# Integration Word Problem

1. Jan 30, 2007

### snowJT

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

$$Area = 4\pi R^2$$

3. 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??

2. Jan 30, 2007

### Dick

3. Jan 30, 2007

### Sympathy

this is precalculus math?

4. Jan 30, 2007

### snowJT

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

5. Jan 30, 2007

### Dick

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

6. Jan 31, 2007

### HallsofIvy

Staff Emeritus
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?

7. Jan 31, 2007

### snowJT

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

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

??

8. Feb 1, 2007

### HallsofIvy

Staff Emeritus
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:
$$\frac{d R^3}{dt}= \frac{d R^3}{dR}\frac{dR}{dt}$$

9. Feb 1, 2007

### Schrodinger's Dog

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