Calculating Raindrop Radius at Different Times Using Integration

[snowJT]

Homework Statement

A spherical raindrops evaporates at a rate proportional to its surface area. If its radius is 3mm, and 1 hour later has been reduced to 2mm, find an expresssion for the raduis of the raindrops at anytime.

Homework Equations

Volume = \frac{4}{3}\pi R^3

Area = 4\pi R^2

The Attempt at a Solution

\frac{d}{dt}(\frac{4}{3}\pi R^3) = -k (4\pi R^2)

4\pi R^3 \frac{d}{dt} = -k 4\pi R^2

\frac{R dR}{dt} = -k

then... how do I plug in 2 and 3 mm??

 

[Dick]

There's already a problem with your DE. d(R(t)^3)/dt=?

 

[Sympathy]

this is precalculus math?

 

[snowJT]

I thought it was just basic integration? Sorry, I'll try and see if this can get moved then or something

 

[Dick]

It reduces to the world's easiest differential equation. It's pretty basic, if not precisely 'pre-calc'.

 

[HallsofIvy]

I thought it was just basic integration? Sorry, I'll try and see if this can get moved then or something

Well, it is basic differentiation- but that is "calculus", not "pre-calculus"!

In any case, go back and look at your work again. What is the derivative of R³ with respect to R? Using the chain rule, what is the derivative of R³ with respect to t?

 

[snowJT]

= 4\pi\int R^3\frac{d}{dt}

= \frac{4\pi R^4}{t}

??

 

[HallsofIvy]

= 4\pi\int R^3\frac{d}{dt}

= \frac{4\pi R^4}{t}

??

I agree: ?? That first line makes no sense. If "d/dt" is the derivative operator, you have to have something to differentiate! Also an integral has to have a "dt" or "dx" so you will know what the variable of integration is.

If this is in response to my previous question "what is the derivative of R3 with respect to t", that was prompted by your
\frac{d}{dt} \frac{4}{3}\pi R^3= 4\pi R^3 \frac{d}{dt}
which, again, makes no sense- what is that final d/dt applied to?

Use the chain rule:
\frac{d R^3}{dt}= \frac{d R^3}{dR}\frac{dR}{dt}

 

[Schrodinger's Dog]

https://www.physicsforums.com/showthread.php?t=154042

I'm not typing all that out again

This is essentially how it is derived and this question asks about how to integrate the volume of half a sphere, but really it's asking pretty much the same thing, one is calculus, the other is calculus I never learned how it was derived until about two months ago. Talking about circles and spheres at the time and started playing around with the figures and what do you know

do all that with t, and then integrate it and there you have it.

\int_3^2 \int 4 \pi r^2/t =\int_3^2 \frac{4}{3}(\frac{\pi r^3}{t}) dt=

r between the value of 2 & 3; then make the equation =r; that's how I'd do it, something like that?