac{96}{π h^{2}}$

Rate of change of water level = $\frac{d h}{d t} = \frac{96}{π h^{2}}$

When water level is 4 cm $\frac{d h}{d t} = \frac{96}{π 4^{2}} = \frac{6}{π} c m / \sec$

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An inverted cone has a depth of 40 cm and a base of radius 5cm .water is poured into it at a rate of 3/2 cubic centimetres per minute.find the rate at which the level of water in the cone is rising when the depth is 4 cm ..