Rate of Change: Show dh/dt Proportional to h^(1/2)

In summary, The conversation discusses a rate of change question involving a cylindrical tank with fluid flowing out and a constant cross section. The volume of fluid remaining in the tank is represented by V and the rate of flow is proportional to the square root of V. The problem is to show that the depth of fluid, h, satisfies the differential equation dh/dt = -kh^(1/2), where k is a positive constant. The conversation also mentions using the chain rule to differentiate the static equation and connect the rates of change.
  • #1
padraig
10
0
Sorry to bore you all with this rate of change question, I really hate this area of maths and struggle to get my hea round it. Here it is, I'd be greatful for any help you have to offer:

Fluid flows out of a cylindrical tank with constant corss section. At the time t minutes (t is greater of equal to 0), the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows, in m^3min^-1, is proportional to the square root of V. Show that the depth h metre's of fluid in the tank satisfies the differential equation:

dh/dt = -kh^(1/2) where k is a +ve constant

Cheers

Pat
 
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  • #2
dV/dt=-aV^(1/2), by assumption.

V(t)=pi*R^(2)*h(t)=b*h(t) (where R is the radius)
Rearranging a bit, and introducing k yields the desired result
 
  • #3
thanks v much, great help

Pat
 
  • #4
In other words, the whole point of "rate of change problems" is to use a static equation (typically a geometry formula or something like that) that has no time variable, t, in it and then use the chain rule to differentiate both sides of the equation- giving a formula connecting the rates of change of the quantities.
 

What does the rate of change represent in this equation?

The rate of change, dh/dt, represents the instantaneous change in the variable h with respect to time. It is a measure of how quickly the value of h is changing at a specific point in time.

How is the rate of change related to the variable h?

The rate of change, dh/dt, is directly proportional to the square root of h, as represented by the equation dh/dt ∝ h^(1/2). This means that as the value of h increases, the rate of change also increases.

What does it mean for dh/dt to be proportional to h^(1/2)?

This means that as the value of h increases, the rate of change also increases, but at a decreasing rate. In other words, the rate of change is not constant, but rather it decreases as h increases.

How can this equation be applied in real-world situations?

This equation can be applied in various situations where there is a changing quantity that is proportional to the square root of another quantity. For example, it can be used to model the growth rate of a population or the decay rate of a radioactive substance.

What is the significance of the exponent 1/2 in this equation?

The exponent 1/2 represents the square root function, which is a mathematical operation that is used to find the side length of a square with a given area. In this equation, it indicates that the rate of change is proportional to the square root of the variable h.

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