Question

Water is running into a conical vessel $15cm$ deep and $5cm$ in radius, at the rate of $0.1c{m}^3/s$, when the water is $6cm$ deep, find at what rate is
i) the water level is rising
ii) the water surface area increasing
iii) the welted surface of the vessel increasing.

Answer 1

Let $v$ be the volume of water in the cone i.e., the volume of the water cone $C{A}^{\prime }{B}^{\prime }$ at any time $t$

Let $C{O}^{\prime }=hO{A}^{\prime }=rC{A}^{\prime }=1$

$\tan \alpha =\frac{OA}{CO}=\frac{5}{15}=\frac{1}{3}$

$\tan \alpha =\frac{O^{\prime }{A}^{\prime }}{C{O}^{\prime }}=\frac{r}{h}$

$\frac{1}{3}=\frac{r}{h}$

$v=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi {\left(\frac{h}{3}\right)}^{2}h=\frac{\pi }{2\pi }{h}^3$

$\dfrac {dv}{dt}=\dfrac {3}{2\pi}\pi h^2 \dfrac {}dh{}dt

$

$0.1=\frac{3\pi }{27}{h}^2\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{2.7}{3\pi h^2}$

${\left(\frac{dh}{dt}\right)}_{h=6}=\frac{2.7}{3\pi \left(36\right)}=\frac{1}{40\pi }$

Water level is rising $=\frac{1}{40\pi }cm/sec$

$\left(II\right)A=\pi r^2=\pi \frac{h^2}{9}\Rightarrow \frac{dA}{dt}=\frac{2\pi h}{9}\times \frac{dh}{dt}$

$\frac{dA}{dt}=\frac{1}{30}c{m}^2/sec$

$\left(III\right)I^2=h^2+r^2\Rightarrow h^2+\frac{h^2}{9}=\frac{10h^2}{9}$

$I=\sqrt{10}/3h$

$S=\pi rl=\pi \left(\frac{h}{3}\right)\left(\frac{\sqrt{1}0h}{3}\right)$

$S=N9\sqrt{20}{h}^2\Rightarrow \frac{ds}{dt}=\frac{2\pi \sqrt{10}h}{9}\frac{dh}{dt}$

$\frac{ds}{dt}=\frac{\sqrt{10}}{30}c{m}^2/sec$