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fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Water Pouring into Cone Problem Plz Help!

Consider a water tank shaped like an inverted right circular cone of heaight 1m and top radius 1m. Let h(t) denote the height of the water level of the tank in metres with time in days. Water flow into the tank at a constant rate of 0.01 m^3 per day. Water evaporates from the tank at a rate of 0.01 A in m^3 per day where A is the area of the water surface. When h = 0.2, how fast is the water level rising?

The book says that we say the Volume = 0.01t - 0.01(pi)h²*t

I get why 0.01 times the time will give us the added volume because it is at a constant rate. But how can the subtracting volume be equal to 0.01 times pi times h² times time? Becuz the subtracting part is not constant, so how can multiplying it by a constant time give us the correct answer? Unless the book is wrong.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Water Pouring into Cone Problem Plz Help!

Am I missing something here?

h(t) is a function of time denoting the height of the water level in the tank.
We're told that h = 0.2, and thus the height is a constant, independent of time. Therefore, the water level does not rise.

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Why did the vector cross the road?
It wanted to be normal.

Luke T

Guest

Re: Water Pouring into Cone Problem Plz Help!

im pretty sure the water level rises because when they solved it, they took the derivative of h(t), which was not zero.

Here's how they solved it:

(pi/3)h³ = 10 * 10-³ * t - 0.01 * (pi) * h² * t

when h = 0.2:  t = ((pi/3) * 0.2³)/(10-² - 0.01 * pi * 0.2²) = about 1

(pi)h²(dh/dt) = 0.01 - 0.01 * pi * h² - 0.01 * pi * h * t * (dh/dt)

dh/dt = 0.0633 m/day