Integrating Height as a Function of Time: Help Needed

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In summary: CH/8\mu. So the integral would be:K=8\muH-CH/8\mu.In summary, the goal is to find height as a function of time; however, each time the integration is attempted, the result is an equation that relates height to both time and height. However, by finding the integrals for h(t)=-CH/8\mu and solving for h(t), it is possible to find height as a function of time.
  • #1
sicjeff
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I am having difficulty in integrating this one. My objective is to find height as a function of time; however each time I actually do the integration, I end up with being able to find height as a function of both time and height. I have attempted using Maple to find a unique solution but it produces a (Lambar delta) (possible spelling error). Here is my integration.

int(t)=int([8(mu)(h)]/[2(sigma)(g)(r)-(rho)(g)(h)(r^2)] dh

I keep getting non real answers when I plug in values. Sigma is a surface tension but it keeps getting caught into natural logs with a negative value(impossible)

Any help would be quite appreciated.
 
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  • #2
What are you trying to do here??
Is it to find the anti-derivative:
[tex]I=\int\frac{8\mu{h}}{2\sigma{gr}-\rho{gh}r^{2}}dh[/tex]
If so, what are constants, and what is variable quantities with respect to which variables?
 
  • #3
the only variables are time and height. The others are constants.
 
  • #4
What is r?
 
  • #5
a non changing radius.
I will define all variables
t=time
h=height
r=radius
mu= fluid viscocity
sigma=surface tension
g=gravitational force
rho= density
 
  • #6
Well, and how does t enter the picture.

Whatever do you mean by int(t)??
 
  • #7
the integral of t from t=0 to t so basically the left hand side of the equation reduces to t.
 
  • #8
The expression [itex]\int_{0}^{t}tdt[/itex]?
First of all, your notation is a meaningless conjunction of integration variable and limit of integration, nor would it in this abuse of notation equal t.

Are you sure your left-hand side shouldn't be [itex]\int_{0}^{t}d\tau=t[/itex]?
 
  • #9
right I forgot to say with what respect I was integrating the left hand side, but I am sure that the left hand side reduces to t.

I agree with the first equation that you posted except the left hand side should be t instead of I. the limits of integration from the right side are h= 0 to h=h(final)
 
  • #10
Okay, so what your ACTUAL equation should be is:
[tex]t=\int_{0}^{h(t)}\frac{8\mu{H}}{2\sigma{gr}-\rho{gH}r^{2}}dH[/tex]

That, at the very least, is a mathematically meaningful expression.
Is that what you meant?
 
  • #11
yes. I'm sorry, but I don't know how to use the equation editor
 
  • #12
Now, you may rewrite your right-hand side as:
[tex]t=K\int_{0}^{h(t)}\frac{h}{1-CH}dH[/tex]
for suitable constants K, C.
Can you perform that integration?
 
  • #13
I'm sorry I don't guess I follow you. That isn't something that I have seen in my undergraduate mathematics.
 
  • #14
Sorry, it should have been:

[tex]t=K\int_{0}^{h(t)}\frac{H}{1-CH}dH[/tex]
for suitable constants K, C.

That's just algebra.
 
  • #15
okay, now I see. Thanks. I can do that integration quite easily.
 
  • #16
You'll get an implicit expression for h(t).
 

Related to Integrating Height as a Function of Time: Help Needed

1. What is the purpose of integrating height as a function of time?

The purpose of integrating height as a function of time is to determine the total change in height over a specified period of time. This can be useful in various scientific fields such as physics and engineering to analyze the motion of objects and calculate their displacement.

2. How is height integrated as a function of time?

Height is integrated as a function of time by using the fundamental theorem of calculus, which states that the integral of a function is equal to the area under its curve. In this case, the height function is represented by a curve, and the integral is calculated by finding the area under this curve.

3. What is the difference between integrating height and differentiating height?

Integrating height and differentiating height are two inverse operations. Integrating height yields the total change in height over a specified time interval, while differentiating height yields the instantaneous rate of change of height at a specific point in time.

4. What are some real-life applications of integrating height as a function of time?

Integrating height as a function of time can be applied in various fields such as physics, engineering, and sports. In physics, it can be used to analyze the motion of objects, while in engineering, it can be used to calculate the displacement of structures. In sports, it can be used to track the height of a jump or the trajectory of a ball.

5. Are there any limitations to integrating height as a function of time?

Yes, there are limitations to integrating height as a function of time. It assumes that the acceleration due to gravity is constant, which may not always be the case in real-life scenarios. Additionally, it does not take into account other factors such as air resistance or friction, which may affect the actual height of an object.

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