How is the Height of Water in a Container Changing Over Time?

Click For Summary
SUMMARY

The discussion focuses on determining the height of water in a container with a radius defined by the function r(z) = 10/sqrt(z), filled at a constant rate of 22π ft³/min. The solution involves integrating the volume formula, leading to the relationship v(H) = 100π ln(H) and the rate of height change dH/dt = 11H/50. The user seeks clarification on transitioning from dH/dt to H(t) and confirms the initial condition that at t=0, the height is zero. The final derived function for height over time is H(T) = 11t²/100.

PREREQUISITES
  • Understanding of calculus, specifically integration and the Fundamental Theorem of Calculus (FTC).
  • Familiarity with related rates in differential calculus.
  • Knowledge of volume calculations for cylindrical shapes.
  • Ability to interpret and manipulate functions and their derivatives.
NEXT STEPS
  • Study the application of the Fundamental Theorem of Calculus in solving differential equations.
  • Explore related rates problems in calculus for better understanding of dynamic systems.
  • Learn about volume and surface area calculations for various geometric shapes.
  • Investigate initial value problems and their significance in differential equations.
USEFUL FOR

Students and educators in calculus, particularly those focusing on applications of integration and related rates, as well as anyone interested in modeling physical systems involving fluid dynamics.

armolinasf
Messages
195
Reaction score
0

Homework Statement



If there is a container with a radius in feet given by r(z)=10/sqrt(z) and it is being filled at a constant rate of 22pi ft^3/min. find a function that gives the height of the water as a function of time.

The Attempt at a Solution



since the radius is a function of the container is a function of z feet above the ground we integrate pir^2 => pi(10/sqrt(z))^2

100\pi\int^{H}_{0}1/z*dz

Applying the FTC we get v(H)=100pi*ln(H) making the v'(H)=100pi/H

We also know that the container is being filled at a rate of 22pi cubic feet per minute, this dv/dt

Since we know dv/dt and dv/dH we can create a related rate problem to solve for dH/dt:

dv/dt=dv/dH *dH/dt => 22pi=100pi/H*dH/dt => 11H/50=dH/dtThe way I interpret 11H/50=dH/dt is that if the height is H=10, for example, then the rate at which the height is increasing 110/50 ft/minute is this a correct interpretation?

My next question is regarding how I can go from dH/dt to H(t), that is if i know the time i can find how high the water level is.

My thinking is that I use: \int^{t}_{0}dH/dt*dH

so if Dh/dt is 11H/50 and I apply the FTC and evaluate on the interval o to T, where T is time in minutes I get H(T)=11t^2/100Does this make sense? or am I going wrong somewhere?

Thanks for the help
 
Physics news on Phys.org
It looks fine to me, though you switched notation from z to H. Also, don't forget your constant of integration. You need an additional piece of information: namely, the height at some initial time. I guess you've just assumed that at t=0 the height is zero.
 
There was one detail I omitted. Namely, that the container starts at z=1 foot, so i switched the notation from z to H to differentiate between the function r(z) and the actual container I'm trying to model which ranges z=H=1 to some height z or H. And yes at t=0 the height is zero.

I'm glad to hear that it makes sense: I really feel like I'm getting a good sense of calculus...
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
5K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K