Spherical raindrop, mass, radius, and time

AI Thread Summary
The discussion centers on finding an expression for the radius of a spherical raindrop as a function of time, r(t), given its initial radius r0 at t=0. The mass of the raindrop is expressed as M = ρ(4/3)πr³, and the rate of change of mass is related to the cube of the radius by the equation dM/dt = Cr³. Participants emphasize the need to differentiate the mass function with respect to time, applying the chain rule since both mass and radius are functions of time. The integration of the differential equation requires proper separation of variables to solve for r(t). The conversation highlights the importance of revisiting fundamental calculus concepts, such as the chain rule, to approach the problem effectively.
Raquel Aduriz
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Homework Statement



Separate variables and integrate to find an expression for r(t), given r0 at t=0

Homework Equations



M=ρ(4/3)πr3, thus V=(4/3)πr3

dM/dt=Cr3 where C is a constant

The Attempt at a Solution



∫dM=∫Cr3dt

M+constant=??

I have no idea how to integrate r because it's a function of t but we're not given the function. I don't think that integrating the left side this way will be very helpful either. Any advice would be much appreciated!
 
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Raquel Aduriz said:

Homework Statement



Separate variables and integrate to find an expression for r(t), given r0 at t=0

Homework Equations



M=ρ(4/3)πr3, thus V=(4/3)πr3

dM/dt=Cr3 where C is a constant

The Attempt at a Solution



∫dM=∫Cr3dt

M+constant=??

I have no idea how to integrate r because it's a function of t but we're not given the function. I don't think that integrating the left side this way will be very helpful either. Any advice would be much appreciated!
You want to express M in terms of r. Once you do that, then dM(r) / dt = Cr3, and you can separate the variables to get r on one side and dt on the other.
 
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Ok... I guess I don't really understand how to do that. Can I do that using just the M equation, or do I need something else?
 
Raquel Aduriz said:
Ok... I guess I don't really understand how to do that. Can I do that using just the M equation, or do I need something else?
You gave an expression for the mass of the raindrop in terms of the radius right there in Section 2 of the template. It's the equation M = ...
 
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Alright, so I have M=ρ(4/3)πr3. What exactly do I do with this? Take the derivative and set it equal to the other? Sorry, I'm sure I'm missing something really obvious.
 
Raquel Aduriz said:
Alright, so I have M=ρ(4/3)πr3. What exactly do I do with this? Take the derivative and set it equal to the other? Sorry, I'm sure I'm missing something really obvious.
This problem is about solving a differential equation by separation of variables. Have you studied how to do this yet?
 
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Yes, but a few years ago so maybe I've forgotten something key. So I have C*r*dt=ρ*4*π*dr. Is that correct?
 
Raquel Aduriz said:
Yes, but a few years ago so maybe I've forgotten something key. So I have C*r*dt=ρ*4*π*dr. Is that correct?
Not quite.

Presumably, as the rain drop falls, it gets bigger as it collects more moisture; therefore, the radius of the drop grows as time passes.

Since the drop is spherical, M(t) = ρ(4/3)π*[r(t)]3, where M and r are written as functions of time.

You are also given the condition that the change in the mass of the drop, dM(t)/dt, at any given time is proportional to the cube of the radius of the drop, or dM(t)/dt = Cr3. At t = 0, r(t) = r(0) = r0.

You should check your differentiation of M(t) w.r.t. time. You might have to use the chain rule here, since M is a function of r, but r is a function of t.
 
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I got it! Can't believe I forgot about the chain rule. Thank you so much for your help!
 
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