Optimizing Water Flow: Finding Maximum Mass and Rate of Change

In summary: You can also solve it by raising both sides to the 5/4 power, or by using a calculator to find the inverse of the function t^{0.2}.To solve for t^-.2, you can use the same methods as above, but you'll need to take the reciprocal of both sides after raising them to the 5/4 power or using logarithms. Alternatively, you could also rewrite t^-.2 as 1/t^{0.2} and then solve using the same methods as before.
  • #1
lgen0290
16
0

Homework Statement



Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 6.00t 0.8 - 3.35t + 23.00, with t 0, m in grams, and t in seconds.
(a) At what time is the water mass greatest?

(b) What is that greatest mass?

(c) In kilograms per minute, what is the rate of mass change at t = 2.00 s?

(d) In kilograms per minute, what is the rate of mass change at t = 5.00 s?

Homework Equations


The first derivative would be 4.8t^-.2-335


The Attempt at a Solution



a)I set the derivative equal to zero and figured t to be 2.3375, but it says that's wrong.
b)I assume I'd plug a into the original and try tht, but I can't get a.
c and d)I tried to put 2 and 5 into t as the origional, but they are not right.
 
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  • #2
a) is, as you said, the solution for t when

[tex]\frac{dm}{dt}=0.[/tex]

However, the solution to

[tex]t^{0.2}=\left[\frac{4.8}{3.35}\right][/tex]

is not 2.3375, check your algebra.

b), as you said is [tex]m(t)[/tex] for the solution above
c) and d) can both be found by substituting the times into the expression for [tex]\frac{dm}{dt}.[/tex]
 
  • #3
Thanks. Where did the 3.35 come from? How would I solve t^-.2?
 
  • #4
The 3.35 is from the original expression for m, unless you've mistyped it. I used logarithms and the change of base rule to solve for t.
 

1. What is a derivative in relation to water flow?

In the context of water flow, a derivative is a mathematical concept that represents the rate of change of a quantity with respect to another. In this case, it can be used to calculate the rate at which water is flowing in a given direction.

2. How is the derivative of water flow calculated?

The derivative of water flow is typically calculated using the fundamental equation for fluid mechanics, which is known as the Navier-Stokes equation. This equation takes into account various variables such as the fluid's viscosity, pressure, and velocity to determine the derivative of water flow.

3. Can the derivative of water flow be negative?

Yes, the derivative of water flow can be negative. This indicates that the rate of change of water flow is decreasing, meaning that the flow is slowing down.

4. How does the derivative of water flow affect the flow rate?

The derivative of water flow is directly related to the flow rate. A higher derivative value indicates a faster rate of change, which translates to a higher flow rate. Conversely, a lower derivative value means a slower rate of change and a lower flow rate.

5. What are some real-world applications of understanding the derivative of water flow?

Understanding the derivative of water flow is crucial in various fields, including hydrology, civil engineering, and environmental science. It can be used to design efficient water distribution systems, predict and prevent flooding, and model the impact of human activities on water resources.

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