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Question Number 132440 by liberty last updated on 14/Feb/21

$$\mathrm{A}\mathrm{vessel}\mathrm{containing}\mathrm{water}\mathrm{has}\mathrm{the}$$$$\mathrm{shape}\mathrm{of}\mathrm{and}\mathrm{inverted}\mathrm{right}\mathrm{circular}$$$$\mathrm{cone}\mathrm{with}\mathrm{base}\mathrm{radius}2\mathrm{m}\mathrm{and}\mathrm{height}5\mathrm{m}$$$$\mathrm{The}\mathrm{water}\mathrm{flows}\mathrm{from}\mathrm{the}\mathrm{apex}$$$$\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{at}\mathrm{a}\mathrm{constant}\mathrm{rate}$$$$\mathrm{of}0.2{\mathrm{m}}^3/\min .\mathrm{Find}\mathrm{the}\mathrm{rate}\mathrm{at}$$$$\mathrm{which}\mathrm{the}\mathrm{water}\mathrm{level}\mathrm{is}\mathrm{dropping}$$$$\mathrm{when}\mathrm{the}\mathrm{depth}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{is}$$$$4\mathrm{m}.$$

Answered by bemath last updated on 14/Feb/21

$$\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{at}\mathrm{any}$$$$\mathrm{time}\mathrm{t}\mathrm{is}\mathrm{V}=\frac{1}{3}{\pi }^2\mathrm{h}$$$$\mathrm{We}\mathrm{are}\mathrm{given}\frac{\mathrm{dV}}{\mathrm{dt}}=-0.2\mathrm{and}\mathrm{asked}$$$$\mathrm{to}\mathrm{find}\frac{\mathrm{dh}}{\mathrm{dt}}\mathrm{when}\mathrm{h}=4.\frac{\mathrm{dV}}{\mathrm{dt}}\mathrm{is}\mathrm{negative}$$$$\mathrm{the}\mathrm{volume}\mathrm{is}\mathrm{decreasing}.$$$$\mathrm{from}\mathrm{proportional}\mathrm{sides}\frac{\mathrm{r}}{\mathrm{h}}=\frac{2}{5}$$$$\mathrm{r}=\frac{2\mathrm{h}}{5}\Rightarrow \mathrm{V}=\frac{4\pi {\mathrm{h}}^3}{75}\mathrm{and}\frac{\mathrm{dV}}{\mathrm{dh}}=\frac{4\pi {\mathrm{h}}^2}{25}$$$$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{dV}}{\mathrm{dh}}\times \frac{\mathrm{dh}}{\mathrm{dt}};-0.2=\frac{4\pi {\mathrm{h}}^2}{25}\times \frac{\mathrm{dh}}{\mathrm{dt}}$$$$\frac{\mathrm{dh}}{\mathrm{dt}}=-\frac{5}{4\pi {\mathrm{h}}^2}\mathrm{when}\mathrm{h}=4$$$$:\frac{\mathrm{dh}}{\mathrm{dt}}{\mid }_4=-\frac{5}{64\pi }\approx -0.0249\mathrm{m}/\min$$

Commented by bemath last updated on 14/Feb/21

Commented by liberty last updated on 16/Feb/21

$$\mathrm{nice}$$

Answered by mr W last updated on 14/Feb/21

$$h_0=5m$$$$r_0=2m$$$$V_0=\frac{\pi r_0^2{h}_0}{3}$$$$\frac{V}{V_0}={\left(\frac{h}{h_0}\right)}^3$$$$V={\left(\frac{h}{h_0}\right)}^3{V}_0$$$$\frac{dV}{dt}=3{\left(\frac{h}{h_0}\right)}^2\frac{V_0}{h_0}\times \frac{dh}{dt}=\pi r_0^2{\left(\frac{h}{h_0}\right)}^2\frac{dh}{dt}$$$$\Rightarrow \frac{dh}{dt}=\frac{\frac{dV}{dt}}{\pi r_0^2{\left(\frac{h}{h_0}\right)}^2}=\frac{-0.2}{\pi \times 2^2\times {\left(\frac{4}{5}\right)}^2}=-\frac{5}{64\pi }$$$$\approx -0.025m/min=-1.49m/h$$

Commented by otchereabdullai@gmail.com last updated on 14/Feb/21

$$\mathrm{fantastic}$$

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