Finding a differential equation of a growing object

In summary: Just as you can't get rid of mass (r). So it is alright to leave it as it is in the final answer. In summary, the differential equation for the change in radius (r) of a round object over time can be expressed as dr/dt = c4π(3r/4πD)^{2/3}, where c is a constant and D is the density of the object. The equation was derived by first expressing the surface area (A) in terms of radius (R) and then substituting it into the equation dr/dt = cA. The final equation includes only r terms and constants, with the density (D) remaining as a constant in the equation.
  • #1
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Homework Statement



Mass , r, of a round object grows in proportions with its SA. The density of the object is constant. Find the differential equation for the change in 'r' of the object over time. Arbitrary constants can be combined but final answer must depend on only r and constants.

Homework Equations



Mass is proportional to volume.

The Attempt at a Solution



I have,

density (D) = r/V (r= mass)
r = (D)*V This shows mass is proportional to volume.

SA = 4πR^2
V = (4/3)πR^3

Isolate R in SA equation.
Lets call SA = A

A = 4πR^2
r = (A/4π)^0.5

Sub into V

V = (4/3)π(A/4π)^1.5

dR/dt = r*V = r*(4/3)π(A/4π)^1.5

However I still have A in the differential equation.

Am I on the right track?
 
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  • #2
I am a bit confused. I did not get the last equation.. Isn't it dimensionally inconsistent? On the left you have length over time and on the right you have mass times volume. Are you not supposed to find dr/dt? Why do you write dR/dt? I thought of starting like:
[tex]\frac{dr}{dt}=cA[/tex]
Since it says "proportional" I guess we should include a constant of proportionality, c.
 
  • #3
r is a really bad choice for a mass of a sphere, that looks like the radius. Okay, you cannot change that.

r = (A/4π)^0.5
Where does that come from? It is not true.
Edit: Ah, you mean R here. Well, there goes the r confusion...

You can express that A in the last equation in terms of R or r.

dR/dt = r*V
Why?

Hint: This problem gets much easier if you express everything in terms of the radius.
 
  • #4
You're right, my mistake. It's suppose to be dr/dt. If the equation is dr/dt = cA , that'll still leave us with a variable R(radius) without any r (mass) terms.
 
  • #5
Yes, but what if you isolate R in the equation of volume? Then you also have r = D*V, so I think this is the solution.
 
  • #6
So, V = (4/3)πR^3, then R = (3V/(4π))^(1/3). But what can I do with R? How can I incorporate it into the DE with only r terms and constants?

I wrote dr/dt = r*V but I think I meant to write dr/dt = r/V because that equals to density which is the only info given + not knowing if the eqn made sense.
 
  • #7
You actually need to first use [itex]A=4πR^2[/itex]. Then:
[itex]\frac{dr}{dt}=cA[/itex]
[itex]\frac{dr}{dt}=c4πR^2[/itex]
[itex]\frac{dr}{dt}=c4π(\frac{3V}{4π})^{2/3}[/itex]
[itex]\frac{dr}{dt}=c4π(\frac{3r}{4πD})^{2/3}[/itex]
 
  • #8
That makes sense. But one question, density (D) is still in the equation. Is there any way to turn that into terms of r? or do we just combine D with a constant?
 
  • #9
Since it is a constant the expression in terms r should also be a constant, and since r is variable, another variable will appear to make the expression constant. Besides, since D is a constant, is it not allowed to be in the equation?
 
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  • #10
D is constant, that's fine. The final answer depends on D, there is no way to get rid of it.
 
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Related to Finding a differential equation of a growing object

1. What is a differential equation?

A differential equation is a mathematical equation that relates the rate of change of a variable to the value of the variable itself. It is often used to model systems that involve change over time, such as the growth of an object.

2. How do you find a differential equation for a growing object?

The first step is to identify the variables involved in the growth of the object, such as the initial size, rate of growth, and any external factors that may affect the growth. Then, using known mathematical relationships, you can write an equation that relates the rate of change of the object's size to these variables.

3. Can you give an example of a differential equation for a growing object?

One example is the logistic growth equation, which is often used to model population growth. It is written as dP/dt = kP(N-P), where P is the population size, t is time, k is a constant, and N is the maximum possible population size. This equation describes how the population size changes over time, taking into account the initial population size and the maximum capacity of the environment.

4. Are there different types of differential equations for growing objects?

Yes, there are several types of differential equations that can be used to model the growth of objects. The type of equation used depends on the specific characteristics of the object and the factors that influence its growth. Some common types include linear, exponential, and logistic growth equations.

5. How are differential equations used in science?

Differential equations are used in many fields of science to model and understand the behavior of systems that involve change. They are particularly useful in studying physical processes, such as growth, decay, and motion, as well as in fields like biology, economics, and engineering.

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