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A-vessel-containing-water-has-the-shape-of-and-inverted-right-circular-cone-with-base-radius-2m-and-height-5m-The-water-flows-from-the-apex-of-the-cone-at-a-constant-rate-of-0-2-m-3-min-Find-the-r

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Question Number 132440 by liberty last updated on 14/Feb/21
A vessel containing water has the shape of and inverted right circular cone with base radius 2m and height 5m The water flows from the apex of the cone at a constant rate of 0.2 m^3 /min. Find the rate at which the water level is dropping when the depth of the water is 4m.
$$\mathrm{A}\:\mathrm{vessel}\:\mathrm{containing}\:\mathrm{water}\:\mathrm{has}\:\mathrm{the}\: \\ $$$$\mathrm{shape}\:\mathrm{of}\:\mathrm{and}\:\mathrm{inverted}\:\mathrm{right}\:\mathrm{circular} \\ $$$$\mathrm{cone}\:\mathrm{with}\:\mathrm{base}\:\mathrm{radius}\:\mathrm{2m}\:\mathrm{and}\:\mathrm{height}\:\mathrm{5m} \\ $$$$\mathrm{The}\:\mathrm{water}\:\mathrm{flows}\:\mathrm{from}\:\mathrm{the}\:\mathrm{apex} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{cone}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{rate}\: \\ $$$$\mathrm{of}\:\mathrm{0}.\mathrm{2}\:\mathrm{m}^{\mathrm{3}} /\mathrm{min}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{at} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{water}\:\mathrm{level}\:\mathrm{is}\:\mathrm{dropping} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{depth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{water}\:\mathrm{is} \\ $$$$\mathrm{4m}. \\ $$
Answered by bemath last updated on 14/Feb/21
The volume of the water at any time t is V=(1/3)π^2 h We are given (dV/dt) = −0.2 and asked to find (dh/dt) when h=4 . (dV/dt) is negative the volume is decreasing . from proportional sides (r/h)=(2/5) r= ((2h)/5) ⇒V=((4πh^3 )/(75)) and (dV/dh)=((4πh^2 )/(25)) (dV/dt)=(dV/dh)×(dh/dt) ; −0.2 = ((4πh^2 )/(25))×(dh/dt) (dh/dt) = −(5/(4πh^2 )) when h=4 : (dh/dt)∣_4 = −(5/(64π)) ≈ −0.0249 m/min
$$\mathrm{The}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{water}\:\mathrm{at}\:\mathrm{any} \\ $$$$\mathrm{time}\:\mathrm{t}\:\mathrm{is}\:\mathrm{V}=\frac{\mathrm{1}}{\mathrm{3}}\pi^{\mathrm{2}} \mathrm{h} \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{given}\:\frac{\mathrm{dV}}{\mathrm{dt}}\:=\:−\mathrm{0}.\mathrm{2}\:\mathrm{and}\:\mathrm{asked} \\ $$$$\mathrm{to}\:\mathrm{find}\:\frac{\mathrm{dh}}{\mathrm{dt}}\:\mathrm{when}\:\mathrm{h}=\mathrm{4}\:.\:\frac{\mathrm{dV}}{\mathrm{dt}}\:\mathrm{is}\:\mathrm{negative} \\ $$$$\mathrm{the}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{decreasing}\:. \\ $$$$\mathrm{from}\:\mathrm{proportional}\:\mathrm{sides}\:\frac{\mathrm{r}}{\mathrm{h}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\:\mathrm{r}=\:\frac{\mathrm{2h}}{\mathrm{5}}\:\Rightarrow\mathrm{V}=\frac{\mathrm{4}\pi\mathrm{h}^{\mathrm{3}} }{\mathrm{75}}\:\mathrm{and}\:\frac{\mathrm{dV}}{\mathrm{dh}}=\frac{\mathrm{4}\pi\mathrm{h}^{\mathrm{2}} }{\mathrm{25}} \\ $$$$\:\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{dV}}{\mathrm{dh}}×\frac{\mathrm{dh}}{\mathrm{dt}}\:;\:−\mathrm{0}.\mathrm{2}\:=\:\frac{\mathrm{4}\pi\mathrm{h}^{\mathrm{2}} }{\mathrm{25}}×\frac{\mathrm{dh}}{\mathrm{dt}} \\ $$$$\:\frac{\mathrm{dh}}{\mathrm{dt}}\:=\:−\frac{\mathrm{5}}{\mathrm{4}\pi\mathrm{h}^{\mathrm{2}} }\:\mathrm{when}\:\mathrm{h}=\mathrm{4} \\ $$$$:\:\frac{\mathrm{dh}}{\mathrm{dt}}\mid_{\mathrm{4}} \:=\:−\frac{\mathrm{5}}{\mathrm{64}\pi}\:\approx\:−\mathrm{0}.\mathrm{0249}\:\mathrm{m}/\mathrm{min} \\ $$
Commented by bemath last updated on 14/Feb/21
Commented by liberty last updated on 16/Feb/21
nice
$$\mathrm{nice} \\ $$
Answered by mr W last updated on 14/Feb/21
h_0 =5 m r_0 =2 m V_0 =((πr_0 ^2 h_0 )/3) (V/V_0 )=((h/h_0 ))^3 V=((h/h_0 ))^3 V_0 (dV/dt)=3((h/h_0 ))^2 (V_0 /h_0 )×(dh/dt)=πr_0 ^2 ((h/h_0 ))^2 (dh/dt) ⇒(dh/dt)=((dV/dt)/(πr_0 ^2 ((h/h_0 ))^2 ))=((−0.2)/(π×2^2 ×((4/5))^2 ))=−(5/(64π)) ≈−0.025 m/min =−1.49 m/h
$${h}_{\mathrm{0}} =\mathrm{5}\:{m} \\ $$$${r}_{\mathrm{0}} =\mathrm{2}\:{m} \\ $$$${V}_{\mathrm{0}} =\frac{\pi{r}_{\mathrm{0}} ^{\mathrm{2}} {h}_{\mathrm{0}} }{\mathrm{3}} \\ $$$$\frac{{V}}{{V}_{\mathrm{0}} }=\left(\frac{{h}}{{h}_{\mathrm{0}} }\right)^{\mathrm{3}} \\ $$$${V}=\left(\frac{{h}}{{h}_{\mathrm{0}} }\right)^{\mathrm{3}} {V}_{\mathrm{0}} \\ $$$$\frac{{dV}}{{dt}}=\mathrm{3}\left(\frac{{h}}{{h}_{\mathrm{0}} }\right)^{\mathrm{2}} \frac{{V}_{\mathrm{0}} }{{h}_{\mathrm{0}} }×\frac{{dh}}{{dt}}=\pi{r}_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{{h}}{{h}_{\mathrm{0}} }\right)^{\mathrm{2}} \frac{{dh}}{{dt}} \\ $$$$\Rightarrow\frac{{dh}}{{dt}}=\frac{\frac{{dV}}{{dt}}}{\pi{r}_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{{h}}{{h}_{\mathrm{0}} }\right)^{\mathrm{2}} }=\frac{−\mathrm{0}.\mathrm{2}}{\pi×\mathrm{2}^{\mathrm{2}} ×\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{2}} }=−\frac{\mathrm{5}}{\mathrm{64}\pi} \\ $$$$\approx−\mathrm{0}.\mathrm{025}\:{m}/{min}\:=−\mathrm{1}.\mathrm{49}\:{m}/{h} \\ $$
Commented by otchereabdullai@gmail.com last updated on 14/Feb/21
fantastic
$$\mathrm{fantastic} \\ $$

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