Water is being poured into a cylinder container at a constant rate. The cylinder has a diameter of 4cm and a height of 11cm. Find the rate of which the volume is changing if the water level is raising at a rate of 1.5 cm/sec.
How do you set it up
Please show work
thanks
How do you set it up
Please show work
thanks
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The volume V is the product of the circular area (A) and the cm-height (h), V = Ah.
(1) A = (Pi)(d/2)^2, where d = the diameter = 4 cm, so --->
A = 4(Pi) cm^2 = 12.566 cm^2 (approx)
(2) Then V = (12.566 cm^2)(h cm) = 12.566 h cm^2, remember the h is in cm
(3) dV/dt = (12.566 cm^2) dh/dt = (12.566 cm^2)(1.5 cm/sec) = 18.850 cm^3/sec
Note: The cylinder will overflow after 0.5836 sec.
(1) A = (Pi)(d/2)^2, where d = the diameter = 4 cm, so --->
A = 4(Pi) cm^2 = 12.566 cm^2 (approx)
(2) Then V = (12.566 cm^2)(h cm) = 12.566 h cm^2, remember the h is in cm
(3) dV/dt = (12.566 cm^2) dh/dt = (12.566 cm^2)(1.5 cm/sec) = 18.850 cm^3/sec
Note: The cylinder will overflow after 0.5836 sec.
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V = πr^2 * h
dV/dt = πr^2 * dh/dt
Plug in both the radius and the rate of change of the height.
dV/dt = 4π * 1.5cm/sec
dV/dt = 6π cm/sec
dV/dt = πr^2 * dh/dt
Plug in both the radius and the rate of change of the height.
dV/dt = 4π * 1.5cm/sec
dV/dt = 6π cm/sec
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more like an algebraic form