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If water is poured into an inverted hollow cone whose semi-vertical angle is 30° , then the its depth (measured along axis) increases at the
rate of 1cm/sec, the rate of which the volume of water increases when the depth is 24 cm.

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Solution

$\begin{matrix} V = \frac{1}{3} π r^{2} h \\ \tan θ = \frac{r}{h} \\ V = \frac{π}{3} \times h^{2} \tan^{2} θ h \\ \frac{d v}{d t} = \frac{π}{3} \tan^{2} θ 3 h^{2} \frac{d h}{d t} \\ = π h^{2} \tan^{2} θ \frac{d h}{d t} \\ = π (24)^{2} \tan^{2} 30^{\circ} (1) {cm}^{3} / \sec \\ = π (24) (24) \times \frac{1}{3} \\ = 192 π {cm}^{3} / \sec \end{matrix}$

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Rate Measure

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If water is poured into an inverted hollow cone whose semi-vertical angle is 30° , then the its depth (measured along axis) increases at the rate of 1cm/sec, the rate of which the volume of water increases when the depth is 24 cm.

V=13πr2htan⁡θ=rhV=π3×h2tan2⁡θhdvdt=π3tan2⁡θ3h2dhdt=πh2tan2⁡θdhdt=π(24)2tan2⁡30∘(1)cm3/sec= π(24)(24)×13=192πcm3/sec

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If water is poured into an inverted hollow cone whose semi-vertical angle is 30° , then the its depth (measured along axis) increases at the
rate of 1cm/sec, the rate of which the volume of water increases when the depth is 24 cm.