Respuesta :
The conic water vessel of height 40 cm and radius 8 cm filled at 20 cm^3/s gives;
- The rate at which the water level is rising when the water level is 12 cm from the vertex is approximately 1.1 cm/s.
- When the vessel is one-quarter filled the rate at which the water level is rising is approximately 0.21 cm/s.
How can the water level rate be found?
The volume of the vessel, v = π•r^2•h/3
[tex] \frac{h}{r} = \frac{40}{8} = 5[/tex]
h = 5•r
Therefore;
v = π•r^2•(h)/3 = π•(h/5)^2•h/3
v = π•h^3/75
By chain rule of differentiation, we have;
[tex] \frac{dv}{dh} = \frac{dv}{dt} \times \frac{dt}{dh} [/tex]
Which gives;
[tex]π• \frac{ {h}^{2} }{25} = 20 × \frac{dt}{dh} [/tex]
[tex] \frac{dh}{dt} = \frac{20}{\pi \frac{ {h}^{2} }{25} } [/tex]
When the height is 12 cm from the vertex, we have;
[tex] \frac{dh}{dt} = \frac{20}{\pi \frac{ {12}^{2} }{25} } = 1.1 [/tex]
- The rate at which the water level is rising when the water level is 12 cm from the vertex is approximately 1.1 cm/s.
When the vessel is one-quarter filled, we have;
v = π•h^3/75
π•(40)^3/(75×4) = π•h^3/75
10•(40)^2 = 16,000 = h^3
h = (16,000)^(1/3) = 25 (approx)
Which gives;
[tex] \frac{dh}{dt} = \frac{20}{\pi \frac{ {25}^{2} }{25} } = 0.21 [/tex]
When the vessel is one-quarter filled the rate at which the water level is rising is approximately 0.21 cm/s.
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