Calculus Related Rates Question
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Calculus Related Rates Question

[From: ] [author: ] [Date: 11-12-17] [Hit: ]
566 cm^2)(h cm) = 12.566 h cm^2,(3) dV/dt = (12.566cm^2) dh/dt = (12.566 cm^2)(1.5 cm/sec) = 18.......
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

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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.

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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

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more like an algebraic form
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