How Fast is the Water Level Dropping in the Tank?

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SUMMARY

The discussion focuses on calculating the rate at which the water level is dropping in a cylindrical tank with a fixed radius of 0.8 m, where water is leaking at a rate of 0.2 m³/min. By applying the formula for the volume of a cylinder, V = πr²h, and differentiating it, the relationship between the change in volume and the change in height is established. The derived equation shows that the rate of change of the water level, dh/dt, is approximately 0.0995 m/min. Additionally, a hypothetical scenario is posed regarding solving for volume when both height and radius rates of change are provided.

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  • Understanding of calculus, specifically differentiation
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  • Basic understanding of units of measurement for volume and height
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Mitchtwitchita
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Water is leaking out of a cylindrical tank , with circular base radius 0.8 m at the rate of 0.2 m3/min. The water level is falling at ...? ..m/min. What is the question mark? (Hint: the radius does not change)

V=(pi)r^2h, dV/dt = 0.2, r = 0.8 ?

Can anyone give me a hand in getting this one started?
 
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Differentiate both sides of V= (pi)r^2 h, remembering that "the radius does not change". Since h is the "water level", "the rate at which the water level is falling" is dh/dt.
 
So, dV/dt = (pi)r^2*dh/dt
0.2 = (pi)(o.8)^2*dh/dt
dh/dt =0.0995 ?

Hypothetically, how would you solve for volume if the rates of change of the height and radii are given, as well as the radius and height when the derivative would seem to cancel the h in the equation?
 

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